Đề kiểm tra Giữa kì 2 Toán 6 có đáp án (Đề 3)

a) $\left(\frac{9}{16}-\frac{5}{8}+\frac{3}{4}\right):\frac{11}{32}$

$=\left(\frac{9}{16}-\frac{10}{16}+\frac{12}{16}\right):\frac{11}{32}$

$=\left(\frac{9-10+12}{16}\right):\frac{11}{32}$

$=\frac{11}{16}:\frac{11}{32}$

$=\frac{11}{16}.\frac{32}{11}$

$=\frac{11.32}{16.11}$

$=\frac{\mathrm{11.16.2}}{16.11}=2$

b) $\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}+\frac{3}{13}.\frac{-5}{9}$

$=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}$

$=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)$

$=\frac{5}{9}.\left(\frac{7+9-3}{13}\right)$

$=\frac{5}{9}.\frac{13}{13}=\frac{5}{9}.1=\frac{5}{9}$

c) $\frac{-3}{7}+\frac{-22}{26}+\left(\frac{2}{-13}+\frac{3}{7}+{2021}^0\right)$

$=\frac{-3}{7}+\frac{-22}{26}+\left(\frac{2}{-13}+\frac{3}{7}+1\right)$

$=\frac{-3}{7}+\frac{-22}{26}+\frac{2}{-13}+\frac{3}{7}+1$

$=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{-22}{26}+\frac{2}{-13}\right)+1$

$=\left(\frac{-3+3}{7}\right)+\left(\frac{-22}{26}+\frac{-4}{26}\right)+1$

$=\frac{0}{7}+\frac{\left(-22\right)+\left(-4\right)}{26}+1$

$=0+\frac{-26}{26}+1$

$=0+(-1)+1=0$

a) $\frac{7}{8}+x=\frac{3}{5}$

$x=\frac{3}{5}-\frac{7}{8}$

$x=\frac{24}{40}-\frac{35}{40}$

$x=\frac{24-35}{40}$

$x=\frac{-11}{40}$

b) $\frac{2}{3}\left(2x-1\right)=\frac{-4}{21}$

$$2x-1=\frac{-4}{21}:\frac{2}{3}$$

$$2x-1=\frac{-4}{21}.\frac{3}{2}$$

$$2x-1=\frac{-2}{7}$$

$$2x=\frac{-2}{7}+1$$

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

$$x=\frac{5}{7}:2$$

$$x=\frac{5}{7}.\frac{1}{2}$$

$$x=\frac{5}{14}$$

a) Ta có ba điểm A, B, C thuộc tia Ox và OA < OB (3 < 6) nên A nằm giữa O và B.

b) Vì A nằm giữa O và B nên AO + BA = OB

Thay số: 3 + AB = 6

AB = 6 – 3

AB = 3cm

Vì A nằm giữa O và B; OA = AB = $\frac{OB}{2}=\frac{6}{2}=3cm$ nên A là trung điểm của OB

Chứng minh rằng

 A = $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{10}^2}<1$

Ta có: $\frac{1}{2^2}=\frac{1}{2.2}<\frac{1}{1.2}$

$\frac{1}{3^2}=\frac{1}{3.3}<\frac{1}{2.3}$

$\frac{1}{4^2}=\frac{1}{4.4}<\frac{1}{3.4}$

…

$\frac{1}{{10}^2}=\frac{1}{10.10}<\frac{1}{9.10}$

Nên $\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{{10}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}$

A <$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}$

Ta lại có: $\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}$ = 1 - $\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}$

Vì $\frac{9}{10}<1$ nên A < 1