(x-8)+(x-4)=x+1
Giải giúp mình nha <3
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\(x^5-x^3+x^2-1=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x+1\right)^2\left(x^2-x+1\right)\)
1)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(x\times\left(x+2\right)-3\times\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\times\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)
nhân chéo lên, ngại chết đc
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
Hệ số góc bằng 2 `=> 3a+4 = 2 <=> a=-2/3`
Tung độ gốc bằng 1 `=> -b+3=1 <=> b=2`
`=> y=-2/3 x +2`
a)Ta có:5/3.x^2-1/2.x^2y
=(5/3-1/2).x^2y
= 7/6.x^2y(Bậc 3)
b)Ta có: 7/6.(-2)^2(-1)
= 7/6.4.(-1)
= 7/6.(-4)
=-28/6
a, - A=\(\dfrac{5}{3}\).x2.y-\(\dfrac{-1}{2}\).x2.y
=\(\dfrac{13}{6}\).x2.y
- Bậc= 3.
b, A=\(\dfrac{13}{6}\).(-2)2.(-1)
=\(\dfrac{13}{6}\).4.(-1)
=\(\dfrac{-26}{3}\)
\(\Rightarrow\)x-8+x-4-x-1=0
\(\Rightarrow\)x-13=0
\(\Rightarrow\)x=13
học tốt nha\(\left(x-8\right)+\left(x-4\right)=x+1\)
\(\Leftrightarrow x-8+x-4=x+1\)
\(\Leftrightarrow2x-12=x+1\)
\(\Leftrightarrow2x-x=12+1\Leftrightarrow x=13\)
Vậy \(x=13\)