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\(1,x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(2,\left(x+2\right)\left(x-3\right)-x-2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)
\(3,36x^2-49=0\)
\(\Leftrightarrow\left(6x\right)^2-7^2=0\)
\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}6x-7=0\\6x+7=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{6}\\x=\frac{7}{6}\end{cases}}\)
Chúc bn học giỏi nhoa!!!
Ta có : x2 - x = 0
=> x(x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
1) \(4x^3-36x=0\)
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}4x=0\\x^2-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=\pm3\end{cases}}\)
2) \(\left(3x-5\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow9x^2-30x+25-x^2-2x-1=0\)
\(\Leftrightarrow8x^2-32x+24=0\)
\(\Leftrightarrow8\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow8\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=1\end{cases}}\)
\(a,4x^3-36x=0\)
\(\Rightarrow4x\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow4x=0\) hoặc \(x+3=0\) hoặc \(x-3=0\)
\(\Rightarrow x\in\left\{0;-3;3\right\}\)
Vậy.....
\(b,\left(3x-5\right)^2-\left(x+1\right)^2=0\)
\(\Rightarrow\left(4x-4\right)\left(2x-6\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}}\)
Vậy...
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a, (x+3)2 - ( 2x + 1 ).( x+3)=0 b, x3-12x2+36x =0
=> (x+3).(x+3-2x-1) => x(x2-12x+36) = 0
=>(x+3).(-x+2) => x(x-6)2 = 0
=> x+3=0 <=> x=-3 => x=0 <=> x=0
-x+2=0 <=> x=-2 x-6= 0 <=> x=6
\(a,x^2-5x=0\)
\(x.\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\Rightarrow x=5\end{cases}}\)
vậy x=0 hay x=5
\(b,x^2-x=0\)
\(x.\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\Rightarrow x=1\end{cases}}\)
vậy x=0 hay x=1
\(c,36x^2-49=0\)
\(\Rightarrow36x^2=49\)
\(x^2=\frac{49}{36}=\frac{7^2}{6^2}=\frac{\left(-7\right)^2}{\left(-6\right)^2}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{6}\\x=-\frac{7}{6}\end{cases}}\)
vậy x=\(\frac{7}{6}hayx=-\frac{7}{6}\)
a) \(x^2-36=0\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)
b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)
\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)
\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)
36x - x2 = 0
<=> x(36 - x) = 0
<=> x = 0 hoặc 36 - x = 0
<=> x = 0 hoặc x = 36
Vậy x = 0 hoặc x = 36
ung ho minh len 200 nha