\(a,3x^2+5x+2=0\\ \Leftrightarrow\left(3x^2+3x\right)+\left(2x+2\right)=0\\ \Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\end{matrix}\right.\)

b, ĐKXĐL\(x\ne\pm\dfrac{2}{3}\)

\(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\\ \Leftrightarrow\dfrac{\left(3x+2\right)^2}{\left(3x+2\right)\left(3x-2\right)}-\dfrac{6\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\dfrac{9x^2}{\left(3x+2\right)\left(3x-2\right)}=0\\ \Leftrightarrow\dfrac{9x^2+12x+4-18x+12-9x^2}{\left(3x+2\right)\left(3x-2\right)}=0\\ \Leftrightarrow-6x+16=0\\ \Leftrightarrow x=\dfrac{8}{3}\left(tm\right)\)

giúp mình với ạ câu nào cũng được

<=>4x-8=0 

<=>4x=8 

=.x=2(nhan)

1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

Giải phương trình $x^2-4x+6=\sqrt{2x^2-5x+3}+\sqrt{-3x^2+9x-5}$ - Phương trình - hệ phương trình - bất phương trình - Diễn đàn Toán học

$\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}$

ĐKXĐ: \(x\ne\frac{2}{3};x\ne-\frac{2}{3}\)

\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}-\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)

\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)-9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)

\(\Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\)

\(\Leftrightarrow16-6x=0\)

\(\Leftrightarrow6x=16\)

\(\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)

Vậy \(S=\left\{\frac{8}{3}\right\}\)

\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\) (1)

đk: \(x\ne\pm\frac{2}{3}\)

(1)\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\frac{9x^2-6x+16}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow9x^2-6x+16=9x^2\)

\(\Leftrightarrow16-6x=0\)

\(\Leftrightarrow x=\frac{8}{3}\)(thỏa mãn đkxđ)

vậy:...................

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

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

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

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

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

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

Vậy pt có tập nghiệm \(S=\left\{-\dfrac{2}{3};-\dfrac{3}{2}\right\}\)

Phương trình trên tương đương:

(5-x)(2+3x)=(2-3x)(2+3x)

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

Đặt 2+3x làm nhân tử chung rồi giải pt tích rồi kết luận

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

\(< =>\left(5-x\right)\left(2+3x\right)=2^2-\left(3x\right)^2\)

\(< =>\left(5-x\right)\left(2+3x\right)-\left(2-3x\right)\left(2+3x\right)=0\)

\(< =>\left(2+3x\right)\left(5-x-2+3x\right)=0\)

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

\(< =>\left[{}\begin{matrix}2+3x=0\\2x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

S=\(\left\{\dfrac{-2}{3},\dfrac{-3}{2}\right\}\)

\(a,\dfrac{x+1}{5}-\dfrac{2}{x}\)

\(=\dfrac{x\left(x+1\right)-2.5}{5x}=\dfrac{x^2+x-10}{5x}\)

\(b,\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)

\(=\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)

a. \(\dfrac{x+1}{5}\)-\(\dfrac{2}{x}\)=\(\dfrac{x\left(x+1\right)-2.5}{5x}\)=\(\dfrac{x^2+x-10}{5x}\)

b. \(\dfrac{x+y}{9x}:\dfrac{x+y}{3x}\)=\(\dfrac{x+y}{9x}.\dfrac{3x}{x+y}=\dfrac{1}{3}\)