a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)

\(\Leftrightarrow40x=46+9+25=80\)

\(\Leftrightarrow x=2\)

b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)

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

\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)

\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)

\(\Leftrightarrow2x=-2+6-5=-1\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(1+\left(2x-1\right)\left(4x-5\right)-4x^2=0\)

\(1+8x^2-14x+5-4x^2=0\)

\(4x^2-14x+6=0\)

\(2x^2-7x+3=0\)

\(2x^2-6x-x+3=0\)

\(2x\left(x-3\right)-\left(x-3\right)=0\)

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

\(2x-1=0\) hoặc \(x-3=0\)

\(2x=1\) hoặc \(x-3=0\)

\(x=\frac12\) hoặc \(x=3\)

Vậy \(x=\frac12\) hoặc \(x=3\)

1 + (2x - 1)(4x - 5) - 4x² = 0

1 + 8x² - 10x - 4x + 5 - 4x² = 0

4x² - 14x + 6 = 0

4x² - 2x - 12x + 6 = 0

(4x² - 2x) - (12x - 6) = 0

2x(2x - 1) - 6(2x - 1) = 0

(2x - 1)(2x - 6) = 0

2x - 1 = 0 hoặc 2x - 6 = 0

*) 2x - 1 = 0

2x = 1

*) 2x - 6 = 0

2x = 6

x = 6 : 2

x = 3

Vậy:

a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

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

\(\Leftrightarrow3x\left(x-1\right)=\left(x-1\right)^2\Leftrightarrow\left(x-1\right)\left(x-1-3x\right)=0\)

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

b) \(\Leftrightarrow x^3-7x^2+14x-8=0\)

\(\Leftrightarrow x^3-2x^2-5x^2+10x+4x-8=0\)

\(\Leftrightarrow x^2\left(x-2\right)-5x\left(x-2\right)+4\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-5x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=4\end{matrix}\right.\)

c) \(3x^2=4x\Leftrightarrow x\left(3x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

d) \(\Leftrightarrow x^2-6x-7=0\)

\(\Leftrightarrow x^2-6x+9-16=0\)

\(\Leftrightarrow\left(x-3\right)^2-16=0\Leftrightarrow\left(x-7\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

Bài này chỉ cần phá ngoặc là xong 

a, PT <=> \(-x^3-4x+8=15\)

\(x^3+4x+7=0\)( vô nghiệm )

b, PT <=> \(24x+25=49\)

\(x=1\)

b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)

=>3x+21=2

=>x=-19/3

d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)

\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)

=>8x=8

hay x=1

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

b, \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{y^2-xy-xy+x^2}{\left(xy-x^2\right)\left(y^2-xy\right)}=\frac{x^2+y^2}{xy^3-xyxy-xyxy+x^3y}\)Tu rut gon tiep

c, tt

d, cx r

a) \(\frac{1}{x}-\frac{1}{x+1}=\frac{x+1}{x\left(x+1\right)}-\frac{x}{x\left(x+1\right)}\)

\(=\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)

b) \(\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}\)

\(=\frac{y}{xy\left(y-x\right)}-\frac{x}{xy\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

c) \(\frac{9x-3}{4x-1}-\frac{3x}{1-4x}=\frac{9x-3}{4x-1}+\frac{3x}{4x-1}\)

\(=\frac{9x-3+3x}{4x-1}=\frac{6x-3}{4x-1}\)