\(\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2+x+2x+2-x^2-5x+3x+15=0\)

\(\Leftrightarrow x+17=0\)

\(\Leftrightarrow x=-17\)

(x+2)(x+1)-(x-3)(x+5)=0

\(\Leftrightarrow\) (x²+x+2x+2)-(x²+5x-3x-15)=0

\(\Leftrightarrow\)x²+x+2x+2-x²-5x+3x+15=0

\(\Leftrightarrow\)x+17=0

\(\Rightarrow\)x=-17

     \(x^3-2x^2-x+2=0\)

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

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

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

Tìm được \(x\in\left\{2;1;-1\right\}\)

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

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

Mà \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\) (2)

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

Bài 1:

Đặt biểu thức trên là A

Ta có:\(A=\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-3\right)=x^2-x-2-\left(x^2-x-6\right)\)

                                                                                      \(=x^2-x-2-x^2+x+6=4\)

Vậy biểu thức A không phụ thuộc vào biến x (đpcm)

Bài 2:

a)\(\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)

\(\Leftrightarrow x^2-3x-10+x-x^2+2=15\)

\(\Leftrightarrow-2x-8=15\)

\(\Leftrightarrow-2x=23\)\(\Leftrightarrow x=\frac{-23}{2}\)

Vậy...................................................................................

câu b) tương tự câu a) thôi,bạn tự làm đi nhé

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, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)

\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)

\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)

b, Ta có : \(\left(x+5\right)^2-9x-45=0\)

\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)

TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)

c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)

\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)

\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )

e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)

TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )

TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)

\(\Rightarrow25\left(x+1\right)^4-26\left(x+1\right)^2+1=0\Leftrightarrow25\left(x+1\right)^4-25\left(x+1\right)^2-\left(\left(x+1\right)^2-1\right)=0\)

\(\Leftrightarrow25\left(x+1\right)^2.\left(\left(x+1\right)^2-1\right)-\left(\left(x+1\right)^2-1\right)=0\)

\(\Leftrightarrow\left(\left(x+1\right)^2-1\right).\left(25\left(x+1\right)^2-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2-1=0\\25\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0,-2\\x=-\frac{4}{5},-\frac{6}{5}\end{cases}}}\)

\(x^2+x-1=0\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\frac{5}{4}=0\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{\sqrt{5}}{2}\\x+\frac{1}{2}=\frac{-\sqrt{5}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}-1}{2}\\x=\frac{-\sqrt{5}-1}{2}\end{cases}}}\)

a, <=> x.(x-3)+5.(x-3) = 0

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

<=> x-3=0 hoặc x+5=0

<=> x=3 hoặc x=-5

Vậy ........

b, ĐKXĐ : x khác 1 và 2

pt <=> x^2-1 = 0

<=> (x-1).(x+1) = 0

<=> x-1 = 0 hoặc x+1 = 0

<=> x=-1 ( vì x khác 1 và 2 )

Vậy x=-1

k mk nha

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 = \(- 2\)

tách hết ra là xong

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

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

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

\(\Rightarrow-x-4=0\)

\(\Rightarrow-x=4\)

\(\Rightarrow x=-4\)