1. \(\left(x-4\right)^2-25=0\)
<=> (x-4+5).(x-4-5) = 0
<=> (x+1)(x-9) = 0
<=> \(\left[\begin{matrix}x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1;9}
2. \(\left(2x-1\right)^2+\left(2-x\right)\left(2x-1\right)=0\)
<=> (2x-1)(2x-1+2-x) = 0
<=> (2x-1)(x+1) = 0
<=> \(\left[\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}2x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0.5\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = {-1 ; 0,5}
3. \(x^2+6x+9=4x^2\)
<=> \(\left(x+3\right)^2-4x^2=0\)
<=> (x+3+2x)(x+3-2x) = 0
<=> (3x+3)(3-x) = 0
<=> \(\left[\begin{matrix}3x+3=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=-3\\x=3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = {-1 ; 3}
4. (2x-5)(x+11) = (5-2x)(2x+1)
<=> (2x-5)(x+11) = - (2x-5)(2x+1)
<=> x + 11 = -2x - 1
<=> x+2x = -12
<=> 3x = -12
<=> x = -4
Vậy phương trình có một nghiệm duy nhất là x = -4
5. \(2x^2+5x+3=0\)
<=> \(2x^2+2x+3x+3=0\)
<=> \(2x\left(x+1\right)+3\left(x+1\right)=0\)
<=> \(\left(x+1\right)\left(2x+3\right)=0\)
<=> \(\left[\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = { -1 ; -3/2 }
1) (x-4)^2-25=0
<=> (x-4+5)(x-4-5)=0
\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
2) (2x-1)2+(2-x)(2x-1)=0
<=> (2x-1)(2+2-x)=0
<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=4\end{matrix}\right.\)
3) x^2+6x+9=4x^2
<=> 3x^2 -6x-9=0
<=> x^2 -2x -3=0
<=> x^2 -3x+x-3=0
<=> x(x-3)+(x-3)=0
<=> (x-3)(x+1)=0
=>\(\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
4) (2x-5)(x+11)=(5-2x)(2x+1)
-(5-2x)(x+11)-(5-2x)(2x+1)=0
(5-2x)(x+11+2x+1)=0
=>\(\left[\begin{matrix}x=\frac{5}{2}\\x=-4\end{matrix}\right.\)
5)2x^2+5x+3=0
2x^2+2x+3x+3=0
2x(x+1)+3(x+1)=0
(x+1)(2x+3)=0
=>\(\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)
a) \(\left(x-4\right)^2-25=0\)
=> (x-4)2 = 25
Suy ra \(\left\{\begin{matrix}x-4=5\\x-4=-5\end{matrix}\right.\)
=> \(\left\{\begin{matrix}x=5+4\\x=-5+4\end{matrix}\right.\)
=>\(\left\{\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2. (2x-1)2 + (2-x)(2x-1)=0
\(\Rightarrow\left(2x-1\right)\left(2x-1+2-x\right)=0\)
\(\Rightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\x=-1\end{matrix}\right.\)
3. x2+6x+9=4x2
3x2-6x-9 =0
3(x2 - 2x -3) =0
\(3\left(x^2-3x+x-3\right)=0\)
(x-3)(x+1) =0
\(\left\{\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
4. (2x-5)(x+11)=(5-2x)(2x+1)
(2x-5)(x+11) + (2x-5)(2x+1)=0
(2x-5)(3x+12) =0
\(\left\{\begin{matrix}2x-5=0\\3\left(x+4\right)=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=2,5\\x=-4\end{matrix}\right.\)
5. 2x2+5x+3=0
(2x2+2x)+(3x+3)=0
2x(x+1) +3(x+1) =0
(x+1)(2x+3) =0
\(\left\{\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)
\(\left\{\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)
