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Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
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`(x-2):(x-1)=(x+4)(x+7)`
\(< =>\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\left(x\ne1;x\ne-7\right)\)
`=>(x-2)(x+7)=(x+4)(x-1)`
`<=>x^2 +7x-2x-14=x^2 -x+4x-4`
`<=>x^2 +5x-14-x^2 -3x+4=0`
`<=>2x-10=0`
`<=>2x=10`
`<=>x=5(tm)`
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a: \(\dfrac{x-3}{5-x}=\dfrac{5}{7}\left(x\ne5\right)\)
=>7(x-3)=5(5-x)
=>7x-21=25-5x
=>12x=46
=>x=23/6
b: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)(ĐKXĐ: \(x\notin\left\{1;-7\right\}\))
=>(x-2)(x+7)=(x+4)(x-1)
=>\(x^2+5x-14=x^2+3x-4\)
=>5x-14=3x-4
=>2x=10
=>x=5(nhận)
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Ta có
x+4/x+7=x-2/x-1=x+4-(x-2) / x+7-(x-1)
=x+4-x+2/x+7-x+1
=x-x+(4+2)/x-x+(7+1)
=3/4
x+4/x+7=3/4=>(x+4)x4=(x+7)x3=>4x+16=3x+21=>4x-3x=21-16=>x=5
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\(\frac{x-2}{x-1}=\frac{x+4}{x-7}\)
(x - 2)(x - 7) = (x + 4)(x - 1)
x2 - 7x - 2x + 14 = x2 - x + 4x - 4
x2 - 9x + 14 = x2 + 3x - 4
x2 - 9x - x2 - 3x = - 4 - 14
- 12x = - 18
x = \(\frac{3}{2}\)
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\(\dfrac{x-1}{7}+\dfrac{x-2}{3}+\dfrac{x-3}{5}+\dfrac{x-4}{2}=6\\ =>\left(\dfrac{x-1}{7}-1\right)+\left(\dfrac{x-2}{3}-2\right)+\left(\dfrac{x-3}{5}-1\right)+\left(\dfrac{x-4}{2}-2\right)=0\\ =>\dfrac{x-8}{7}+\dfrac{x-8}{3}+\dfrac{x-8}{5}+\dfrac{x-8}{2}=0\\ =>\left(x-8\right)\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{2}\right)=0\\ =>x-8=0\\ =>x=8\)
\(\dfrac{x}{4}\) = \(\dfrac{2}{7}\)
\(x\) = \(\dfrac{2}{7}\) \(\times\) 4
\(x\) = \(\dfrac{8}{7}\)