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\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
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1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\dfrac{3}{2}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{3}{2}\)
\(x=\dfrac{13}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{8}{x}=13+\frac{3}{7}=\frac{94}{7}\)\(\Leftrightarrow x=\frac{7x8}{94}=\frac{56}{94}=\frac{28}{47}\)
\(\frac{8}{x}=\frac{3}{7}+13\)
\(\frac{8}{x}=\frac{94}{7}\)
\(x=8:\frac{94}{7}\)
\(x=\frac{28}{47}\)
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Bài 2:
\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)
Bài 1:
a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)
nên x=13/8-7/8=6/8=3/4
b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)
nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Thiếu đề
b) ` 4/7 xx 8/13 + 8/13 xx 3/7`
= ` 8/13 xx ( 4/7 + 3/7)`
` = 8/13 xx 7/7`
` = 8/13 xx 1`
` = 8/13`
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x x - 4/7 = 15/20
(hình như sai đề, tại vì lớp 4 chưa học lũy thừa)
x * 2/7 -2/3 =1/2
x* 2/7 = 1/2 + 2/3
x* 2/7 = 7/6
x= 7/6 : 2/7
x= 49/12
x : 5/4 + 4/5 =2
x: 5/4 = 2 - 4/5
x: 5/4 = 6/5
x= 6/5 * 5/4
x= 3/2
4/3 x 7/8 x 5
= 7/6 x 5
= 35/6
8/9 x 6/5 x 2/3
= 16/15 x 2/3
= 32/45
4/7 : 5/7 x 10/11
= 4/5 x 10/11
= 8/11
8/13 : 7/13 : 4
= 8/7 : 4
= 2/7
x=5(nếu x thuộc z)
X =5 chứ đâu