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

3x²+3x-2x²-4x=1+x²

3x²+3x-2x²-4x-x²=1

x=-1

vậy............

`x^4+3x^2-2=0`

Đặt `x^2=t(t>=0)`

`pt<=>t^2+3t-2=0`

`<=>t^2+3t+9/4=17/4`

`<=>(t+3/2)^2=17/4`

`<=>t+3/2=sqrt{17}/2(do \ t>=0=>t+3/2>=3/2)`

`<=>t=(sqrt{17}-3)/2`

`<=>x^2=(sqrt{17}-3)/2`

`<=>x=+-sqrt{(sqrt{17}-3)/2}`

a: \(=\dfrac{x-2x-1}{x+1}=\dfrac{-\left(x+1\right)}{x+1}=-1\)

b: \(=\dfrac{2+2x}{x\left(x+1\right)}=\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2}{x}\)

c: \(=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

Bài 1: 

a: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)

Bài 2: 

a: \(4x^2+4x+1\)

b: \(9x^2+9x+\dfrac{9}{4}\)

Bài 2:

a: \(A+B=\dfrac{2}{3}x^5y^5z^3-\dfrac{1}{3}x^5y^5z^3=\dfrac{1}{3}x^5y^5z^3\)

\(B-A=-\dfrac{1}{3}x^5y^5z^3-\dfrac{2}{3}x^5y^5z^3=-x^5y^5z^3\)

b: \(C+D=4x^{10}y^8+\dfrac{5}{2}x^{10}y^8=\dfrac{13}{2}x^{10}y^8\)

\(C-d=4x^{10}y^8-\dfrac{5}{2}x^{10}y^8=\dfrac{3}{2}x^{10}y^8\)

Đề là gì bn nhỉ

(4x - 3)(4x + 3) - 8(2 - 3x)²

= 16x² - 9 - 8(4 - 12x + 9x²)

= 16x² - 9 - (32 - 96x + 72x²)

= 16x² - 9 - 32 + 96x - 72x²

= 16x² - 72x² + 96x - 9 - 32

= -56x² + 96 - 41

= -(-56x² + 96 - 41)

= 56x² - 96 + 41

a) 

$|3x-2|=2x\Rightarrow x\geq 0$.

Xét 2 TH:

TH1: $x\geq \frac{2}{3}$ thì pt trở thành:

$3x-2=2x\Leftrightarrow x=2$ (thỏa mãn)

TH2: $0\leq x< \frac{2}{3}$ thì pt trở thành:

$2-3x=2x\Leftrightarrow x=\frac{2}{5}$ (thỏa mãn)

b) 

PT $\Rightarrow x\geq 0$

$\Rightarrow |4+2x|=4+2x$. PT trở thành:

$4+2x=4x\Leftrightarrow x=2$ (thỏa mãn)

c) 

Xét các TH sau:

TH1: $x\geq \frac{3}{2}$. Khi đó, pt trở thành:

$2x-3=-x+21$

$\Leftrightarrow x=8$ (thỏa mãn)

TH2: $x< \frac{3}{2}$. Khi đó, pt trở thành:

$3-2x=-x+21$

$\Leftrightarrow x=-18$ (thỏa mãn)

d) 

Từ PT suy ra $x-2\geq 0\Leftrightarrow x\geq 2(*)$

Khi đó: $|3x-1|=3x-1$. PT trở thành:
$3x-1=x-2$

$\Leftrightarrow 2x=-1<0\Rightarrow x<0$ (mâu thuẫn với $(*)$)

Vậy PT vô nghiệm.

Đầy tiên ta đi rút gọn biểu thức.

Có : $A = (3x+5).(2x-1) + (4x-1).(3x+2)$

$ = 6x^2 + 7x - 5 + 12x^2 + 5x - 2$

$ = 18x^2 + 12x-7$

Vì $|x| = 2$ nên $x = 2$ hoặc $x=-2$

Với $x=2$ ta có : $A = 18.2^2 + 12.2-7 = 89$

Với $x=-2$ ta có : $A = 18.(-2)^2 + 12.(-2) - 7 = 41$