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20 tháng 12 2021
bn ơi đừng đùa dai nữa bn trung học cơ sở mà ko ccongj đc hơn cả cá em lớp 1
nhắc lần thứ N @congtibaocao
DL
1
NM
15 tháng 9 2023
\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)
\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)
HT
1 tháng 9 2015
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{1\frac{1}{6}-0,875+0,7}+\frac{6}{7}\)
= \(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}.\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}+\frac{6}{7}\)
= \(\frac{1}{2}.\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}+\frac{6}{7}\)
= \(\frac{1}{2}.\frac{1}{\frac{7}{2}}+\frac{6}{7}\)
= \(\frac{1}{2}.\frac{2}{7}+\frac{6}{7}\)
= \(\frac{1}{7}+\frac{6}{7}\)
= 1
13 tháng 7 2021
\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)
\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)
\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)
\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)
\(=\frac{3}{13}.\frac{39}{54}+1\)
\(=\frac{1}{6}+1\)
\(=\frac{7}{6}\)
\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)
\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)
\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)
\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)
\(=\frac{5}{6}+\frac{7}{13}-\frac{2}{9}\)
\(=\frac{195+126-52}{234}\)
\(=\frac{269}{234}\)
13 tháng 7 2021
\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)
\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)
\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)
\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)
\(=\frac{3}{13}.\frac{39}{54}+1\)
\(=\frac{1}{6}+1=\frac{1}{6}+\frac{6}{6}\)
\(=\frac{7}{6}\)
\(\frac{-7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{-7}{9}.\frac{2}{13}+\frac{-7}{9}.\frac{11}{13}+\frac{-2}{9}\)
\(=\frac{-7}{9}.\left(\frac{2}{13}+\frac{11}{13}\right)+\frac{-2}{9}\)
\(=\frac{-7}{9}.1+\frac{-2}{9}\)
\(=\frac{-7}{9}+\frac{-2}{9}\)
\(=\frac{-9}{9}=-1\)
\(\frac{2}{13}.\frac{2}{7}.5\)
\(=\frac{2.2.5}{13.7}\)
\(=\frac{20}{91}\)
\(\frac{1}{5}.\frac{11}{12}.\frac{21}{6}\)
\(=\frac{11.21}{5.12.6}\)
\(=\frac{231}{360}=\frac{77}{120}\)
19 tháng 7 2021
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
NM
Nguyễn Minh Quang
Giáo viên
19 tháng 7 2021
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
TA
11 tháng 10 2017
\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
1/7 . 1/3 + 1/7 . 1/2 - 1/7
=1/7(1/3+1/2-1)
=1/7 (1/3-1/2)
=1/7 (-1/6)=-1/42
\(\dfrac{1}{7}.\dfrac{1}{3}+\dfrac{1}{7}.\dfrac{1}{2}-\dfrac{1}{7}\)
\(\dfrac{1}{7}.\dfrac{1}{3}+\dfrac{1}{7}.\dfrac{1}{2}-\dfrac{1}{7}.1\)
\(\dfrac{1}{7}\left(\dfrac{1}{3}+\dfrac{1}{2}-1\right)\)
\(\dfrac{1}{7}\cdot-\dfrac{1}{6}\)
\(-\dfrac{1}{42}\)