a, ( 2x - 3 )²- (2x + 1)² = -3

4x²-12x+9-4x²+4x-1=-3

-8x-1=-3

-8x=-2

x=\(\frac{1}{4}\)

b, (5x - 1) ² - (5x + 4)(5x - 4) = 7

25x²-10x+1-25x²+16=7

-10x+17=7

-10x=-10

x=1

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

x²-10x+25+x²-9-2x²-4x-2=0

-14x+14=0

-14(x-1)=0

=>x-1=0

x=1

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

\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)

\(\Leftrightarrow-16x+8=-3\)

\(\Leftrightarrow-16x=-11\)

\(\Leftrightarrow x=\frac{11}{16}\)

b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)

\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)

\(\Leftrightarrow-10x+17=7\)

\(\Leftrightarrow-10x=-10\)

\(\Leftrightarrow x=1\)

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

\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)

\(\Leftrightarrow-14x-18=0\)

\(\Leftrightarrow-14x=18\)

\(\Leftrightarrow x=-\frac{9}{7}\)

#H

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

\(\Leftrightarrow8:\left|x-2\right|=2\)

\(\Leftrightarrow\left|x-2\right|=8:2\)

\(\Leftrightarrow\left|x-2\right|=4\)

Xét trường hợp 1: \(x-2=4\)

\(\Rightarrow x=4+2\)

\(\Rightarrow x=6\)

Xét trường hợp 2: \(x-2=-4\)

\(\Rightarrow x=-4+2\)

\(\Rightarrow x=-\left(4-2\right)\)

\(\Rightarrow x=-2\)

Vậy \(x=6\) hoặc \(x=-2\)

b)

bài vách ngọc ngà và bài cà phê ko đường

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

f) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60=5\)

\(\Leftrightarrow30x=60\)

\(\Rightarrow x=2\)

g) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x=14\)

\(\Rightarrow x=\frac{7}{13}\)

A=x^4+3x^3-5x^2+7

B=x^2+4x^2+2x+1=5x^2+2x+1

A-B=x^4+3x^3-5x^2+7-5x^2-2x-1

=x^4+3x^3-10x^2-2x+6

=0 bạn nha