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7 tháng 8 2022
a: \(\Leftrightarrow x^4-24x^2+144=0\)
\(\Leftrightarrow\left(x^2-12\right)^2=0\)
hay \(x=\pm2\sqrt{3}\)
b: \(\Leftrightarrow x^4-2x^2+12x-8=0\)
\(\Leftrightarrow\left(x^2-2x+4\right)\left(x^2+2x-2\right)=0\)
\(\Leftrightarrow x^2+2x-2=0\)
hay \(x\in\left\{-1+\sqrt{3};-1-\sqrt{3}\right\}\)
18 tháng 8 2020
a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
23 tháng 2 2020
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1
AH
Akai Haruma
Giáo viên
12 tháng 5 2020
i)
$I=x^4+4x^3-x^2-14x+6$
$=(x^4+4x^4+4x^2)-5x^2-14x+6$
$=(x^2+2x)^2-6(x^2+2x)+9+x^2-2x-3$
$=(x^2+2x-3)^2+(x^2-2x+1)-4$
$=(x-1)^2(x+3)^2+(x-1)^2-4$
$=(x-1)^2[(x+3)^2+1]-4\geq -4$
Vậy $I_{\min}=-4$ khi $(x-1)^2[(x+3)^2+1]=0\Leftrightarrow x=1$
k)
$K=x^4+2x^3-10x^2-16x+45$
$=(x^4+2x^3+x^2)-11x^2-16x+45$
$=(x^2+x)^2-12(x^2+x)+x^2-4x+45$
$=(x^2+x)^2-12(x^2+x)+36+(x^2-4x+4)+5$
$=(x^2+x-6)^2+(x-2)^2+5$
$=[(x-2)(x+3)]^2+(x-2)^2+5$
$=(x-2)^2[(x+3)^2+1]+5\geq 5$
Vậy $K_{\min}=5$ khi $(x-2)^2[(x+3)^2+1]=0\Leftrightarrow x=2$
AH
Akai Haruma
Giáo viên
12 tháng 5 2020
g)
$G=x^4+4x^3+10x^2+12x+11$
$=(x^4+4x^3+4x^2)+6x^2+12x+11$
$=(x^2+2x)^2+6(x^2+2x)+11$
Đặt $x^2+2x=t$. Khi đó $t=x^2+2x=(x+1)^2-1\geq -1\Rightarrow t+1\geq 0$
$\Rightarrow G=t^2+6t+11=(t+1)^2+4(t+1)+7\geq 7$
Vậy $G_{\min}=7$ khi $t=-1\Leftrightarrow (x+1)^2=0\Leftrightarrow x=-1$
h)
$H=x^4-6x^3+x^2+24x+18$
$=(x^4-6x^3+9x^2)-8x^2+24x+18$
$=(x^2-3x)^2-8(x^2-3x)+18$
$=(x^2-3x)^2-8(x^2-3x)+16+2$
$=(x^2-3x-4)^2+2\geq 2$
Vậy $H_{\min}=2$ khi $x^2-3x-4=0\Leftrightarrow x=4$ hoặc $x=-1$
3 tháng 10 2022
a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)
=>(-2x+12)(4x+12)=0
=>x=-3 hoặc x=6
b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)
=>\(x\simeq0.93\)
d: =>-4x+28+11x=-x+3x+15
=>7x+28=2x+15
=>5x=-13
=>x=-13/5
e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)
=>-9x=-3x+5
=>-6x=5
=>x=-5/6
Giúp mik vs mn ơi!
b: \(=\dfrac{x^4-x^3-2x^3+2x^2+x^2-x}{x-1}=x^3-2x^2+x\)