Question

giúp em giải bài này với ạ

```latex
\begin{align*}
\text{a) } & (x-3)^2 - (x-3)(3-x^2) = 0 \\
\text{b) } & 4(x+1)^2 + (2x-1)^2 - 8(x-1)(x+1) = 11 \\
\text{c) } & (x-3)^2 - (x+3)(x-5) = 3x + 4 \\
\text{d) } & 3(x-5)^2 + 2x(x-5) = 0 \\
\text{e) } & (2x-1)^2 - (x+3)^2 = 0 \\
\text{f) } & 4(2+3x)(3x-2) - (6x+1)^2 = 7 \\
\text{g) } & 3(2x-1)^2 - 6x(2x-3) = 6 \\
\text{h) } & (2x+3)^2 - 4(x-1)^2 = 16 \\
\end{align*}
```

Answer 1

a: \(\left(x-3\right)^2-\left(x-3\right)\left(3-x^2\right)=0\)

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

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

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

=>\(\left[\begin{array}{l}x-3=0\\ x+3=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-3\\ x=2\end{array}\right.\)

b: \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

=>\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

=>\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

=>4x=-2

=>\(x=-\frac24=-\frac12\)

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

=>\(x^2-6x+9-\left(x^2-5x+3x-15\right)=3x+4\)

=>\(x^2-6x+9-\left(x^2-2x-15\right)=3x+4\)

=>\(x^2-6x+9-x^2+2x+15=3x+4\)

=>-4x+24=3x+4

=>-7x=-20

=>\(x=\frac{20}{7}\)

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

=>\(\left(x-5\right)\left\lbrack3\left(x-5\right)+2x\right\rbrack=0\)

=>(x-5)(3x-15+2x)=0

=>(x-5)(5x-15)=0

=>5(x-5)(x-3)=0

=>(x-5)(x-3)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=3\end{array}\right.\)

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

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

=>(x-4)(3x+2)=0

=>\(\left[\begin{array}{l}x-4=0\\ 3x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-\frac23\end{array}\right.\)

f: \(4\left(3x+2\right)\left(3x-2\right)-\left(6x+1\right)^2=7\)

=>\(4\left(9x^2-4\right)-36x^2-12x-1=7\)

=>\(36x^2-16-36x^2-12x-1=7\)

=>-12x-17=7

=>-12x=24

=>x=-2

g: \(3\left(2x-1\right)^2-6x\left(2x-3\right)=6\)

=>\(3\left(4x^2-4x+1\right)-6x\left(2x-3\right)=6\)

=>\(12x^2-12x+3-12x^2+18x=6\)

=>6x+3=6

=>6x=3

=>\(x=\frac36=\frac12\)

h: \(\left(2x+3\right)^2-4\left(x-1\right)^2=16\)

=>\(4x^2+12x+9-4\left(x^2-2x+1\right)=16\)

=>\(4x^2+12x+9-4x^2+8x-4=16\)

=>20x+5=16

=>20x=11

=>\(x=\frac{11}{20}\)