Tìm X giúp mk vs :
( x -1 ).( x2 + 1 ) =0
Cảm ơn !!!!!!!!!!!!!!!
K
Khách
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30 tháng 9 2016
a) \(\left(2x-1\right)^2-25=0\)
\(\left(2x-1\right)^2=0+25=25\)
\(\left(2x-1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-1=5\\2x-1=-5\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}2x=6\\2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\end{array}\right.\)
b) \(8x^3-50x=0\)
\(2x\left(4x^2-25\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=0\\4x^2-25=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-\frac{5}{2}\end{array}\right.\end{array}\right.\)
CH
Cô Hoàng Huyền
Admin
VIP
4 tháng 8 2017
Do \(x\left(x+1\right)⋮2\Rightarrow\left(y^2+1\right)⋮2\Rightarrow\) y2 là số lẻ hay y là số lẻ.
Ta đặt \(y=2k+1\left(k\in Z\right)\), khi đó \(x\left(x+1\right)=\left(2k+1\right)^2+1\)
\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)-\left(2k+1\right)^2=\frac{5}{4}\)
\(\Leftrightarrow4\left(x+\frac{1}{2}\right)^2-4\left(2k+1\right)^2=5\Leftrightarrow\left[\left(2x+1-4k-2\right)\right]\left[\left(2x+1+4k+2\right)\right]=5\)
\(\Leftrightarrow\left(2x-4k-1\right)\left(2x+4k+3\right)=5\)
Tới đây ta tìm được các cặp (x, k), từ đó suy ra các cặp (x,y)
14 tháng 4 2022
a: M(1)=-3
=>1-2m+m=3
=>1-m=3
hay m=-2
b: M(x)=x2+4x-2
Đặt M(x)=0
=>x2+4x+4-6=0
=>(x+2)2=6
hay \(x\in\left\{\sqrt{6}-2;-\sqrt{6}-2\right\}\)
9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
RW
2
T
2 tháng 6 2017
\(\left(2x-1\right)^3-8\left(x-1\right)\left(x^2+x+1\right)+12x^2=2x+1\)
\(\Leftrightarrow8x^3-12x^2+6x-1-8\left(x^3-1\right)+12x^2-2x-1=0\)
\(\Leftrightarrow4x+6=0\)
\(\Leftrightarrow2\left(2x+3\right)=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\frac{-3}{2}\)
\(\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Leftrightarrow x=1}\)
Vậy........
( x - 1 ).( x2 + 1 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x^2=-1\left(v\text{ô}l\text{í}\right)\end{cases}}\)
Vậy x = 1