Tìm x biết
x-3/27=3/x-2
Giup mk với ạ
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\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)
\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)
\(=-\dfrac{4}{9}x^7.y^3.z\)
`(-x^3zy)(2/3yx^2)^2`
`=-4/9x^3zy.y^2x^4`
`=-4/9x^{3+4}.y^{1+2}z`
`=-4/9x^7y^3z`
\(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow x-3=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)\left[1-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
___________
\(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot\dfrac{1}{2}\cdot x^2+3\cdot\left(\dfrac{1}{2}\right)^2\cdot x+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{4}\)
\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)
`đk:x ne 0,-2`
`a)D=(x/(x+2)+(8x+8)/(x^2+2x)-(x+2)/x):((x^2-x-3)/(x^2+2x)+1/x)`
`=((x^2+8x+8-x^2-4x-4)/(x(x+2))):((x^2-x-3+x+2)/(x(x+2)))`
`=(4x+4)/(x(x+2)):(x^2-1)/(x(x+2))`
`=(4x+4)/(x^2-1)(x ne +-1)`
`=4/(x-1)`
`b)x(x-2)-(x-2)=0`
`<=>(x-2)(x-1)=0`
Vì `x ne 1=>x-1 ne 0`
`=>x-2=0<=>x=2`
`=>D=4/(2-1)=4`
`c)D<0`
Mà `4>0`
`=>x-1<0`
`=>x<1`
Kết hợp đkxđ:
`=>x<1,x ne 0,x ne -2`
`d)D=2`
`<=>4/(x-1)=2`
`<=>2/(x-1)=1`
`<=>x-1=2`
`<=>x=3(tm)`
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=-\dfrac{54}{5}\)
\(\dfrac{x}{2}=-\dfrac{54}{5}\Rightarrow x=-\dfrac{54}{5}.2=-\dfrac{108}{5}\)
\(\dfrac{y}{3}=-\dfrac{54}{5}\Rightarrow y=-\dfrac{54}{5}.3=-\dfrac{162}{5}\)
Vậy \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)
nên \(\dfrac{2x}{4}=\dfrac{3y}{9}\)
mà 2x-3y=54
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=\dfrac{-54}{5}\)
Do đó: \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)
\(x=\dfrac{7}{25}+\dfrac{-1}{5}=\dfrac{7}{25}-\dfrac{1}{5}=\dfrac{2}{25}.\\ x=\dfrac{5}{11}+\dfrac{4}{-9}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{1}{99}.\\ \dfrac{5}{9}-\dfrac{x}{-1}=\dfrac{-1}{3}\Leftrightarrow\dfrac{5}{9}+x=-\dfrac{1}{3}.\Leftrightarrow x=-\dfrac{8}{9}.\)
\(x=\dfrac{7}{25}+-\dfrac{1}{5}=>\dfrac{7}{25}+-\dfrac{5}{25}=>x=\dfrac{2}{25}\)
\(x=\dfrac{5}{11}+\dfrac{4}{-9}=>\dfrac{-45}{-99}+\dfrac{44}{-99}=>x=\dfrac{-1}{-99}=\dfrac{1}{99}\)
\(\dfrac{5}{9}-\dfrac{x}{-1}=-\dfrac{1}{3}=>-\dfrac{1}{3}-\dfrac{5}{9}=>\dfrac{x}{-1}=-\dfrac{8}{9}=>x=-\dfrac{8}{9}\)