a) Ta có: \(\left(x-3\right)=9\left(2x+7\right)\)
\(\Leftrightarrow x-3-9\left(2x+7\right)=0\)
\(\Leftrightarrow x-3-18x-63=0\)
\(\Leftrightarrow-17x-66=0\)
\(\Leftrightarrow-17x=66\)
hay \(x=\frac{-66}{17}\)
Vậy: \(x=\frac{-66}{17}\)
b) Ta có: \(81\left(4x^2+4x+1\right)=121x^2\)
\(\Leftrightarrow324x^2+324x+81-121x^2=0\)
\(\Leftrightarrow203x^2+324x+81=0\)
\(\Leftrightarrow203x^2+261x+63x+81=0\)
\(\Leftrightarrow29x\left(7x+9\right)+9\left(7x+9\right)=0\)
\(\Leftrightarrow\left(7x+9\right)\left(29x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+9=0\\29x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-9\\29x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-9}{7}\\x=\frac{-9}{29}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-9}{7};\frac{-9}{29}\right\}\)
a) ( x-3) =9 (2x+ 7 )
x-3= 18x + 63
x - 18x = 63 + 3
-17x = 66
x = -66/17
b) 81 (4x^2 + 4x +1 ) = 121x^2
324 x^2 + 324x + 81 = 121x^2
324 x^2 + 324x + 81 - 121 x^2 = 0
203 x^2 + 261x + 63x + 81 = 0
29x ( 7x + 9 ) + 63x +81 = 0
(7x + 9 ) (29 x + 9 ) = 0
7x + 9 = 0 hoặc 29x + 9 = 0
7x = -9 29x = -9
X = -9/7 x =-9/29
a) \((x-3)=9(2x+7)\)
\(⇔x-3=18x+63\)
\(⇔x-18x=63+3\)
\(⇔-17x=66\)
\(⇔x=-\dfrac{66}{17}\)
Vậy pt có 1 nghiệm là \(x=-\dfrac{66}{17}\)
b) \(81(4x^2+4x+1)=121x^2\)
\(⇔9^2.(2x+1)^2=(11x)^2\)
\(⇔[9.(2x+1)]^2=(11x)^2\)
\(⇔(18x+9)^2=(11x)^2\)
\(⇔(18x+9)^2-(11x)^2=0\)
\(⇔(29x+9)(7x+9)=0 \)
\(⇔\left[\begin{array}{} 29x+9=0\\ 7x+9=0 \end{array}\right.\)
\(⇔\left[\begin{array}{} x=-\dfrac{9}{29}\\ x=-\dfrac{9}{7} \end{array}\right.\)
Vậy pt có tập nghiệm là S={\(-\dfrac{9}{29};-\dfrac{9}{7}\)}
