Bài tập tuần 3 – Toán lớp 7 – THCS Dịch Vọng Hậu

LUYỆN TẬP: NHÂN, CHIA SỐ HỮU TỈ

PHIẾU BÀI TẬP TUẦN 3 – TRƯỜNG THCS DỊCH VỌNG HẬU

Bài 1: Thực hiện phép tính (hợp lí nếu có thể):

a) $\left( -\frac{28}{19} \right).\left( \frac{-38}{14} \right)$

b) $\left( -\frac{-21}{16} \right).\left( -\frac{24}{7} \right)$

c) $\frac{12}{25}.\frac{23}{7}-\frac{12}{25}.\frac{12}{7}$

d) $13\frac{2}{7}:\left( \frac{-8}{9} \right)+2\frac{5}{7}:\left( \frac{-8}{9} \right)$

e) $\left( \frac{-6}{11} \right).\frac{7}{10}.\left( \frac{11}{-6} \right).\left( -20 \right)$

f) $\left( -1\frac{1}{2} \right):\frac{3}{4}.\left( -4\frac{1}{2} \right)$

g) $21-3\frac{3}{4}:\left( \frac{3}{8}-\frac{1}{6} \right)$

h) $\left( 2-\frac{3}{2} \right)\left( 2-\frac{4}{3} \right)\left( 2-\frac{5}{4} \right)\left( 2-\frac{6}{5} \right)$

i) $\frac{1}{2011}+\frac{2012.2010}{2011}-2012$

k) $\frac{5}{4}-\frac{5}{4}:\left( \frac{3}{8}-\frac{1}{6} \right):\frac{11}{12}$

l) $\frac{3}{5}:\left( -\frac{-1}{15}-\frac{1}{6} \right)+\frac{3}{5}:\left( \frac{-1}{3}-1\frac{1}{15} \right)$

m) $\left( \frac{-3}{4}+\frac{2}{5} \right):\frac{3}{7}+\left( \frac{3}{5}+\frac{-1}{4} \right):\frac{3}{7}$

n) $\left( 1-\frac{1}{{{2}^{2}}} \right)\left( 1-\frac{1}{{{3}^{2}}} \right)\left( 1-\frac{1}{{{4}^{2}}} \right)…….\left( 1-\frac{1}{{{100}^{2}}} \right)$

0) $2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}$

Bài 2: Tìm x, biết:
a) $1\frac{1}{2}x+\frac{-4}{5}=4$

b) $\frac{2}{3}x+\frac{-1}{2}x=\frac{-5}{12}$

c) $\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}$

d) $\left( 4\frac{2}{3}-2x \right).2\frac{1}{4}=1\frac{1}{2}$ e) $\frac{3}{5}+\frac{4}{9}:x=\frac{2}{3}$

f) $\frac{7}{35}:\left( x-\frac{1}{3} \right)=-\frac{2}{25}$

g) $\frac{3}{5}\left( \frac{5}{3}-x \right)=\frac{3}{5}\left( \frac{2}{3}-\frac{1}{9} \right)$

h) $\frac{3}{4}\left( x-8 \right)=\frac{5}{7}\left( 14-\frac{7}{2} \right)$

i) $3\left( x-\frac{1}{2} \right)-5\left( x+\frac{3}{5} \right)=-x+\frac{1}{5}$ k) $\left( 2x-1 \right)\left( x+\frac{2}{3} \right)=0$

l) $\displaystyle \frac{x+4}{2011}+\frac{x+3}{2012}+\frac{x+2}{2013}+\frac{x+1}{2014}$

m) $\left( x+1 \right)\left( x+5 \right)>0$