tìm x
(2/3.x-5/6)^2=4/25
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\(x\) \(\times\) \(\dfrac{4}{5}\) = 15
\(x\) = 15 : \(\dfrac{4}{5}\)
\(x\) = \(\dfrac{75}{4}\)
\(\dfrac{12}{25}\) \(\times\) \(x\) = \(\dfrac{4}{10}\)
\(x\) = \(\dfrac{4}{10}\) : \(\dfrac{12}{25}\)
\(x\) = \(\dfrac{5}{6}\)
1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
Tìm xx biết: \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19
Trả lời: x=x=
a, Ta có :\(-30+\left(25-x\right)=-1\)
\(\Leftrightarrow\left(-30\right)+25-x=-1\)
\(\Leftrightarrow25-x=\left(-1\right)-\left(-30\right)\)
\(\Leftrightarrow25-x=29\\ \Leftrightarrow x=25-29\)
\(\Leftrightarrow x=\left(-4\right)\)
Vậy \(x=-4\)
b,Ta có :\(\left(x+5\right)+\left(x-9\right)=x+2\)
\(\Leftrightarrow x+5+x-9=x+2\)
\(\Leftrightarrow2.x+\left(5-9\right)=x+2\)
\(\Leftrightarrow\) \(2.x+\left(-4\right)=x+2\)
\(\Leftrightarrow2.x-x=4+2\)
\(\Leftrightarrow x=6\)
Vậy \(x=6\)
a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
\(\left(\frac{2}{3}x-\frac{5}{6}\right)^2=\frac{4}{25}\)
\(\Leftrightarrow\left(\frac{2}{3}x-\frac{5}{6}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{5}{6}=\frac{2}{5}\\\frac{2}{3}x-\frac{5}{6}=-\frac{2}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{37}{30}\\\frac{2}{3}x=\frac{13}{30}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{37}{20}\\x=\frac{13}{20}\end{cases}}\)
\(\left(\frac{2}{3}.x-\frac{5}{6}\right)^2=\left(\frac{2}{5}\right)^2\)
\(\frac{2}{3}.x-\frac{5}{6}=\frac{2}{5}\)
\(\frac{2}{3}.x=\frac{2}{5}+\frac{5}{6}\)
\(\frac{2}{3}.x=\frac{37}{30}\)
\(x=\frac{37}{30}:\frac{2}{3}\)
\(x=\frac{37}{20}\)
vậy \(x=\frac{37}{30}\)