Xin lỗi vì tớ hỏi 2 câu rồi ạ! Mong các bạn thông cảm
\(\frac{1}{4x^2-12x+9}-\frac{3}{9-4x^2}=\frac{4}{4x^2+12x+9}\)
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ĐKXĐ: \(x\ne\pm\frac{3}{2}\)
\(\frac{1}{\left(2x-3\right)^2}+\frac{3}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x+3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{\left(2x-3\right)^2}-\frac{1}{\left(2x-3\right)\left(2x+3\right)}+\frac{4}{\left(2x-3\right)\left(2x+3\right)}-\frac{4}{\left(2x-3\right)^2}=0\)
\(\Leftrightarrow\frac{1}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)-\frac{4}{2x-3}\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{2x-3}-\frac{4}{2x+3}\right)\left(\frac{1}{2x-3}-\frac{1}{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)
\(\frac{12x^2+30x-21}{16x^2-9}-\frac{3x-7}{3-4x}=\frac{6x+5}{4x+3}\)
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
a/\(\Leftrightarrow\left(12x^2+12x+11\right)\left(y^2-2y+2\right)=\left(4x^2+4x+3\right)\left(5y^2-10y+9\right)\)
\(\Leftrightarrow12x^2y^2-24x^2y+24x^2+12xy^2-24xy+24x+11y^2-22y+22=20x^2y^2-40x^2y+36x^2+20xy^2-40xy+36x+15y^2-30y+36\)
Có sai đề ko cậu
\(\frac{1}{4x^2-12x+9}-\frac{3}{9-4x^2}=\frac{4}{4x^2+12x+9}\)
\(\Leftrightarrow\frac{-1}{\left(3-2x\right)^2}-\frac{3}{\left(3-2x\right)\left(3+2x\right)}=\frac{4}{\left(2x+3\right)^2}\)
\(\Leftrightarrow-4x^2-12x-9-27+12x^2-16x^2+48x-36=0\)
\(\Leftrightarrow-8x^2+36x-72=0\)
Rút -4 ra ngoài \(\Leftrightarrow2x^2-9x+18=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\x-6=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=3\\x=6\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=6\end{cases}\left(tmđk\right)}\)