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a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4
(x+7)(x-9)=0
=>\(\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=-7\\x=9\end{cases}}\)
|2x-5| -7 =22
=> |2x-5|=29
=>\(\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\)
=>\(\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\)
=>\(\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)
(x-2)(5-x)=0
=>\(\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
\(\left(x+7\right)\left(x-9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)
a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
a) Ta có: \(x+\left(-5\right)=-\left(-7\right)\)
\(\Leftrightarrow x-5=7\)
\(\Leftrightarrow x=7+5=12\)
Vậy: x=12
b) Ta có: \(-5\cdot\left|x\right|=-30\)
\(\Leftrightarrow\left|x\right|=-30:\left(-5\right)=6\)
hay \(x\in\left\{6;-6\right\}\)
Vậy: \(x\in\left\{6;-6\right\}\)
c) Ta có: \(2x+20=-22\)
\(\Leftrightarrow2x=-22-20=-42\)
hay x=-21
Vậy: x=-21
d) Ta có: \(\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-6\right\}\)
\(\Rightarrow2\left(x-5\right)=14\\ \Rightarrow x-5=7\\ \Rightarrow x=12\)
\(x-\frac{10}{3}=\frac{7}{15}.\frac{3}{5}\)
\(x-\frac{10}{3}=\frac{7}{25}\)
\(x=\frac{7}{25}+\frac{10}{3}=\frac{271}{75}\)
Vậy x = 271/75
\(x-\frac{10}{3}\)\(=\)\(\frac{7}{15}\)\(.\)\(\frac{3}{5}\)
\(x-\frac{10}{3}\)\(=\) \(\frac{7}{25}\)
\(x\) \(=\)\(\frac{7}{25}\)\(+\)\(\frac{10}{3}\)
\(x\) \(=\) \(\frac{271}{75}\)
22 - ( x + 5) : 5 = 2
(x +5 ) : 5 = 22- 2
(x + 5) : 5 = 20
x + 5 = 20 x 5
x + 5 = 100
x = 100 - 5
x = 95