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

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

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

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

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

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

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~

<=> 7x - 2 = |4x + 3/4| (1)

(1) <=> 7x - 2 = 4x + 3/4 <=> 3x = 11/4 <=> x = 11/12 (TM >2/7)

Vậy, PT có nghiệm x = 11/12.

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.

x³-x²=4x²-8x+4

<=>x²(x-1)=4(x²-2x+1)

<=>x²(x-1)=4(x-1)²

<=>x²(x-1)-4(x-1)²=0

<=>(x-1)(x²-4x+4)=0

<=> (x-1)(x-2)²=0

<=>x-1=0 hoặc x-2=0

<=>x=1 hoặc x=2

x³ - x² = 4x² - 8x + 4

x³ - x² - 4x² + 8x - 4 = 0

x²(x - 1) - 4(x - 1)² = 0

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

=> x - 1 = 0 hoặc x - 2 = 0 hoặc x + 2 = 0

=> x = 1 hoặc x = + 2

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

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\\x+2=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=-2\end{array}\right.\)

x³-x²=4x²-8x+4

<=>x²(x-1)=4(x²-2x+1)

<=>x²(x-1)=4(x-1)²

<=>x²(x-1)-4(x-1)²=0

<=>(x-1)(x²-4x+4)=0

<=> (x-1)(x-2)²=0

<=>x-1=0 hoặc x-2=0

<=>x=1 hoặc x=2

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2-30x+11x-33=0\)

\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

Vậy \(x\in\left\{3;-\frac{11}{10}\right\}.\)

           Bài làm :

Ta có :

\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)

\(\Leftrightarrow24x^2+7x-6-\left(4x^2+23x+28\right)=10x^2+3x-1\)

\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x-33=0\)

\(\Leftrightarrow10x^2-30x+11x-33=0\)

\(\Leftrightarrow10x\left(x-3\right)+11\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(10x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\10x+11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)

Vậy x=3 hoặc x=-11/10