Đề thi Cuối kì học kỳ 2 Toán 6 có đáp án (Đề 1)

Hướng dẫn giải:

a) $$25\%-1\frac{1}{4}+0,2:\frac{6}{5}$$

$$=\frac{25}{100}-\frac{5}{4}+\frac{1}{5}:\frac{6}{5}$$

$$=\frac{1}{4}-\frac{5}{4}+\frac{1}{5}.\frac{5}{6}$$

$$=\left(\frac{1}{4}-\frac{5}{4}\right)+\frac{1.5}{5.6}$$

$$=\frac{-4}{4}+\frac{1}{6}$$

$$=-1+\frac{1}{6}$$

$$=\frac{-6}{6}+\frac{1}{6}$$

$$=\frac{-5}{6}$$

b) $$\frac{8}{9}+\frac{1}{9}.\frac{2}{9}+\frac{1}{9}.\frac{7}{9}$$

$$=\frac{8}{9}+\left(\frac{1}{9}.\frac{2}{9}+\frac{1}{9}.\frac{7}{9}\right)$$

$$=\frac{8}{9}+\frac{1}{9}.\left(\frac{2}{9}+\frac{7}{9}\right)$$

$$=\frac{8}{9}+\frac{1}{9}.\frac{9}{9}$$

$$=\frac{8}{9}+\frac{1}{9}.1$$

$=\frac{8}{9}+\frac{1}{9}$

$=\frac{9}{9}$

= 1.

c) $\frac{5}{39}\cdot \left(7\frac{4}{5}\cdot 1\frac{2}{3}+8\frac{1}{3}\cdot 7\frac{4}{5}\right)$

$=\frac{5}{39}<span></span>\cdot \left[7\frac{4}{5}\cdot \left(1\frac{2}{3}+8\frac{1}{3}\right)\right]$

$=\frac{5}{39}\cdot \frac{39}{5}\left(1+\frac{2}{3}+8+\frac{1}{3}\right)$

$=\frac{5.39}{39.5}\cdot \left(9+\frac{2}{3}+\frac{1}{3}\right)$

$=1\cdot \left(9+\frac{3}{3}\right)$

= 9 + 1

= 10.

Hướng dẫn giải

a) $$\frac{-5}{12}:x=\frac{4}{15}$$

$$x=\frac{-5}{12}:\frac{4}{15}$$

$$x=\frac{-5}{12}.\frac{15}{4}$$

$x=\frac{-5.15}{12.4}$

$x=\frac{-\mathrm{5.3.5}}{\mathrm{3.4.4}}$

$$x=-\frac{25}{16}.$$

Vậy $$x=\frac{-25}{16}$$.

b) $$\frac{3}{4}x-\frac{2}{3}x=\frac{2}{7}.\frac{1}{6}+\frac{5}{7}.\frac{1}{6}$$

$$\left(\frac{3}{4}-\frac{2}{3}\right)x=\frac{1}{6}.\frac{2}{7}+\frac{1}{6}.\frac{5}{7}$$

$$\left(\frac{9}{12}-\frac{8}{12}\right)x=\frac{1}{6}\left(\frac{2}{7}+\frac{5}{7}\right)$$

$\frac{1}{12}.x=\frac{1}{6}.\frac{7}{7}$
$$\frac{1}{12}x=\frac{1}{6}$$

$$x=\frac{1}{6}:\frac{1}{12}$$

$x=\frac{1}{6}.\frac{12}{1}$

x = 2.

Vậy x = 2.

c) 4x – (3 + 5x) = 14

4x – 3 – 5x = 14

4x – 5x = 14 + 3

(4 – 5).x = 17

–x = 17

x = –17.

Vậy x = –17.

Hướng dẫn giải

Số học sinh giỏi của lớp là: $8:\frac{2}{3}=12$ (học sinh)

Số học sinh khá của lớp là: 12 : 80% = 15 (học sinh)

Số học sinh trung bình của lớp là: $\frac{7}{9}.(15+12)=21$ (học sinh)

Số học sinh lớp $$6A$$ là: 12 + 15 + 21 = 48 (học sinh).

Ta có:  $A=\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{4^2}+\genfrac{}{}{0ex}{}{1}{6^2}+...+\genfrac{}{}{0ex}{}{1}{{100}^2}$

               $=\genfrac{}{}{0ex}{}{1}{2^2}\left(1+\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}\right)$.

Mặt khác: $\genfrac{}{}{0ex}{}{1}{2^2}=\genfrac{}{}{0ex}{}{1}{2.2}<\genfrac{}{}{0ex}{}{1}{1.2}=\genfrac{}{}{0ex}{}{2-1}{1.2}=\genfrac{}{}{0ex}{}{2}{1.2}-\genfrac{}{}{0ex}{}{1}{1.2}=1-\genfrac{}{}{0ex}{}{1}{2}$

                  $\genfrac{}{}{0ex}{}{1}{3^2}=\genfrac{}{}{0ex}{}{1}{3.3}<\genfrac{}{}{0ex}{}{1}{2.3}=\genfrac{}{}{0ex}{}{3-2}{2.3}=\genfrac{}{}{0ex}{}{3}{2.3}-\genfrac{}{}{0ex}{}{2}{2.3}=\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{3}$

                  ………………..

                  $\genfrac{}{}{0ex}{}{1}{{50}^2}=\genfrac{}{}{0ex}{}{1}{50.50}<\genfrac{}{}{0ex}{}{1}{49.50}=\genfrac{}{}{0ex}{}{50-49}{49.50}=\genfrac{}{}{0ex}{}{50}{49.50}-\genfrac{}{}{0ex}{}{49}{49.50}=\genfrac{}{}{0ex}{}{1}{49}-\genfrac{}{}{0ex}{}{1}{50}$

Do đó  $\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}<1-\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{3}+...+\genfrac{}{}{0ex}{}{1}{49}-\genfrac{}{}{0ex}{}{1}{50}$

Suy ra $\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}<1-\genfrac{}{}{0ex}{}{1}{50}$

           $\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}<\genfrac{}{}{0ex}{}{49}{50}<\genfrac{}{}{0ex}{}{50}{50}=1$

           $\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}<1$

Từ đó ta có: $1+\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}<1+1=2$

$$A=\genfrac{}{}{0ex}{}{1}{2^2}\left(1+\genfrac{}{}{0ex}{}{1}{2^2}+\genfrac{}{}{0ex}{}{1}{3^2}+...+\genfrac{}{}{0ex}{}{1}{{50}^2}\right)<\genfrac{}{}{0ex}{}{1}{4}.2=\genfrac{}{}{0ex}{}{1}{2}$$.

Vậy $A<\genfrac{}{}{0ex}{}{1}{2}.$