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1
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: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
CM
25 tháng 8 2019
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
KV
9 tháng 12 2023
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
DT
18 tháng 2 2022
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
18 tháng 2 2022
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
a/ \(\Rightarrow9\left(16x^2+24x+9\right)=16\left(9x^2-30x+25\right)\)
\(\Rightarrow144x^2+216x+81=144x^2-480x+400\)
\(\Rightarrow696x=319\Rightarrow x=\frac{11}{24}\)