\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)

\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)

\(\Leftrightarrow x^2-9=5\)

\(\Leftrightarrow x^2=14\)

\(x=\sqrt{14}\)

.....

a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)

Vậy phương trình có tập nghiệm S = { 0 }

b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)

- Ta có :  \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)

Vậy phương trình có tập nghiệm S = { 3 }

\(\frac{12x^2+30x-21}{16x^2-9}-\frac{3x-7}{3-4x}=\frac{6x+5}{4x+3}\)

ĐKXĐ: \(x\ne\pm\frac{3}{4}\)

\(< =>\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}+\frac{3x-7}{4x-3}=\frac{6x+5}{4x+3}\)

\(=>12x^2+30x-21+\left(3x-7\right)\left(4x+3\right)=\left(6x+5\right)\left(4x-3\right)\)

\(< =>12x^2+30x-21+12x^2-19x-21=24x^2+2x-15\)

\(< =>24x^2+11x-42=24x^2+2x-15\)

\(< =>24x^2+11x-42-24x^2-2x+15=0\)

\(< =>9x-27=0\)

\(< =>x=3\left(TM\right)\)

Tập nghiệm phương trình \(S=\left\{3\right\}\)

\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)-\(\frac{3x-7}{3-4x}\)=\(\frac{6x+5}{4x+3}\)

\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)+\(\frac{\left(3x-7\right)\left(4x+3\right)}{\left(4x-3\right)\left(4x+3\right)}\)=\(\frac{\left(6x+5\right)\left(4x-3\right)}{\left(4x-3\right)\left(4x+3\right)}\)

12x²+30x-21+12x²-28x+9x-21=24x²+20x-18x-15

12x²+12x²-24x²+30x-28x+9x-20x+18x=21+21-15

-9x =27

x =\(\frac{27}{-9}\)

x =-3

\(\dfrac{6x+5}{12x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+10x-7}{16x^2-9}\)

\(\Leftrightarrow\dfrac{6x+5}{3\left(4x+3\right)}-\dfrac{3x-7}{3\left(4x-3\right)}=\dfrac{12x^2+30x-21}{3\left(4x-3\right)\left(4x+3\right)}\)

\(\Leftrightarrow\left(6x+5\right)\left(4x-3\right)-\left(3x-7\right)\left(4x+3\right)=12x^2+30x-21\)

\(\Leftrightarrow24x^2-18x+20x-15-\left(12x^2+9x-28x-21\right)=12x^2+30x-21\)

\(\Leftrightarrow24x^2+2x-15-12x^2+19x+21=12x^2+30x-21\)

=>31x+6=30x-21

=>x=-27

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

b: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

\(a,ĐK:x\ge0\\ PT\Leftrightarrow4\sqrt{x}-2\sqrt{x}+3\sqrt{x}=12\\ \Leftrightarrow5\sqrt{x}=12\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\\ \Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\\ b,PT\Leftrightarrow\sqrt{\left(2x-3\right)^2}=7\Leftrightarrow\left|2x-3\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=7\\3-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)