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

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

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

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

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

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

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~

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

đầu bài là tìm x ạ

\(a,< =>2\left(x-4\right)^2+4x\left(x-4\right)=0< =>\left(x-4\right)\left(2x-8+4x\right)=0\)\(< =>\left(x-4\right)\left(6x-8\right)=0< =>\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

b,\(< =>4x\left(x-2\right)=0< =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

c,\(< =>3x\left(x+2\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d,\(< =>4x^2\left(2+x\right)=0< =>\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a) Thực hiện rút gọn VT = -2x – 64

Giải phương trình -2x – 64 = 0 thu được x = -32.

b) Thực hiện rút gọn VT = -62 x +12

Giải phương trình -62x + 12 = -50 thu được x = 1.

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

a) 3x² + 8x + 4 = 0

=> 3x² + 6x + 2x +  4 = 0

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

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

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

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

b) 4x² - 4x - 3 = 0

=> 4x² - 6x + 2x - 3 = 0

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

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

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

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

\(a,3x^2+8x+4=0\) 

\(\Rightarrow3x^2+6x+2x+4=0\) 

\(\Rightarrow3x\left(x+2\right)+2\left(x+2\right)=0\) 

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

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

Vậy....