Lần sau đăng thì chia thành nhiều câu hỏi nhé

\(16^2-9.\left(x+1\right)^2=0\)

\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)

\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)

\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)

\(\left[13-3x\right].\left[19+3x\right]=0\)

\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)

KL:..............................

Nhiều câu hỏi mà bn ??

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

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

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

\(a,\)\(x^4-4x^3+4x^2=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

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

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

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

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

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)

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

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

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

\(x+1=4\)

\(x=3\)

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Rightarrow\left(x^3+2^3\right)-x^3-2x=15\)

\(\Rightarrow x^3+8-x^3-2x=15\)

\(\Rightarrow8-2x=15\)

=>2x=8-15=-7

=>x=\(\frac{-7}{2}\)

\(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

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

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

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

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

=>x²-1=0 hoặc -3x²=0

+)Nếu x²-1=0

=>x²=1

=>x=-1 hoặc x=1

+)Nếu -3x²=0

=>3x²=0

=>x²=0

=>x=0

Vậy x=-1 hoặc x=1 hoặc x=0

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

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

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

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

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

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

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

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 

=> \(\left(x-\frac{1}{2}\right)^2=0\)

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

1) \(x^3-x^2=4x^2-8x+4\)

\(\Leftrightarrow x^3-x^2-4x^2+8x-4=0\)

\(\Leftrightarrow x^2-5x^2+8x-4=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....

b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........

c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......

d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......

a) (2x - 3)² = (x + 5)²

=> 4x² - 12x + 9 = x² + 10x + 25

=> 4x² - 12x + 9 - (x² + 10x + 25) = 0

=> 3x² - 22x - 16 = 0

=> 3x² - 24x + 2x - 16 = 0

=> 3x(x - 8) + 2(x - 8) = 0

=> (3x + 2)(x - 8) = 0

=> \(\orbr{\begin{cases}3x+2=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=8\end{cases}}\)

b) x²(x - 1) - 4x² + 8x - 4 = 0

=> x²(x - 1) - (2x  - 2)² = 0

=> x²(x - 1) - [2(x- 1)]² = 0

=> x²(x - 1) - 4(x - 1)² = 0

=> (x - 1)(x² - 4(x - 1) = 0

=> (x - 1)(x² - 4x + 4) = 0

=> (x - 1)(x - 2)² = 0

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

c) x² + 7x + 12 = 0

=> x² + 3x + 4x + 12 = 0

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

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

=> \(\orbr{\begin{cases}x+4=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-3\end{cases}}\)

d) x² + 3x - 18 = 0

=> x² + 6x - 3x - 18 = 0

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

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

=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

e) x(x + 6) - 7x - 42 = 0

=> x(x + 6) - 7(x + 6) = 0

=> (x - 7)(x + 6) = 0

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

1. ( 2x - 3 )² = ( x + 5 )²

<=> ( 2x - 3 )² - ( x + 5 )² = 0

<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0

<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0

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

<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

2. x²( x - 1 ) - 4x² + 8x - 4 = 0

<=> x²( x - 1 ) - ( 4x² - 8x + 4 ) = 0

<=> x²( x - 1 ) - 4( x² - 2x + 1 ) = 0

<=> x²( x - 1 ) - 4( x - 1 )² = 0

<=> ( x - 1 )[ x² - 4( x - 1 ) ] = 0

<=> ( x - 1 )( x² - 4x + 4 ) = 0

<=> ( x - 1 )( x - 2 )² = 0

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

3. x² + 7x + 12 = 0

<=> x² + 3x + 4x + 12 = 0

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

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

<=> \(\orbr{\begin{cases}x+3=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=-4\end{cases}}\)

4. x² + 3x - 18 = 0

<=> x² - 3x + 6x - 18 = 0

<=> x( x - 3 ) + 6( x - 3 ) = 0

<=> ( x - 3 )( x + 6 ) = 0

<=> \(\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

5. x( x + 6 ) - 7x - 42 = 0

<=> x( x + 6 ) - 7( x + 6 ) = 0

<=> ( x + 6 )( x - 7 ) = 0

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