\(9x^2-1=\left(3x+1\right)\cdot\left(2x-3\right)\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\cdot\left(2x-3\right)=0 \\ \Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\\\Leftrightarrow \left(3x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-2\end{matrix}\right.\\ \)
1. \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{-1}{3};-2\right\}\)
2. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow\left(2x-1\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-1-7\right)\left(2x-1+7\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{4;-3\right\}\)
3. \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(\Leftrightarrow\left(5x-3-4x+7\right)\left(5x-3+4x-7\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{-4;\dfrac{10}{9}\right\}\)
4. \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+28x+49=9\left(x^2+4x+4\right)\)
\(\Leftrightarrow4x^2+28x+49=9x^2+36x+36\)
\(\Leftrightarrow\left(4x^2-9x^2\right)+\left(28x-36x\right)=36-49\)
\(\Leftrightarrow-5x^2-8x=-13\)
\(\Leftrightarrow-5x^2-8x+13=0\)
\(\Leftrightarrow-5x^2+5x-13x+13=0\)
\(\Leftrightarrow-5x\left(x-1\right)-13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-13}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{1;\dfrac{-13}{5}\right\}\)
1.
Ta có: 9x2-1=(3x+1)(2x-3)
\(\Leftrightarrow\)(3x-1)(3x+1)-(3x+1)(2x-3)=0
\(\Leftrightarrow\)(3x+1)(3x-1-2x+3)=0
\(\Leftrightarrow\)(3x+1)(x+2)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy:phương trình đã cho có tập nghiệm là: S=\(\left\{\dfrac{-1}{3};-2\right\}\)
b)Ta có: (2x-1)2=49
\(\Leftrightarrow\)(2x-1)2-49=0
\(\Leftrightarrow\)(2x-1-7)(2x-1+7)=0
\(\Leftrightarrow\)(2x-8)(2x+6)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
Vậy:phương trình đã cho có tập nghiệm là: S=\(\left\{4;-3\right\}\)
3. Ta có: (5x-3)2-(4x-7)2=0
\(\Leftrightarrow\)(5x-3-4x+7)(5x-3+4x-7)=0
\(\Leftrightarrow\)(x+4)(9x-10)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)
Vậy:phương trình đã cho có tập nghiệm là: S=\(\left\{-4;\dfrac{10}{9}\right\}\)
4. Ta có: (2x+7)2=9(x+2)2
\(\Leftrightarrow\)(2x+7)2-\(\left[3\left(x+2\right)\right]^2\)=0
\(\Leftrightarrow\)(2x+7-3x-6)(2x+7+3x+6)=0
\(\Leftrightarrow\)(-x+1)(5x+13)=0
\(\Leftrightarrow\left[{}\begin{matrix}-x+1=0\\5x+13=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-13}{5}\end{matrix}\right.\)
Vậy:phương trình đã cho có tập nghiệm là: S=\(\left\{1;\dfrac{-13}{5}\right\}\)
b) 
d) 
f) 
h) 
k) 
n) 
