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

Ta có \(\Delta=5^2+4.2.3=49,\sqrt{\Delta}=7\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+7}{4}=\frac{1}{2}\\x=\frac{-5-7}{4}=-3\end{cases}}\)

Ta có : \(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\)\(x+2=0\)

hoặc   \(x-1=0\)

hoặc   \(x^2+x+6=0\)

\(\Leftrightarrow\) \(x=-2\)(tm)

hoặc    \(x=1\)(tm)

hoặc   \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)

a: 3x-2=2x-3

=>x=-1

b: 2x+3=5x+9

=>-3x=6

=>x=-2

c: 5-2x=7

=>2x=-2

=>x=-2

d: 10x+3-5x=4x+12

=>5x+3=4x+12

=>x=9

e: 11x+42-2x=100-9x-22

=>9x+42=78-9x

=>18x=36

=>x=2

f: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

\(3x\left(-2x^3+7x^2+5x-1\right)=-6x^4+21x^3+15x^2-3x\)

A+B+C

=2x^3-5x^2+x-7+x^2-2x+6+C

=2x^3-4x^2-x-1-x^3+4x^2-1

=x^3-x-2

Ta có : \(\frac{\left(5x+3\right)\left(3x+11\right)}{4}-\frac{x-7}{12}=0\)

=> \(\frac{3\left(5x+3\right)\left(3x+11\right)}{12}-\frac{x-7}{12}=0\)

=> \(3\left(5x+3\right)\left(3x+11\right)-\left(x-7\right)=0\)

=> \(3\left(15x^2+9x+55x+33\right)-x+7=0\)

=> \(45x^2+27x+165x+99-x+7=0\)

=> \(45x^2+191x+106=0\)

=> \(45x^2+2.\sqrt{45}x.\frac{191}{2\sqrt{45}}+\frac{191^2}{\left(2\sqrt{45}\right)^2}-\frac{17401}{180}=0\)

=> \(\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}\right)^2-\left(\sqrt{\frac{17401}{180}}\right)^2=0\)

=> \(\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}\right)\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}\right)=0\)

=> \(\left[{}\begin{matrix}x\sqrt{45}+\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}=0\\x\sqrt{45}+\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x\sqrt{45}=-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}\\x\sqrt{45}=-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}}{\sqrt{45}}\\x=\frac{-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}}{\sqrt{45}}\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là \(\left[{}\begin{matrix}x=\frac{-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}}{\sqrt{45}}\\x=\frac{-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}}{\sqrt{45}}\end{matrix}\right.\) .

ĐKXĐ: \(x\ne\dfrac{2}{5}\)

\(\dfrac{2x}{5x-2}+\dfrac{4}{5x-2}\) \(=\dfrac{2x+4}{5x-2}\)

b, TH1 2.3X-2+5X-6=0

=>6x-4+5x-6=0

=>11x+-10=0=>11x=0+10==>x=\(\frac{10}{11}\)

c,

hình như sai đề k tìm ra đáp án bạn ơi 

ko đâu ak, mk hỏi lại rồi, nhưng mà sai thì nhờ bn sửa vs nhé hjhj