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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 $x=-3$
NG
a,x3+3x2+3x+1
b,x2+6x+9
c,-x3+9x2-27x+27
d,x2+4x+4
k,10x-25-x2
f,(x+y)2-9x2
g,8x3+42x2y+16xy2+6xy+y3
2
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) \(x^3+3x^2+3x+1=x^2+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x-1\right)^3\)
b) \(x^2+6x+9=x^2+2\cdot3\cdot x+3^2=\left(x+3\right)^2\)
c) \(-x^3+9x^2-27x+27\)
\(=-\left(x^3-9x^2+27x-27\right)\)
\(=-\left(x^3-3\cdot3\cdot x^2+3\cdot3^2\cdot x-3^3\right)=-\left(x-3\right)^3\)
d) \(x^2+4x+4=x^2+2\cdot2\cdot x+2^2=\left(x+2\right)^2\)
k) \(10x-25-x^2=-x^2+10x-25=-\left(x^2-10x+25\right)\)
\(=-\left(x^2-2\cdot5\cdot x+5^2\right)=-\left(x-5\right)^2\)
f) \(\left(x+y\right)^2-9x^2=\left(x-y\right)^2-\left(3x\right)^2=\left[\left(x-y\right)-3x\right]\left[\left(x-y\right)+3x\right]\)
\(=\left(x-y-3x\right)\left(x-y+3x\right)=\left(-2x-y\right)\left(4x-y\right)\)
19 tháng 10 2021
\(a,=\left(x-2\right)^2\\ b,=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ c,=\left(1-2x\right)\left(1+2x+4x^2\right)\\ d,=\left(x+1\right)^3\\ e,Sửa:\left(x+y\right)^2-9x^2=\left(x+y-3x\right)\left(x+y+3x\right)\\ =\left(y-2x\right)\left(4x+y\right)\\ f,=\left(x+3\right)^2\\ g,=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\\ h,=8\left(x^3-\dfrac{1}{64}\right)=8\left(x-\dfrac{1}{4}\right)\left(x^2+\dfrac{1}{4}x+\dfrac{1}{16}\right)\)
19 tháng 10 2021
a) \(\left(x-2\right)^2\)
b) \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
c) \(\left(1-2x\right)\left(1+2x+4x^2\right)\)
d) \(\left(x+1\right)^3\)
e) \(\left(x+y-3\sqrt{x}\right)\left(x+y+3\sqrt{x}\right)\)
f) \(\left(x+3\right)^2\)
g) \(-\left(x-5\right)^2\)
h) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
17 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(2x - 1)^2 - 25 = 0`
`<=> (2x - 1)^2 = 25`
`<=> (2x - 1)^2 = (+-5)^2`
`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, `S = {-2; 3}`
`b,`
`8x^3 - 50x = 0`
`<=> x(8x^2 - 50) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)
Vậy, `S = {-5/2; 0; 5/2}.`
KV
17 tháng 7 2023
a) (2x - 1)² - 25 = 0
(2x - 1)² - 5² = 0
(2x - 1 - 5)(2x - 1 + 5) = 0
(2x - 6)(2x + 4) = 0
2x - 6 = 0 hoặc 2x + 4 = 0
*) 2x - 6 = 0
2x = 6
x = 3
*) 2x + 4 = 0
2x = -4
x = -2
Vậy x = -2; x = 3
b) 8x³ - 50x = 0
2x(4x² - 25) = 0
2x[(2x)² - 5²] = 0
2x(2x - 5)(2x + 5) = 0
2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0
*) 2x = 0
x = 0
*) 2x - 5 = 0
2x = 5
x = 5/2
*) 2x + 5 = 0
2x = -5
x = -5/2
Vậy x = -5/2; x = 0; x = 5/2
12 tháng 10 2023
a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c: \(2x-1-x^2\)
\(=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)
d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)
g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)
\(=\left(5-x\right)\left(5+3x\right)\)
h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)
\(=\left(x+1+2x-1\right)\left(x+1-2x+1\right)\)
\(=3x\left(-x+2\right)\)
i: \(=x^2y^2-4xy+4-3\)
\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)
k: \(=y^2-\left(x-1\right)^2\)
\(=\left(y-x+1\right)\left(y+x-1\right)\)
l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)
m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)
6 tháng 8 2021
b)x2-2x+1=4
⇔(x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c)x2-4x+4=9
⇔ (x-2)2=9
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d)4x2-4x+1=4
⇔ (2x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
e)x2-2x-8=0
⇔ x2-4x+2x-8=0
⇔ x(x-4)+2(x-4)=0
⇔(x-4)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
f)9x2-6x-8=0
⇔ 9x2-12x+6x-8=0
⇔ 3x(3x-4)+2(3x-4)=0
⇔ (3x-4)(3x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
HN
1 tháng 7 2021
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
DD
Đoàn Đức Hà
Giáo viên
28 tháng 5 2022
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
a, x2 - 10x = -25 b, 4x2 - 4x = -1 c, 8x3 +12x2 +6x+1=0
=>x2-10x+25=0 =>(2x)2-2.2x.1+1=0 =>(2x+1)3=0
=>(x-5)2=0 =>(2x-1)2=0 =>2x+1=0
=>x-5=0 =>2x-1=0 =>x = -1/2
=>x=5 =>x=1/2
a.x2-10x=-25
x2-10x+25=0
(x-5)2=0
x=5