Tìm x
x/5+x/2=42,35
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a: x(x+5)^2=x+5
=>(x+5)(x^2+5x-1)=0
=>x+5=0 hoặc x^2+5x-1=0
=>\(x\in\left\{-5;\dfrac{-5+\sqrt{29}}{2};\dfrac{-5-\sqrt{29}}{2}\right\}\)
b: x(x-2)=(x-2)
=>(x-2)(x-1)=0
=>x=2 hoặc x=1
\(x^2\left(x-5\right)+x-5=0\)
\(\Rightarrow x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
\(x^2\left(x-5\right)+x-5=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
Bài 1:
a: x/-2=-18/x
=>x2=36
=>x=6 hoặc x=-6
b: x/2+x/5=17/10
=>7/10x=17/10
hay x=17/7
`x xx 6/7=5/14`
`=>x=5/14:6/7`
`=>x=5/14xx7/6`
`=>x=35/84`
`=>x=5/12`
Vậy `x=5/12`
__
`x:2/3=4/9`
`=>x=4/9xx2/3`
`=>x=8/27`
Vậy `x=8/27`
__
`x-1/4=3/2`
`=>x=3/2+1/4`
`=>x=6/4+1/4`
`=>x=7/4`
Vậy `x=7/4`
__
`x+4/5=8/9`
`=>x=8/9-4/5`
`=>x=40/45-36/45`
`=>x=4/45`
Vậy `x=4/45`
\(x\cdot\dfrac{6}{7}=\dfrac{5}{14}\)
\(x\) \(=\dfrac{5}{14}:\dfrac{6}{7}\)
\(x\) \(=\dfrac{5}{12}\)
\(x:\dfrac{2}{3}=\dfrac{4}{9}\)
\(x\) \(=\dfrac{4}{9}\cdot\dfrac{2}{3}\)
\(x\) \(=\dfrac{8}{27}\)
\(x-\dfrac{1}{4}=\dfrac{3}{2}\)
\(x\) \(=\dfrac{3}{2}+\dfrac{1}{4}\)
\(x\) \(=\dfrac{7}{4}\)
\(x+\dfrac{4}{5}=\dfrac{8}{9}\)
\(x\) \(=\dfrac{8}{9}-\dfrac{4}{5}\)
\(x\) \(=\dfrac{4}{45}\)
\(\frac{x}{5}+\frac{x}{2}=42,35\)
\(\Rightarrow\frac{2x}{10}+\frac{5x}{10}=42,35\)
\(\Rightarrow\frac{7x}{10}=42,35\)
\(\Rightarrow7x=42,35\times10=423,5\)
\(\Rightarrow x=423.5:7=60,5\)
\(\frac{x}{5}+\frac{x}{2}=42,35 \)
\(x\cdot\frac{1}{5}+x\cdot\frac{1}{2}=42,35\)
\(x\cdot0,2+x\cdot0,5=42,35\)
\(x\cdot\left(0,2+0,5\right)=42,35\)
\(x\cdot0,7=42,35\)
\(x=42,35:0,7\)
\(x=60,5.\)
Vậy \(x=60,5.\)