g x 9/10 = 2/9
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![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a)`
`x + 10 = 20`
`=> x = 20 -10`
`=> x = 10`
Vậy, `x = 10`
`b)`
`2 * x + 15 = 35`
`=> 2x = 35 - 15`
`=> 2x = 20`
`=> x = 20 \div 2`
`=> x = 10`
Vậy, `x = 10`
`c)`
`3 * ( x + 2 ) = 15`
`=> x + 2 = 15 \div 3`
`=> x + 2 = 5`
`=> x = 5 - 2`
`=> x = 3`
Vậy, `x = 3`
`d)`
`10 * x + 15 * 11 = 20 * 10`
`=> 10x + 165 = 200`
`=> 10x = 200 - 165`
`=> 10x = 35`
`=> x = 35 \div 10`
`=> x = 3,5`
Vậy,` x = 3,5`
`e)`
`4 * ( x + 2 ) = 3 * 4`
`=> x + 2 = 12 \div 4`
`=> x + 2 = 3`
`=> x = 3 - 2`
`=> x = 1`
Vậy,` x = 1`
`f)`
`33 x + 135 = 26 * 9`
`=> 33x + 135 = 234`
`=> 33x = 234 - 135`
`=> 33x = 99`
`=> x = 99 \div 33`
`=> x = 3`
Vậy, `x = 3`
`g)`
`2 * x + 15 + 16 + 17 = 100`
`=> 2x + 48 = 100`
`=> 2x = 100 - 48`
`=> 2x = 52`
`=> x = 52 \div 2`
`=> x =26`
`h)`
`2 * (x + 9 + 10 + 11) = 4 . 12 . 25`
`=> 2 * (x + 9 + 10 + 11) = 4*25*12`
`=> 2 * (x + 9 + 10 + 11) = 100*12`
`=> x + 9 + 10 + 11 = 100*12 \div 2`
`=> x + 30 = 600`
`=> x = 600 - 30`
`=> x = 570`
Vậy, `x = 570.`
a) \(x+10=20\Leftrightarrow x=10\)
b) \(2x+15=35\Leftrightarrow2x=20\Leftrightarrow x=10\)
c) \(3.\left(x+2\right)=15\Leftrightarrow x+2=5\Leftrightarrow x=3\)
d) \(10x+15.11=20.10\Leftrightarrow10x+165=200\Leftrightarrow10x=35\Leftrightarrow x=\dfrac{35}{10}=\dfrac{7}{2}\)
e) \(4.\left(x+2\right)=3.4\Leftrightarrow x+2=3\Leftrightarrow x=1\)
f) \(35x+135=26.9\Leftrightarrow35x=234-135\Leftrightarrow35x=99\Leftrightarrow x=\dfrac{99}{35}\)
g) \(2x+15+16+17=100\Leftrightarrow2x+48=100\Leftrightarrow2x=52\Leftrightarrow x=26\)
h) \(2.\left(x+9+10+11\right)=4.12.25\)
\(\Leftrightarrow x+30=2.12.25\)
\(\Leftrightarrow x=600-30\)
\(\Leftrightarrow x=570\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(G=x^4+10x^3+10x^2+10x+10\)
\(=x^4+10\left(x^3+x^2+x+1\right)\)
\(=\left(-9^4\right)+10\left[\left(-9\right)^3+\left(-9\right)^2+-9+1\right]\)
\(=6561+10\cdot-656\)
\(=6561-6560\)
\(=1\)
Thay `x=-9` vào biểu thức G:
`G=(-9)^4+10.(-9)^3+10.(-9)^2+10.(-9)+10`
`=6561-7290+810-90+10`
`=1`
![](https://rs.olm.vn/images/avt/0.png?1311)
e)10:(x-2)=2
x-2=10:2
x=5+2
x=7
g)(70-x):8=9
70-x=8*9
x=72-70
x=2
h)(x-10)-30=47
x-10=47+30
x=77+10
x-87
![](https://rs.olm.vn/images/avt/0.png?1311)
9 x 2 = 18
9 x 5 = 45
9 x 8 = 72
9 x 10 = 90
2 x 9 = 18
5 x 9 = 45
8 x 9 = 72
10 x 9 = 90
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
b: \(5x+9⋮x+2\)
=>\(5x+10-1⋮x+2\)
=>\(-1⋮x+2\)
=>\(x+2\in\left\{1;-1\right\}\)
=>\(x\in\left\{-1;-3\right\}\)
c: \(2x+11⋮x+3\)
=>\(2x+6+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
d: \(4x+9⋮2x+1\)
=>\(4x+2+7⋮2x+1\)
=>\(7⋮2x+1\)
=>\(2x+1\in\left\{1;-1;7;-7\right\}\)
=>\(2x\in\left\{0;-2;6;-8\right\}\)
=>\(x\in\left\{0;-1;3;-4\right\}\)
e: \(6x+7⋮3x+1\)
=>\(6x+2+5⋮3x+1\)
=>\(5⋮3x+1\)
=>\(3x+1\in\left\{1;-1;5;-5\right\}\)
=>\(3x\in\left\{0;-2;4;-6\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)
g: \(10x+13⋮5x+1\)
=>\(10x+2+11⋮5x+1\)
=>\(11⋮5x+1\)
=>\(5x+1\in\left\{1;-1;11;-11\right\}\)
=>\(5x\in\left\{0;-2;10;-12\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Rightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy x = 2
b) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Rightarrow-3x=-48\)
\(\Leftrightarrow x=16\)
Vậy x = 16
c) \(\dfrac{1}{9}=\dfrac{-2x}{10}\)
\(\Rightarrow-18x=10\)
\(\Leftrightarrow x=-\dfrac{5}{9}\)
Vậy \(x=-\dfrac{5}{9}\)
d) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{x}-5=\dfrac{-9}{x}+2\)
\(\Leftrightarrow\dfrac{3-5x}{x}=\dfrac{-9+2x}{x}\)
\(\Rightarrow3-5x=-9+2x\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy \(x=\dfrac{12}{7}\)
e) ĐKXĐ: \(x\ne0\)
\(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy \(x=\pm4\)
a) Ta có: \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Leftrightarrow-x=\dfrac{6\cdot\left(-8\right)}{3}=-16\)
hay x=16
Vậy: x=16
![](https://rs.olm.vn/images/avt/0.png?1311)
\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
\(a,|x-\frac{1}{2}|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)
\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)
\(\frac{28}{27}=x:\frac{3}{7}\)
\(x=\frac{4}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)
\(=\dfrac{5}{7}\times1\)
\(=\dfrac{5}{7}\)
b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)
\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)
\(=1+0\)
\(=1\)
c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)
\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)
\(=\dfrac{5}{7}+\dfrac{2}{7}\)
\(=1\)
d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)
\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)
\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)
\(=\dfrac{1}{2}+1\)
\(=\dfrac{3}{2}\)
e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)
\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)
\(=\dfrac{4}{5}+\dfrac{12}{10}\)
\(=\dfrac{4}{5}+\dfrac{6}{5}\)
\(=\dfrac{10}{5}\)
\(=2\)
g) \(362\times728+326\times272\)
\(=326\times\left(728+272\right)\)
\(=326\times1000\)
\(=326000\)