a) x³ - 7x + 6 = 0

x³ - x - 6x + 6 = 0

(x³ - x) - (6x - 6) = 0

x(x² - 1) - 6(x - 1) = 0

x(x - 1)(x + 1) - 6(x - 1) = 0

(x - 1)[x(x + 1) - 6] = 0

(x - 1)(x² + x - 6) = 0

(x - 1)(x² - 2x + 3x - 6) = 0

(x - 1)[(x² - 2x) + (3x - 6)] = 0

(x - 1)[x(x - 2) + 3(x - 2)] = 0

(x - 1)(x - 2)(x + 3) = 0

x - 1 = 0 hoặc x - 2 = 0 hoăkc x + 3 = 0

*) x - 1 = 0

x = 1

*) x - 2 = 0

x = 2

*) x + 3 = 0

x = -3

Vậy x = -3; x = 1; x = 2

a: \(x^3-7x+6=0\)

=>\(x^3-x-6x+6=0\)

=>\(x\left(x^2-1\right)-6\left(x-1\right)=0\)

=>x(x-1)(x+1)-6(x-1)=0

=>(x-1)(x^2+x-6)=0

=>(x-1)(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x-1=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-3\\ x=2\end{array}\right.\)

b: \(x^4+4x^2-5=0\)

=>\(x^4+5x^2-x^2-5=0\)

=>\(\left(x^2+5\right)\left(x^2-1\right)=0\)

=>\(x^2-1=0\)

=>\(x^2=1\)

=>\(\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

c: \(x^4+x^3-x^2-x=0\)

=>\(x^3\left(x+1\right)-x\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^3-x\right)=0\)

=>\(x\left(x+1\right)^2\cdot\left(x-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x+1=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-1\\ x=1\end{array}\right.\)

d: \(x^2+6x-x-6=0\)

=>x(x+6)-(x+6)=0

=>(x+6)(x-1)=0

=>\(\left[\begin{array}{l}x+6=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-6\\ x=1\end{array}\right.\)

e: \(x^2-4x+5x-20=0\)

=>x(x-4)+5(x-4)=0

=>(x-4)(x+5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-5\end{array}\right.\)

f: \(x^2-10x+2x-20=0\)

=>x(x-10)+2(x-10)=0

=>(x-10)(x+2)=0

=>\(\left[\begin{array}{l}x-10=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10\\ x=-2\end{array}\right.\)

g: \(x^4-x^3-x^2+1=0\)

=>\(x^3\left(x-1\right)-\left(x^2-1\right)=0\)

=>\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x-1\right)=0\)

TH1: x-1=0

=>x=1

TH2: \(x^3-x-1=0\)

=>x≃1,32

h: \(x^5+x^4+x^3+x^2+x+1=0\)

=>\(x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)=0\)

=>\(\left(x^2+x+1\right)\left(x^3+1\right)=0\)

mà \(x^2+x+1=\left(x+\frac12\right)^2+\frac34\ge\frac34>0\forall x\)

nên \(x^3+1=0\)

=>\(x^3=-1\)

=>x=-1

i: \(x^2-9+\left(x+3\right)\left(3x-5\right)=0\)

=>(x-3)(x+3)+(x+3)(3x-5)=0

=>(x+3)(x-3+3x-5)=0

=>(x+3)(4x-8)=0

=>4(x+3)(x-2)=0

=>(x+3)(x-2)=0

=>\(\left[\begin{array}{l}x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=2\end{array}\right.\)

j: \(64x^2-9+8x+3=0\)

=>(8x+3)(8x-3)+(8x+3)=0

=>(8x+3)(8x-3+1)=0

=>(8x+3)(8x-2)=0

=>\(\left[\begin{array}{l}8x+3=0\\ 8x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac38\\ x=\frac28=\frac14\end{array}\right.\)

3.(\(x-5\))\(^2\) + 2\(x\) (\(x-5\)) = 0

(\(x-5\))[3.(\(x-5)\) + 2\(x\)] = 0

(\(x-5)\).[3\(x-15\) + 2\(x\)] = 0

(\(x-5\))[(3\(x\) + 2\(x\)) - 15] = 0

(\(x-5\))[5\(x\) - 15] = 0

\(\left[\begin{array}{l}x-5=0\\ 5x-15=0\end{array}\right.\)

\(\left[\begin{array}{l}x=5\\ 5x=15\end{array}\right.\)

\(\left[\begin{array}{l}x=5\\ x=15:5\end{array}\right.\)

\(\left[\begin{array}{l}x=5\\ x=3\end{array}\right.\)

Vậy \(x\) ∈ {3; 5}

x = 5;5;6

a, 3x ³ - 3x = 0

=> 3x ( x ² - 1 ) = 0

=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

b, x ( x - 2 ) + ( x - 2 ) = 0

=> ( x - 2 ) ( x + 1 ) = 0

=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

c, 5x ( x - 2000 ) - x + 2000 = 0

=> ( x - 2000 ) ( 5x - 1 ) = 0

=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

\(a,x^3-13x=0\)

\(x.\left(x^2-13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)

\(b,2-25x^2=0\)

\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

a, x ³ - 13 x = 0

=> x ( x ² - 13 ) = 0

=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)

b, 2 - 25 x ² = 0

=> 25 x ² = 2

=> x ² = 0,08

=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)

x, x ² - x + \(\frac{1}{4}\)= 0 

=> \(\left(x-\frac{1}{2}\right)^2=0\)

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

Bài giải:

a) x³ – $\frac{1}{4}$x = 0 => x(x² –${\left(\frac{1}{2}\right)}^2$) = 0

=>x(x - $\frac{1}{2}$)(x + $\frac{1}{2}$) = 0

Hoặc x = 0

Hoặc x - $\frac{1}{2}$ = 0 => x = $\frac{1}{2}$

Hoặc x + $\frac{1}{2}$ = 0 => x = -$\frac{1}{2}$

Vậy x = 0; x = -$\frac{1}{2}$; x = $\frac{1}{2}$.

b) (2x – 1)² – (x + 3)² = 0

[(2x - 1) - (x + 3)][(2x - 1) + (x + 3)] = 0

(2x - 1 - x - 3)(2x - 1 + x + 3) = 0

(x - 4)(3x + 2) = 0

Hoặc x - 4 = 0 => x = 4

Hoặc 3x + 2 = 0 => 3x = 2 => x = -$\frac{2}{3}$

Vậy x = 4; x = -$\frac{2}{3}$.

c) x²(x – 3) + 12 – 4x = 0

x²(x – 3) - 4(x -3)= 0

(x - 3)(x²- 2²) = 0

(x - 3)(x - 2)(x + 2) = 0

Hoặc x - 3 = 0 => x = 3

Hoặc x - 2 =0 => x = 2

a ) \(x^3-\dfrac{1}{4}x=0\)

\(\Leftrightarrow\) \(x\left(x^2-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)

Hoặc x = 0

Hoặc \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

Hoặc \(x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{2}\)

b) \((2x - 1 )^2 - (x + 3)^2 = 0\)

\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x-3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

Hoặc \(x-4=0\Rightarrow x=4\)

Hoặc \(3x+2=0\Rightarrow3x=-2\Rightarrow x=-\dfrac{2}{3}\)

c) \(x^2 (x-3) + 12 - 4x = 0\)

\(\Leftrightarrow x^2\left(x-3\right)-\left(4x-12\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-2^2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)

Hoặc \((x - 3) = 0\) \(\Rightarrow\) x = 3

Hoặc \(x - 2 = 0\) \(\Rightarrow\) x = 2

Hoặc \(x + 2 = 0 ​\) \(\Rightarrow\) x = \(- 2\)

a) \(x+5x^2=0\)

<=>\(x\left(1+5x\right)=0\)

+) \(x=0\) (TM)

+)\(1+5x=0\)

<=>\(5x=-1\)

<=>\(x=\dfrac{-1}{5}\) (TM)

Vậy \(x\) có 2 giá trị: \(x=\dfrac{-1}{5}\); \(x=0\)

b)\(x+1=\left(x+1\right)^2\)

<=>\(x+1-\left(x+1\right)^2=0\)

<=>\(\left(x+1\right)\left(1-x-1\right)=0\)

<=>\(\left(x+1\right)\left(-x\right)=0\)

+)\(x+1=0\)

<=>\(x=-1\) (TM)

+)\(-x=0\)

<=>\(x=0\) (TM)

Vậy \(x\) có 2 giá trị : \(x=-1\); \(x=0\)

c) \(x^3+x=0\)

<=> \(x\left(x^2+1\right)=0\)

+) \(x=0\) (TM)

+) \(x^2+1=0\)

<=>\(x^2=-1\)

Ta có: \(x^2\) >= 0, \(-1< 0\). Mà vế trái = vế phải

=> \(x^2=-1\) ( Vô nghiệm)

Vậy \(x=0\)

a) \(x+5x^2=0\)

\(x\left(1+5x\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(1+5x=0\)

\(\Leftrightarrow x=0\) hoặc \(x=\dfrac{-1}{5}\)

b) \(x+1=\left(x+1\right)^2\)

\(\Leftrightarrow x+1-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)-x=0\)

\(\Leftrightarrow x+1=0\) hoặc \(-x=0\)

\(\Leftrightarrow x=-1\) hoặc \(x=0\)

e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)

\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)

\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)

=> 2 TH

*3x+7=0               *10x-4=0

=>3x=-7               =>10x=4

=>x=-7/3              =>x=4/10=2/5

vậy x=-7/3 hoặc x=2/5

g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)

\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)

\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)

\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)

\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)

=> 2 TH

*-(x+3)=0          *3x-5=0

=>-x=-3            =>3x=5  

=x=3                =>x=5/3

h)\(x^2-x^2+x-1=0\)

\(\Rightarrow0+x-1=0\)

\(\Rightarrow x-1=0\)

=>x=0+1

=>x=1

vậy x=1

k, x(x+ 16) - 7x - 42 = 0

=>x^2+16x-7x-42=0

=>x^2+9x-42=0

vì x^2>0

do đó x^2+9x-42>0

nên o có gt nào của x t/m y/cầu đề bài

m)x^2+7x+12=0

=>x^2+3x++4x+12=0

=>x(x+3)+4(x+3)=0

=>(x+4).(x+3)=0

=>2 TH

=> *x+4=0

=>x=-4

vậy x=-4

*x+3=0

=>x=-3

vậy x=-3

n)x^2-7x+12=0

=>x^2-4x-3x+12=0

=>x(x-4)-3(x-4)=0

=>(x-3).(x-4)=0

=>2 TH

*x-3=0=>x=0+3=>x=3

*x-4=0=>x=0+4=>x=4

vậy x=3 hoặc x=4

a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1

b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1

\(1,x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(2,\left(x+2\right)\left(x-3\right)-x-2=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)-\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}\)

\(3,36x^2-49=0\)

\(\Leftrightarrow\left(6x\right)^2-7^2=0\)

\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}6x-7=0\\6x+7=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{6}\\x=\frac{7}{6}\end{cases}}\)

Chúc bn học giỏi nhoa!!!

Ta có : x² - x = 0

=> x(x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

a,x2+6x-7=0

=>x2+7x-x-7=0

=>(x^2+7x)-(x+7)=0

=>x(x+7)-(x+7)=0 =>(x+7)(x-1)=0

=>\(\orbr{\begin{cases}x+7=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=1\end{cases}}}\)

b, x^3-2x^2-5x+6=0

=>x(x^2-2x-5+6)=0

=>x(x^2-2x+1)=0\(^{\orbr{\begin{cases}x=0\\\left(x-1^2\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

c, 2x^2-5x+3=0

=>2x^2-2x-3x+3=0

\(x^3-19x-30=0\)

\(\Rightarrow x^3+5x^2+6x-5x^2-25x-30=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+2x+3x+6\right)=0\)

\(\Rightarrow\left(x-5\right)[x\left(x+2\right)+3\left(x+2\right)]=0\)

\(\Rightarrow\left(x-5\right)\left(x+3\right)\left(x+2\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-5=0\\x+3=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=-3\\x=-2\end{cases}}\)

a) x³ + 3x² + 3x + 1 = 64

=> (x + 1)³ = 64

=> (x + 1)³ = 4³

=> x + 1 = 4 => x = 3

b) x³ + 6x² + 9x = 4x

=> x³ + 6x² + 9x - 4x = 0

=> x³ + 6x² + 5x = 0

=> x³ + 5x² + x² + 5x = 0

=> x²(x + 5) + x(x + 5) = 0

=> (x + 5)(x² + x) = 0

=> (x + 5)x(x + 1) = 0

=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)

c) 4(x - 2)² = (x + 2)²

=> 4(x² - 4x + 4) = x² + 4x + 4

=> 4x² - 16x + 16 = x² + 4x + 4

=> 4x² - 16x + 16 - x² - 4x - 4 = 0

=> 3x² - 20x + 12 = 0

=> 3x² - 18x - 2x + 12 = 0

=> 3x(x - 6) - 2(x - 6) = 0

=> (x - 6)(3x - 2) = 0

=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)

d) x⁴ - 16x² = 0

=> x²(x² - 16) = 0

=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

e) x⁴ - 4x³ + x² - 4x = 0

=> x⁴ + x² - 4x³ - 4x = 0

=> x²(x² + 1) - 4x(x² + 1) = 0

=> (x² - 4x)(x² + 1) = 0

=> x(x - 4)(x² + 1) = 0

=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x² + 1 \(\ge\)1 > 0 \(\forall\)x)

f) x³ + x = 0 => x(x²  + 1) = 0 => x = 0 (vì x² + 1 \(\ge1>0\forall\)x)

\(a,x^3+3x^2+3x+1=64\)

\(\left(x+1\right)^3=64\)

\(\left(x+1\right)^3=4^3\)

\(x+1=4\)

\(x=3\)