Bạn tham khảo cách làm!

`(3x+2)*(x+4)-(3x-1)*(x-5)=0`

\(\Leftrightarrow3x\left(x+4\right)+2\left(x+4\right)-3x\left(x-5\right)+1\left(x-5\right)=0\)

\(\Leftrightarrow3x^2+3x\cdot4+2x+2\cdot4-3x^2+3x\cdot5+x-5=0\)

\(\Leftrightarrow3x^2+12x+2x+8-3x^2+15x+x-5=0\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(12x+2x+15x+x\right)+\left(8-5\right)=0\)

\(\Leftrightarrow30x+3=0\)

\(\Leftrightarrow30x=0-3\)

`=> 30x=-3`

`-> x=-3 \div 30`

`-> x=-1/10 `

\(\dfrac{x+1}{x-1}+\dfrac{x-2}{x+2}+\dfrac{x-3}{x+3}+\dfrac{x+4}{x-4}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x-4\right)+\left(x-2\right)\left(x-1\right)\left(x+3\right)\left(x-4\right)+\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x-4\right)+\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow4x^4+20x-96=0\)

\(\Leftrightarrow4\left(x^4+5x-24\right)=0\)

\(\Leftrightarrow x^4+5x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2,45...\\x=1,94...\end{matrix}\right.\)

Vậy: \(S=\left\{-2,45...;1,94...\right\}\)

x = 0

y = 1 số tự nhiên bất kỳ

ĐKXĐ: \(x\in R\)

\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)

=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)

=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\left(\sqrt{x^4-x^3+2x^2}-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-x^3+2x^2-16}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-2x^3+x^3-2x^2+4x^2-8x+8x-16}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left(3x+1\right)=\dfrac{2\left(x-2\right)\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\)

=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)

=>x-2=0

=>x=2(nhận)

\(3x^2-5x+6=2x\sqrt{x^2-x+2}\)

\(\Leftrightarrow\left[x^2-2x\sqrt{x^2-x+2}+\left(x^2-x+2\right)\right]+\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{x^2-x+2}\right)^2+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{x^2-x+2}\\x-2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.

Bài làm

\(36^2+\frac{1}{x^2}+21x+\frac{7}{2x}-18=0\)

\(\Leftrightarrow\frac{36^2.2.x^2}{2x^2}+\frac{2}{2x^2}+\frac{2.x^2.21x}{2x^2}+\frac{7x}{2x^2}-\frac{2.x^2.18}{2x^2}=0\)

\(\Rightarrow2592x^2+2+42x^3+7x-36x^2=0\)

\(\Leftrightarrow2556x^2+42x^3+7x+2=0\)

tự giải nốt. 

Không có cách khác à bạn? Mình làm cách đấy rồi mà thấy nó dài vl luôn nên đăng nên hỏi coi có cách khác không

Thanks ^^

pt trên \(< =>1296+\frac{2}{2x^2}+\frac{7x}{2x^2}+21x-18=0\)

\(< =>1278+\frac{7x+2}{2x}+21x=0\)

\(< =>1278+\frac{9}{2}=-21x\)

\(< =>\frac{2565}{2}=-21x\)

\(< =>x=\frac{2565}{-42}=-\frac{855}{14}\)

Ko chắc lắm :P

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)