Question

PHÂN TÍCH ĐA THứC THÀNH NHÂN TỬ

TÌM X

a.(2x-1)²-25=0

b.8x²-50x=0

c.(x-2)(x²+2x+7)+2(x²-4)-5(x-2)=0

Answer 1

a) Ta có: \(\left(2x-1\right)^2-25=0\)

hay \(\left(2x-1\right)^2-5^2=0\)

\(\Rightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-2\right\}\)

b) Ta có: \(8x^2-50x=0\Rightarrow x\left(8x-50\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\8x=50\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{50}{8}=\frac{25}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{25}{4}\right\}\)

c) Ta có: \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left[\left(x^2+2x+7\right)+2\left(x+2\right)-5\right]=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)(1)

Ta có: \(x^2+4x+6=x^2+4x+4+2=\left(x+2\right)^2+2\)

mà \(\left(x+2\right)^2\ge0\forall x\)

nên \(\left(x+2\right)^2+2\ge2>0\forall x\)

nên \(x^2+4x+6=0\) là điều vô lý (2)

Từ (1) và (2) suy ra

\(x-2=0\Leftrightarrow x=2\)

Vậy: x=2

Answer 2

a) Ta có: ( 2x - 1 )² - 25 = 0
=> ( 2x - 1 )² - 5² = 0
=> ( 2x - 6 )( 2x + 4 ) = 0
=>\(\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy x = 3 và x = -2

b) 8x² - 50x = 0
=> x( 8x - 50 ) = 0
=>\(\left[{}\begin{matrix}x=0\\8x-50=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\8x=50\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=\frac{25}{4}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\frac{25}{4}\)

c) Ta có: ( x - 2 )( x² + 2x + 7 ) + 2( x² - 4 ) - 5( x - 2 ) = 0
=> ( x - 2 )( x² + 2x + 7 ) + 2( x - 2 )( x + 2 ) - 5( x - 2 ) = 0
=> ( x - 2)( x² + 4x + 6 ) = 0 (*)
Lại có: x² + 4x + 6 = ( x² + 4x + 4 ) + 2
= ( x + 2 )² + 2
Mà ( x + 2 )² \(\ge\) 0 \(\forall\) x
=> x² + 4x + 6 = 0 (Vô lí) (**)
Từ (*) và (**) => x - 2 = 0
=> x = 2