Đề thi giữa kì 2 Toán 6 Cánh diều – Đề 09 (có đáp án)

1.

a) $11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)=\left(11+\frac{3}{13}\right)-\left[\left(2+\frac{4}{7}\right)+\left(5+\frac{3}{13}\right)\right]$

$=11+\frac{3}{13}-\left[7+\frac{4}{7}+\frac{3}{13}\right]=11+\frac{3}{13}-7-\frac{4}{7}-\frac{3}{13}=4-\frac{4}{7}=\frac{24}{7}$;

b) $$\frac{-5}{7}\cdot \frac{2}{11}+\frac{-5}{7}\cdot \frac{9}{11}+\frac{5}{7}=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{5}{7}=\frac{-5}{7}\cdot 1+\frac{5}{7}=0$$.

2.

Số tiền được giảm giá của món đồ chơi là:

$$50000.10\%=5000$$ (đồng);

Số tiền Nam phải trả để mua món đồ chơi đó là:

$$50000--5000=45000$$ (đồng)

Vậy Nam phải trả $$45000$$ đồng để mua món đồ chơi đó.

 Ta có: $A=\frac{2}{5.7}+\frac{5}{7.12}+\frac{7}{12.19}+\frac{9}{19.28}+\frac{11}{28.39}+\frac{1}{39.40}$

$=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{19}+\frac{1}{19}-\frac{1}{28}+\frac{1}{28}-\frac{1}{39}+\frac{1}{39}-\frac{1}{40}$

$=\frac{1}{5}-\frac{1}{40}$$=\frac{8}{40}-\frac{1}{40}=\frac{7}{40}$;

$$B=\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+\frac{1}{119}+\frac{1}{170}$$

$$=\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+\frac{2}{238}+\frac{2}{340}$$

$$=2.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)$$

$$=2.\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)$$

$$=\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)$$

$$=\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{20}\right)$$$$=\frac{2}{3}.\left(\frac{4}{20}-\frac{1}{20}\right)$$$$=\frac{2}{3}.\frac{3}{20}$$$$=\frac{1}{10}$$.

Ta có: $\frac{7}{40}>\frac{4}{40}=\frac{1}{10}$. Do đó $A>B$.