Đề thi giữa kì 1 Toán 7 Cánh Diều (2023-2024) – Đề 5 (có đáp án)

a $\frac{5}{6}+\frac{-11}{30}+\frac{7}{5}$

$=\frac{25}{30}+\frac{-11}{30}+\frac{42}{30}$

$=\frac{56}{30}=\frac{28}{15}$

 b $${\left(\frac{-4}{5}\right)}^2\cdot \frac{57}{2}+\frac{-7}{2}\cdot {\left(\frac{-4}{5}\right)}^2$$

 $$={\left(\frac{-4}{5}\right)}^2\cdot \left(\frac{57}{2}+\frac{-7}{2}\right)$$

$$=\frac{16}{25}\cdot \frac{50}{2}=16.$$

 c $\left(\frac{7}{25}-1\frac{2}{9}\right)-\left(\frac{23}{54}-\frac{18}{25}\right)+\frac{-31}{54}$

 $$=\frac{7}{25}-1\frac{2}{9}-\frac{23}{54}+\frac{18}{25}+\frac{-31}{54}$$

$$=\left(\frac{7}{25}+\frac{18}{25}\right)+\left(\frac{-31}{54}+\frac{-23}{54}\right)-1\frac{2}{9}$$

$$=1+\left(-1\right)-\frac{11}{9}$$$$=\frac{-11}{9}$$.

a $\frac{5}{2}-x=\frac{27}{8}$

    $x=\frac{5}{2}-\frac{27}{8}$

    $x=\frac{20}{8}-\frac{27}{8}$              

     $x=\frac{-7}{8}$

Vậy $x=\frac{-7}{8}.$

 b $\frac{-1}{3}+\frac{4}{10}x=0,2$

    $\frac{4}{10}x=0,2-\left(\frac{-1}{3}\right)$

    $\frac{4}{10}x=\frac{8}{15}$

    $x=\frac{8}{15}:\frac{4}{10}$

    $x=\frac{4}{3}$

Vậy $x=\frac{4}{3}$

 c ${\left(3x+2\right)}^2=\frac{25}{49}$

$${\left(3x+2\right)}^2={\left(\frac{5}{7}\right)}^2={\left(\frac{-5}{7}\right)}^2$$

Trường hợp 1: $$3x+2=\frac{5}{7}$$

$$3x=\frac{5}{7}-2$$

$$3x=\frac{-9}{7}$$

$$x=\frac{-9}{7}:3$$

$$x=\frac{-3}{7}$$

Trường hợp 2: $$3x+2=\frac{-5}{7}$$

$$3x=\frac{-5}{7}-2$$

$$3x=\frac{-19}{7}$$

$$x=\frac{-19}{7}:3$$

$$x=\frac{-19}{21}$$

Vậy $$x\in \left\{\frac{-3}{7};\frac{-19}{21}\right\}.$$

Có: $\widehat{AOH}$ và $\widehat{AOK}$ là 2 góc kề bù

nên $\widehat{AOH}+\widehat{AOK}=180^{\circ }$

$\widehat{AOH}+120^{\circ }=180^{\circ }$

$\widehat{AOH}=180^{\circ }-120^{\circ }$

$\widehat{AOH}=60^{\circ }$

a Thể tích của bể bơi là: $$8\cdot 4\cdot 0,5=16\left({\mathrm{m}}^3\right)$$.

 b Đổi: $$16\left({\mathrm{m}}^3\right)=16000\left(\mathrm{d}{\mathrm{m}}^3\right)=16000\left(l\right)$$

Lượng nước máy bơm bơm vào bể sau $5$ giờ $5$ phút (bằng $305$ phút) là: $$50\cdot 305=15250\left(l\right)$$

Vì: $$15250<16000$$

Vậy nên sau $5$ giờ $5$ phút bể vẫn chưa đầy nước.

$A=\frac{7^{2011}+2}{7^{2012}+2}$suy ra $$7\cdot A=\frac{7^{2012}+14}{7^{2012}+2}=1+\frac{12}{7^{2012}+2}$$

$B=\frac{7^{2023}+2}{7^{2024}+2}$suy ra $7\cdot B=\frac{7^{2024}+14}{7^{2024}+2}=1+\frac{12}{7^{2024}+2}$

Có $$7^{2012}+2<7^{2024}+2$$ suy ra $$\frac{12}{7^{2012}+2}>\frac{12}{7^{2024}+2}$$

Nên $$A>B.$$