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

\(\left[\left(2x\right)^2+2.2x.1+1\right]-4=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}\)

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

\(x^3.\left(x-3\right)-\left(x-3\right)=0\)

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

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

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^2.\left(x-1\right)-4x^2+8x-4=0\)

\(x^2.\left(x-1\right)-\left[\left(2x\right)^2-2.2x.2+2^2\right]=0\)

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

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

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

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

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

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

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

Tham khảo nhé~

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

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

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

d,

x²+3x-18=0

=> x²-3x+6x-18=8

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

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

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

\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

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

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

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

`a, x^2-10x+25=0`

`<=>x^2 -2.x.5+5^2=0`

`<=>(x-5)^2=0`

`<=>x-5=0`

`<=>x=5`

__

`x^2 -8x+16=0`

`<=> x^2 - 2.x.4+4^2=0`

`<=>(x-4)^2=0`

`<=>x-4=0`

`<=>x=4`

__

`x^2-49=0`

`<=>x^2 - 7^2=0`

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

`<=>x-7=0` hoặc `x+7=0`

`<=> x=7` hoặc `x=-7`

__

`4x^2-25=0`

`<=> (2x)^2 -5^2=0`

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

`<=>2x-5=0` hoặc `2x+5=0`

`<=> 2x=5` hoặc `2x=-5`

`<=>x=5/2` hoặc `x=-5/2`

a: =>(x-5)^2=0

=>x-5=0

=>x=5

b: =>(x-4)^2=0

=>x-4=0

=>x=4

c: =>(x-7)(x+7)=0

=>x-7=0 hoặc x+7=0

=>x=7 hoặc x=-7

d: =>(2x-5)(2x+5)=0

=>2x-5=0 hoặc 2x+5=0

=>x=5/2 hoặc x=-5/2

a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-3x=9\)

hay x=-3

b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

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

hay x=1