Bài 2: Tìm x:
a)x3 - 1/4x = 0
b) (2x-1)2 - (x+3)2= 0
c) x2 (x-3)+12-4x=0
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CI
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C
3
AH
Akai Haruma
Giáo viên
28 tháng 8 2021
Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
28 tháng 8 2021
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
9 tháng 12 2021
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
C
3
7 tháng 8 2021
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2021
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
ZH
1
5 tháng 10 2021
\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)
\(a,x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}}\)
\(b,\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{3}\\x=4\end{cases}}\)
\(c,x^2\left(x-3\right)+12-4x=0\)
\(\Leftrightarrow x^2\left(x-3\right)+4\left(3-x\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\pm2\end{cases}}}\)
a) x3 - 14/x = 0
<=> x(x + 1/2)(x - 1/2) = 0
<=> x = 0 hoặc x + 1/2 = 0 hoặc x - 1/2 = 0
x = 0 - 1/2 x = 0 + 1/2
x = -1/2 x = 1/2
=> x = 0 hoặc x = -1/2 hoặc x = 1/2
b) (2x - 1)2 - (x + 3)2 = 0
<=> 3x2 - 10x - 8 = 0
<=> 3x2 + 2x - 12x - 8 = 0
<=> x(3x + 2) - 4(3x + 2) = 0
<=> (3x + 2)(x - 4) = 0
3x + 2 = 0 hoặc x - 4 = 0
3x = 0 - 2 x = 0 + 4
3x = -2 x = 4
x = -2/3
=> x = -2/3 hoặc x = 4
c) x2(x - 3) + 12 - 4x = 0
<=> (x2 - x - 6)(x - 2) = 0
<=> (x - 3)(x + 2)(x - 2) = 0
x - 3 = 0 hoặc x + 2 = 0 hoặc x - 2 = 0
x = 0 + 3 x = 0 - 2 x = 0 + 2
x = 3 x = -2 x = 2
=> x = 3 hoặc x = -2 hoặc x = 2