Tìm x
x-6=(6-x)^2
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a, \(x + 1/6 = -3/8 \)
\(x = -3/8 - 1/6\)
\(x = -13/24\)
Vậy \(x = -13/24\)
b, \(-3/7 - x = 4/5 - 2/3\)
\(-3/7 - x = 2/15\)
\(x = -3/7 - 2/15\)
\(x = -59/105.\)
Vậy \(x = -59/105\)
x+1/6=-3/8
x=-3/8-1/6
x=-13/24
-3/7-x=4/5+-2/3
-3/7-x=2/15
x=-3/7-2/15
x=-59/105
1/2x3/2+1/3x4/2+1/4x5/2+1/5x6/2+.......+2/Xx(X+1)=2011/2013
2/2x3+2/3x4+2/4x5+2/5x6+.....+2/Xx(X+1)=2011/2013
2x(1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(x+1)=2011/2013
1/2x3+1/3x4+1/4x5+1/5x6+....+1/Xx(X+1)=2011/4026
1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+.....+ 1/x-1/x+1=2011/4026
1/2-1/x+1=2011/4026
1/x+1=1/2-2011/4026
1/x+1=1/2013
Suy ra x=2012
`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`
\(x\cdot\dfrac{6}{7}=\dfrac{5}{14}\)
\(x\) \(=\dfrac{5}{14}:\dfrac{6}{7}\)
\(x\) \(=\dfrac{5}{12}\)
\(x:\dfrac{2}{3}=\dfrac{4}{9}\)
\(x\) \(=\dfrac{4}{9}\cdot\dfrac{2}{3}\)
\(x\) \(=\dfrac{8}{27}\)
\(x-\dfrac{1}{4}=\dfrac{3}{2}\)
\(x\) \(=\dfrac{3}{2}+\dfrac{1}{4}\)
\(x\) \(=\dfrac{7}{4}\)
\(x+\dfrac{4}{5}=\dfrac{8}{9}\)
\(x\) \(=\dfrac{8}{9}-\dfrac{4}{5}\)
\(x\) \(=\dfrac{4}{45}\)
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
\(x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{1}{2}\\ x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{3}{6}\\ x-\dfrac{3}{6}=\dfrac{25}{6}\\ =>x=\dfrac{25}{6}+\dfrac{3}{6}=\dfrac{28}{6}=\dfrac{14}{3}\)
\(x-\dfrac{3}{6}=\dfrac{28}{6}-\dfrac{1}{2}\)
\(x-\dfrac{3}{6}=\dfrac{28-3}{6}=\dfrac{25}{6}\)
\(=>x=\dfrac{25}{6}+\dfrac{3}{6}=\dfrac{25+3}{6}=\dfrac{28}{6}=\dfrac{4}{1}\)
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow6-\sqrt{x+1}-x-1=0\\ \Leftrightarrow5-x=\sqrt{x+1}\\ \Leftrightarrow25-10x+x^2=x+1\left(x\le5\right)\\ \Leftrightarrow x^2-11x+24=0\\ \Leftrightarrow\left(x^2-3x\right)-\left(8x-24\right)=0\\ \Leftrightarrow x\left(x-3\right)-8\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=8\left(ktm\right)\end{matrix}\right.\)
Vậy \(x=3\)
\(\left(6-x\right)^2=x-6\)\(< =>\left(6-x\right)^2+6-x=0\)
\(< =>\left(6-x\right)\left(6-x+1\right)=0\)
\(< =>\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
Trả lời:
\(x-6=\left(6-x\right)^2\)
\(\Leftrightarrow\left(x-6\right)-\left(6-x\right)^2=0\)
\(\Leftrightarrow\left(x-6\right)-\left(x-6\right)^2=0\)
\(\Leftrightarrow\left(x-6\right)\left(1-x+6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(7-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}}\)
Vậy x = 6; x = 7 là nghiệm của pt.