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) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)\left(2x+1\right)=7\)

\(4x^2-8x+x-2-4x^2-2x+6x+3=7\)

\(-3x+1=7\)

\(-3x=6\)

\(x=-2\)

c) Bạn dùng "tích trung tỉ bằng tích ngoại tỉ"

d) \(\frac{x-1}{4}=\frac{9}{x-1}\)

=> (x - 1)² = 4 . 9 = 36 = (+ 6)²

=> x - 1 = 6 hoặc x - 1 = -6

=> x = 7 hoặc x = -5

c) \(\frac{3x-5}{2x+1}=\frac{6x-7}{4x+3}\)

=> \(\left(3x-5\right)\left(4x+3\right)=\left(6x-7\right)\left(2x+1\right)\)

=> \(12x^2+9x-20x-15=12x^2+6x-14x-7\)

=> \(12x^2-11x-15=12x^2-8x-7\)

=> \(12x^2-12x^2-15-7=-8x+11x\)

=> \(-22=3x\)

=> \(x=-\frac{22}{3}\)

b) tương tự.

nhân chéo nà bạn \(\left(3x-5\right)\left(4x+3\right)=\left(6x-7\right)\left(2x+1\right)\)

                     nhân đa với đa 

d. 9.4 = ( x - 1 ) ( x- 1)