a) (2x+1)2 - 1/9 = 0
b) (x-2)3 + 27 = 0
c) x2 - x = 0
d) x4 =27x
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AH
Akai Haruma
Giáo viên
2 tháng 10 2021
a.
$x^4-6x^2+9=0$
$\Leftrightarrow (x^2-3)^2=0$
$\Leftrightarrow x^2-3=0$
$\Leftrightarrow x^2=3$
$\Leftrightarrow x=\pm \sqrt{3}$
b.
$8x^3+12x^2+6x-63=0$
$\Leftrightarrow (8x^2+12x^2+6x+1)-64=0$
$\Leftrightarrow (2x+1)^3=64=4^3$
$\Leftrightarrow 2x+1=4$
$\Leftrightarrow x=\frac{3}{2}$
c. $(3-2x)^2-25=0$
$\Leftrightarrow (3-2x)^2-5^2=0$
$\Leftrightarrow (3-2x-5)(3-2x+5)=0$
$\Leftrightarrow (-2-2x)(8-2x)=0$
$\Leftrightarrow -2-2x=0$ hoặc $8-2x=0$
$\Leftrightarrow x=-1$ hoặc $x=4$
AH
Akai Haruma
Giáo viên
2 tháng 10 2021
d.
$6(x+1)^2-2(x+1)^3+2(x-1)(x^2+x+1)=1$
$\Leftrightarrow (x+1)^2[6-2(x+1)]+2(x^3-1)=1$
$\Leftrightarrow (x+1)^2(4-2x)+2x^3-3=0$
$\Leftrightarrow 6x+1=0$
$\Leftrightarrow x=\frac{-1}{6}$
e. $(x-2)^2-(x-2)(x+2)=0$
$\Leftrightarrow (x-2)[(x-2)-(x+2)]=0$
$\Leftrightarrow (x-2)(-4)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
f. $x^2-4x+4=25$
$\Leftrightarrow (x-2)^2=5^2=(-5)^2$
$\Leftrightarrow x-2=5$ hoặc $x-2=-5$
$\Leftrightarrow x=7$ hoặc $x=-3$
23 tháng 10 2021
e: ta có: \(4x^2+4x-6=2\)
\(\Leftrightarrow4x^2+4x-8=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
f: Ta có: \(2x^2+7x+3=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
YY
2
10 tháng 2 2022
a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)
b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)
c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)
d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)
R
10 tháng 2 2022
a)
/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
9 tháng 12 2021
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
8 tháng 1 2021
a) Ta có: \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
b) Ta có: \(x^3-6x^2+11x-6=0\)
\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)
\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={1;2;3}
c) Ta có: \(x^3-x^2-21x+45=0\)
\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)
\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: S={3;-5}
d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên (x-2)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy: S={2;-3}
a)=>\(\left(2x+1\right)^2=\frac{1}{9}\)
\(=>\left(2x+1\right)^2=\frac{1}{3^2}\)
\(=>2x+1=\frac{1}{3}\)
\(=>2x=\frac{1}{3}-1\)
\(=>2x=\frac{-2}{3}\)
\(=>x=\frac{-2}{3}:2\)
\(=>x=\frac{-1}{3}\)
Vậy x = \(-\frac{1}{3}\)
b)\(=>\left(x-2\right)^3=27\)
\(=\left(x-2\right)^3=3^3\)
\(=>x-2=3\)
\(=>x=3+2\)
\(=>x=5\)
Vậy x = 5
c)=>x.x-x=0
TH1:\(\hept{\begin{cases}x.x=0\\x=0\end{cases}}\)\(=>\hept{\begin{cases}x=0\\x=0\end{cases}}\)
TH2:\(x.x=1.x=>x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d)\(x^4=27.x\)
\(=>x^4-27x=0\)
\(=>x^4-\left[\left(3\right)^3.x\right]=0\)
\(=>x^3.x-3^3.x=0\)
\(=>x.\left(x^3-3^3\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x^3-3^3=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x^3=3^3\end{cases}=>\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)
X khong thể bằng (-3) được
Vậy x \(\in\){0;3}
a) Ta có: \(\left(2x+1\right)^2-\frac{1}{9}=0\)
\(\left(2x+1\right)^2=\frac{1}{9}\)
mà \(\frac{1}{9}=\left(\frac{1}{3}\right)^2=\left(-\frac{1}{3}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{1}{3}\\2x+1=-\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-\frac{2}{3}\\2x=-\frac{4}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-\frac{2}{3}\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{3};-\frac{2}{3}\right\}\)
b) (x-2)3 + 27 = 0
(x-2)3 = -27
mà -27=(-3)3
=> x-2=-3
=> x= -1
c)Ta có: x2 - x = 0
x . (x-1) = 0
\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy \(x\in\left\{0;1\right\}\)