Đề thi tuyển sinh vào lớp 10 môn Toán chuyên năm 2021-2022 sở GD&ĐT Lào Cai có đáp án

a) Với: $$\{\begin{array}{l}a>0\\ a\ne \left\{1,2\right\}\end{array}$$

Ta có:

$$A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\left(\frac{a+2}{a-2}\right)=\left(\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\left(\frac{a+2}{a-2}\right)$$

$$A=\left(\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\right):\left(\frac{a+2}{a-2}\right)=2\cdot \left(\frac{a-2}{a+2}\right)=\frac{2a-4}{a+2}=2-\frac{8}{a+2}$$

Để $$A\in \mathbb{Z}\Rightarrow 2-\frac{8}{a+2}\in \mathbb{Z}\Rightarrow a+2\in U\left(8\right)=\left\{\pm 1;\pm 2;\pm 4;\pm 8\right\}$$

Do: $$\{\begin{array}{l}a\in {\mathbb{Z}}^{+}\\ a\ne \left\{1;2\right\}\end{array}\Rightarrow a+2\ge 5\Rightarrow a+2\in \left\{8\right\}\Rightarrow a=6\left(TM\right)$$

Vậy $$a=6\Rightarrow A\in \mathbb{Z}$$

b) Đặt :$$M=x^5-2x^4-2021x^3+3x^2+2018x-2021$$

$$=x^5-2x^4-2020x^3-x^3+2x^2+2020x+x^2-2x-2020-1.$$

$$\begin{array}{l}=x^3\left(x^2-2x^{}-2020\right)-x\left(x^2-2x^{}-2020\right)+\left(x^2-2x-2020\right)-1\\ =\left(x^2-2x-2020\right)\left(x^3-x+1\right)-1\end{array}$$

Mà: $$x=1+\sqrt{2021}\iff x-1=\sqrt{2021}\iff {\left(x-1\right)}^2=2021\iff x^2-2x-2020=0.$$

$$\Rightarrow M=-1$$

1) Gọi vận tốc dự định của xe đạp là: $$x\left(km/h\right);x>0.$$

Vận tốc sau khi tăng tốc là: $$x+3\left(km/h\right).$$

Thời gian dự định là: $$\frac{40}{x}\left(h\right).$$

Quãng đường từ lúc tăng tốc là: $$40-20=20\left(km\right).$$

Thời gian lúc chưa tăng tốc là: $$\frac{20}{x}\left(h\right).$$

Thời gian từ lúc tăng tốc là: $$\frac{20}{x+3}\left(h\right).$$

Theo đề bài ta có: $$\frac{20}{x}+\frac{1}{3}+\frac{20}{x+3}=\frac{40}{x}\iff [\begin{array}{l}x=12\left(TM\right)\\ x=-15\left(KTM\right)\end{array}$$

Vậy vận tốc dự định của xe đạp là: 12 (km/h)

2) a) Ta có: $${\Delta }^{\prime }={\left[-\left(m-1\right)\right]}^2-2m+5=m^2-4m+6={\left(m-2\right)}^2+2>0\forall m$$

Suy ra phương trình luôn có 2 nghiệm phân biệt với mọi m.

b) Theo Vi-et ta có: $$\{\begin{array}{l}{x}_1^{}+x_2^{}=2\left(m-1\right)\\ x_1^{}{x}_2^{}=2m-5\end{array}$$

Do: $$x_1^{};x_2^{}$$ là nghiệm của phương trình nên ta có:

$$\begin{array}{l}\{\begin{array}{l}{x}_{{}^1}^2-2\left(m-1\right)x_1^{}+2m-5=0\\ x_{{}^2}^2-2\left(m-1\right)x_2^{}+2m-5=0\end{array}\\ \Rightarrow \{\begin{array}{l}{x}_{{}^1}^2-2m{x}_1^{}+2x_1^{}+2m-1-4=0\\ x_{{}^2}^2-2m{x}_2^{}+2x_2^{}+2m-1-4=0\end{array}\\ \Rightarrow \{\begin{array}{l}{x}_{{}^1}^2-2m{x}_1^{}+2m-1=4-2x_1^{}\\ x_{{}^2}^2-2m{x}_2^{}+2m-1=4-2x_2^{}\end{array}\end{array}$$

Mà: $$\left(x_1^2-2m{x}_1+2m-1\right)\left(x_2^2-2m{x}_2+2m-1\right)<0$$

$$\begin{array}{l}\iff \left(4-2x_1\right)\left(4-2x_2\right)<0\\ \iff 16-8\left(x_1+x_2\right)+4x_1{x}_2<0\end{array}$$

$$\iff 16-8.2\left(m-1\right)+4\left(2m-5\right)<0\iff 12-8m<0\iff m<\frac{3}{2}$$.

a) Xét $$\Delta AFP$$ và $$\Delta ADF$$ có:

 (đpcm).

b) Vì: AF và AE là 2 tiếp tuyến của $$\left(I\right)\Rightarrow AI$$ là trung trực của FE $$\Rightarrow AI\perp FE$$ tại Q.

$$\Rightarrow A{F}^2=AQ.AI$$ (hệ thức lượng) $$\Rightarrow AQ.AI=AP.AD\left(=A{F}^2\right)\Rightarrow \frac{AP}{AQ}=\frac{AI}{AD}$$.

Xét $$\Delta APQ$$ và $$\Delta AID$$ có: $$\frac{AP}{AQ}=\frac{AI}{AD}\left(cmt\right);\hat{A}Chung$$

 nội tiếp (vì: $$\widehat{AQP}$$là góc ngoài tại đỉnh Q)

Ta có: $$\widehat{A_1}=\widehat{A_2}$$ (vì: AI là tia phân giác)

Xét $$\Delta ABN$$ và $$\Delta BMN$$ có: $$\widehat{B_1}=\widehat{A_2}\left(cmt\right);\hat{N}Chung$$

 (đpcm)

c) Ta có:

Mà: $$\{\begin{array}{l}\widehat{IDP}=\widehat{AQP}\left(cmt\right)\\ \widehat{AQT}=\widehat{IQD}\left(doidinh\right)\end{array}\Rightarrow \widehat{AQP}=\widehat{AQT}\Rightarrow$$ đpcm

d) Gọi $$K$$ là giao điểm của $$AI$$ với

Mà:  đpcm.

a) Dự đoán điểm rơi:

$$x=y=\frac{1}{3}\Rightarrow \frac{1}{x^2}+ax+ax\stackrel{Co-Si}{\ge }3.\sqrt[3]{\frac{1}{x^2}\cdot ax\cdot ax}=3.\sqrt[3]{a^2}\Rightarrow \frac{1}{x^2}=ax\Rightarrow a=27$$

Ta có: $$A=53x+53y+\frac{1}{x^2}+\frac{1}{y^2}=\left(27x+27x+\frac{1}{x^2}\right)+\left(27y+27y+\frac{1}{y^2}\right)-\left(x+y\right)$$

$$\Rightarrow A\stackrel{Co-Si}{\ge }3.\sqrt[3]{27x\cdot 27x\cdot \frac{1}{x^2}}+3.\sqrt[3]{27y\cdot 27y\cdot \frac{1}{y^2}}-\left(x+y\right)=27+27-\left(x+y\right)\ge 54-\frac{2}{3}=\frac{160}{3}$$Dấu “=” xảy ra khi $$x=y=\frac{1}{3}$$

Vậy $$\Rightarrow MinA=\frac{160}{3}\iff x=y=\frac{1}{3}$$

b) Ta có: $$x^4+1\ge 2.\sqrt{x^4.1}=2x^2;y^4+1\ge 2.\sqrt{y^4.1}=2y^2;z^4+1\ge 2.\sqrt{z^4.1}=2z^2$$

$$\Rightarrow x^4+y^4+z^4\ge 2\left(x^2+y^2+z^2\right)-3\Rightarrow VT\ge 2\left(x^2+y^2+z^2\right)-3+x^3+y^3+z^3$$

Tương tự: $$x^3+x\ge 2.\sqrt{x^3.x}=2x^2;y^3+y\ge 2.\sqrt{y^3.y}=2y^2;z^3+z\ge 2.\sqrt{z^3.z}=2z^2$$

$$\begin{array}{l}\Rightarrow x^3+y^3+z^3\ge 2\left(x^2+y^2+z^2\right)-\left(x+y+z\right)\\ \Rightarrow VT\ge 2\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+2\left(x^2+y^2+z^2\right)-3\end{array}$$

$$\begin{array}{l}\Rightarrow VT\ge \left(x^2+y^2+z^2\right)-\left(x+y+z\right)+3\left(x^2+y^2+z^2\right)-3\\ \Rightarrow VT\ge \left(x^2+y^2+z^2\right)-\left(x+y+z\right)+3.3-3\end{array}$$$$\Rightarrow VT\ge \left(x^2+y^2+z^2\right)-\left(x+y+z\right)+6$$

Mà: $$x^2+1\ge 2.\sqrt{x^2.1}=2x;y^2+1\ge 2.\sqrt{y^2.1}=2y;z^2+1\ge 2.\sqrt{z^2.1}=2z$$

$$\Rightarrow x^2+y^2+z^2\ge 2\left(x+y+z\right)-3$$

$$\Rightarrow VT\ge 2\left(x+y+z\right)-3-\left(x+y+z\right)+6=\left(x+y+z\right)+3$$ (đpcm)

a) Ta có:

$$\begin{array}{l}{x}^2-2x+2y^2=2\left(xy+1\right)\\ \iff x^2-2x+2y^2=2xy+2\\ \iff x^2-2xy+y^2+y^2-2x=2\end{array}$$

$$\begin{array}{l}\iff {\left(x-y\right)}^2-2x+y^2-2y+1+2y=3\\ \iff {\left(x-y\right)}^2-2\left(x-y\right)+{\left(y-1\right)}^2=3\\ \iff {\left(x-y\right)}^2-2\left(x-y\right)+1+{\left(y-1\right)}^2=4\end{array}$$

$$\iff {\left(x-y-1\right)}^2+{\left(y-1\right)}^2=4\left(=0^2+2^2\right)$$

$$\iff \{\begin{array}{l}x-y-1=0\\ y-1=2\end{array}\vee \{\begin{array}{l}x-y-1=0\\ y-1=-2\end{array}\vee \{\begin{array}{l}y-1=0\\ x-y-1=-2\end{array}\vee \{\begin{array}{l}y-1=0\\ x-y-1=2\end{array}$$

$$\iff \{\begin{array}{l}x=4\\ y=3\end{array}\vee \{\begin{array}{l}x=0\\ y=-1\end{array}\vee \{\begin{array}{l}y=1\\ x=0\end{array}\vee \{\begin{array}{l}y=1\\ x=4\end{array}$$

Vậy $$\left(x;y\right)=\left(4;3\right);\left(0;-1\right);\left(0;1\right);\left(4;1\right).$$

b) Ta có:

$$\begin{array}{l}{x}^3+y^3-p=6xy-8\\ \iff p=x^3+y^3-6xy+8\\ \iff p={\left(x+y\right)}^3-3xy\left(x+y\right)-6xy+8\end{array}$$

$$\begin{array}{l}\iff p=\left[{\left(x+y\right)}^3+8\right]-3xy\left(x+y+2\right)\\ \iff p=\left(x+y+2\right)\left[{\left(x+y\right)}^2-2\left(x+y\right)+4-3xy\right]\end{array}$$

Do $$p$$ là số nguyên tố nên:

$$[\begin{array}{l}\left(x+y+2\right)=1\\ {\left(x+y\right)}^2-2\left(x+y\right)+4-3xy=1\end{array}\Rightarrow {\left(x+y\right)}^2-2\left(x+y\right)+4-3xy=1$$

(Vì: $$x;y\in {\mathbb{Z}}^{+}\Rightarrow x+y+2\ge 4$$)

$$\begin{array}{l}\Rightarrow {\left(x+y\right)}^2-2\left(x+y\right)+4-3xy=1\\ \iff x^2+2xy+y^2-2x-2y-3xy=-3\\ \iff x^2-xy+y^2-2x-2y=-3\end{array}$$

$$\begin{array}{l}\iff 4x^2-4xy+4y^2-8x-8y=-12\\ \iff {\left(2x-y\right)}^2+3y^2-4\left(2x-y\right)+4-12y+12=4\end{array}$$

$$\iff {\left(2x-y-2\right)}^2+3{\left(y-2\right)}^2=4\left(=1^2+{3.1}^2\right)$$

$$\iff \{\begin{array}{l}2x-y-2=1\\ y-2=1\end{array}\vee \{\begin{array}{l}2x-y-2=1\\ y-2=-1\end{array}\vee \{\begin{array}{l}2x-y-2=-1\\ y-2=1\end{array}\vee \{\begin{array}{l}2x-y-2=-1\\ y-2=-1\end{array}$$

$$\iff \{\begin{array}{l}x=3\\ y=3\end{array}\vee \{\begin{array}{l}x=2\\ y=1\end{array}\vee \{\begin{array}{l}x=2\\ y=3\end{array}\vee \{\begin{array}{l}x=1\\ y=1\end{array}$$

TH1: $$\{\begin{array}{l}x=3\\ y=3\end{array}\Rightarrow p=8\left(KTM\right)$$

TH2: $$\{\begin{array}{l}x=2\\ y=1\end{array}\Rightarrow p=5\left(TM\right)$$

TH3: $$\{\begin{array}{l}x=2\\ y=3\end{array}\Rightarrow p=7\left(TM\right)$$

TH4: $$\{\begin{array}{l}x=1\\ y=1\end{array}\Rightarrow p=4\left(KTM\right)$$

Vì: $$p$$ là số nguyên tố lớn nhất $$\Rightarrow p=7$$

Vậy $$p=7$$ thỏa mãn yêu cầu bài toán.