tìm x trong tỉ lệ thức:
a) (x-1)/(x+5)=6/7
b) x2/6=24/25
c) (x-2)/(x-1) = (x+4)/(x+7)(x khác 1,-7)
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11 tháng 3 2022
a: =>x=-7/6+5/8=-13/24
b: =>x=-14/25-3/4=-131/100
c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)
NT
1
15 tháng 10 2021
\(a,\dfrac{4}{x}=\dfrac{8}{x+1}\left(x\ne0;x\ne-1\right)\Rightarrow4x+4=8x\\ \Rightarrow x=1\\ b,\dfrac{x}{7}=\dfrac{x+16}{35}\Rightarrow35x=7x+112\\ \Rightarrow28x=112\Rightarrow x=4\\ c,\dfrac{6}{x-3}=\dfrac{7}{x-5}\left(x\ne3;x\ne5\right)\Rightarrow6x-30=7x-21\\ \Rightarrow x=-9\\ d,\dfrac{44-x}{3}=\dfrac{x-12}{5}\Rightarrow220-5x=3x-36\\ \Rightarrow8x=256\Rightarrow x=32\)
LH
28 tháng 9 2017
a) \(\frac{x-1}{x+5}=\frac{6}{7}\)
7( x - 1 ) = 6( x + 5 )
7x - 7 = 6x - 30
7x - 6x = -30 + 7
x = -23
b) \(\frac{x^2}{6}=\frac{24}{25}\)
\(x^2.25=6.24\)
\(x^2.25=144\)
\(x^2=\frac{144}{25}\)
\(x=\sqrt{\frac{144}{25}}\)
\(x=\frac{12}{5}\)
c) \(\frac{x-2}{x-4}=\frac{x+4}{x+7}\)
\(\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-4\right)\)
\(x^2+7x-2x-14=x^2-4x+4x-16\)
\(x^2+5x-14=x^2^{ }-16\)
\(x^2-x^2+5x=-16+14\)
\(5x=-2\)\(x=\frac{-2}{5}\)
LH
1
XO
11 tháng 7 2019
a) \(\frac{x-1}{x+5}=\frac{6}{7}\)
\(\Rightarrow7.\left(x-1\right)=6.\left(x+5\right)\)
\(\Rightarrow7x-7=6x+30\)
\(\Rightarrow7x-6x=7+30\)
\(\Rightarrow x=37\)
b) \(\frac{x^2}{6}=\frac{24}{25}\)
\(\Rightarrow25.x^2=24.6\)
\(\Rightarrow25.x^2=144\)
\(\Rightarrow x^2=144:25\)
\(\Rightarrow x^2=\frac{144}{25}\)
\(\Rightarrow x^2=\frac{12^2}{5^2}\)
\(\Rightarrow x^2=\frac{12}{5}^2\)
\(\Rightarrow x=\pm\frac{12}{5}\)
11 tháng 7 2019
\(c,\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
\(\Leftrightarrow(x-2)(x+7)=(x+4)(x-1)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow x^2+5x-14-x^2+3x=-4\)
\(\Leftrightarrow\left[x^2-x^2\right]+5x-3x-14=-4\)
\(\Leftrightarrow2x-14=-4\)
\(\Leftrightarrow2x=10\Leftrightarrow x=5\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
a)
b) \(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Leftrightarrow\left(5x\right)^2=144\)
\(\Leftrightarrow\left(5x\right)^2=12^2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)