a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
d với e thì tách hết ra, tự triệt tiêu là ra kết quả, dễ mà :) @La Thị Thu Phượng
a) \(9x^2-49=0\)
\(\left(3x\right)^2-7^2=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x+7=0\\3x-7=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{7}{3}\\x=\frac{7}{3}\end{array}\right.\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(x^3+27-x\left(x^2-1\right)-27=0\)
\(x^3+27-x^3+x-27=0\)
\(x=0\)
c) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(x^2+2x-x-2-x-2=0\)
\(x^2+4=0\)
\(x^2=-4\) (loại vì \(x^2\ge0\) với mọi x)
d) \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
\(3x^2+2x+x^2+2x+1-4x^2+25=0\)
\(4x+26=0\)
\(4x=-26\)
\(x=-\frac{13}{2}\)
e) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)\left(2x+1\right)=7\)
\(4x^2-8x+x-2-4x^2-2x+6x+3=7\)
\(-3x+1=7\)
\(-3x=6\)
\(x=-2\)