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6 tháng 8 2021
a)x3-x2=0
⇔x2(x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)3x2-5x=0
⇔ x(3x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)
c)x3=x5
⇔ x3(1-x2)=0
⇔ x3(1-x)(1+x)=0
⇔\(\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d)(2x+7)2-4(2x+7)=0
⇔ (2x+7)(2x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
6 tháng 8 2021
a) Ta có: \(x^3-x^2=0\)
\(\Leftrightarrow x^2\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b) Ta có: \(3x^2-5x=0\)
\(\Leftrightarrow x\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{3}\end{matrix}\right.\)
c) Ta có: \(x^3=x^5\)
\(\Leftrightarrow x^5-x^3=0\)
\(\Leftrightarrow x^3\left(x^2-1\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(\left(2x+7\right)^2-4\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x+7\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
LP
2
5 tháng 3 2022
1.\(\left(x+2\right)\left(2x-3\right)=x^2-4\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
2.\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
3.\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4.\(x^3+x^2-12x=0\)
\(\Leftrightarrow x\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)
5 tháng 3 2022
a: \(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: =>(x+1)(x+2)=0
=>x=-1 hoặc x=-2
c: =>(2x+3)(x+1)=0
=>x=-1 hoặc x=-3/2
d: =>x(x+4)(x-3)=0
hay \(x\in\left\{0;-4;3\right\}\)
19 tháng 12 2021
Bài 1:
a: \(=\dfrac{2x^4-8x^3+2x^2+2x^3-8x^2+2x+18x^2-72x+18+56x-15}{x^2-4x+1}\)
\(=2x^2+2x+18+\dfrac{56x-15}{x^2-4x+1}\)
PT
1
D
1
AH
Akai Haruma
Giáo viên
17 tháng 4 2023
Lời giải:
a.
PT $\Leftrightarrow 3x^2+\frac{x}{2}-3x^2+3x+2=0$
$\Leftrightarrow \frac{7}{2}x+2=0$
$\Leftrightarrow \frac{7}{2}x=-2$
$\Leftrightarrow x=-2: \frac{7}{2}=\frac{-4}{7}$
b.
PT $\Leftrightarrow 5x^2-3-5x^2-6x=0$
$\Leftrightarrow -3-6x=0$
$\Leftrightarrow 6x=-3$
$\Leftrightarrow x=\frac{-3}{6}=\frac{-1}{2}$
PH
1
3 tháng 7 2021
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
