a) Ta có: \(x^2\ge0\forall x\)
\(x^2+2\ge2>0\forall x\)
Do đó: 2x+1=0
\(\Leftrightarrow2x=-1\)
hay \(x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
b) Ta có: \(x^2+x+1\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
Do đó: 6-2x=0
⇔2x=6
hay x=3
Vậy: x=3
c) Ta có: \(\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)-2\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(3x+5\right)-2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy: x∈{2;-3}
d) Ta có: \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left[\left(4x-1\right)-4\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\cdot\left(-13\right)=0\)
Vì -13≠0
nên 4x-1=0
⇔4x=1
hay \(x=\frac{1}{4}\)
Vậy: \(x=\frac{1}{4}\)
e) Ta có: \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\)
\(\Leftrightarrow\left[\left(5x-3\right)-\left(4x-7\right)\right]\left[\left(5x-3\right)+\left(4x-7\right)\right]=0\)
\(\Leftrightarrow\left(5x-3-4x+7\right)\left(5x-3+4x-7\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(9x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\9x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\frac{10}{9}\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;\frac{10}{9}\right\}\)
f) Ta có: \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-x-6x+6=0\)
\(\Leftrightarrow x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)
Vậy: x∈{1;6}
g) Ta có: \(x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-x+1\right)=0\)(1)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\)(2)
Từ (1) và (2) suy ra \(\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy: x=-1
Bài làm
a. (2x + 1)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x^2+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x^2=-2\left(vo-li\right)\end{matrix}\right.\)
Vậy x = -1/2 là nghiệm phương trình.
b. (x2 + x + 1)(6 - 2x) = 0
Ta có: x2 + x + 1 > 0
\(\Rightarrow6-2x=0\)
=> 2x = 6
=> x = 3
Vậy x = 3 là nghiệm phương trình.
c. (x – 2)(3x + 5) = (2x – 4)(x + 1)
<=> ( x - 2 )( 3x + 5 ) = 2( x - 2 )( x + 1 )
<=> ( x - 2 )[ ( 3x + 5 ) - ( 2x + 1 ) ] = 0
<=> ( x - 2 )( 3x + 5 - 2x - 1 ) = 0
<=> ( x - 2 )( x - 4 ) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
Vậy tập nghiệm S = { 2; 4 }
d. 16x2 - 8x + 1 = 4(x + 3)(4x - 1)
<=> ( 4x - 1 )2 - 4( x + 3 )( 4x - 1 ) = 0
<=> ( 4x - 1 )( 4x - 1 - 16x + 4 ) = 0
<=> ( 4x - 1 )( -12x + 3 ) = 0
<=> \(\left[{}\begin{matrix}4x-1=0\\-12x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy x = 1/4 là nghiệm ptrình.
e. (5x - 3)2 - (4x - 7)2 = 0
<=> ( 5x - 3 - 4x + 7 )( 5x - 3 + 4x - 7 ) = 0
<=> ( x + 4 )( 9x - 10 ) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\9x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\frac{10}{9}\end{matrix}\right.\)
Vậy tập nghiệm ptrình S = { -4; 10/9 }
f. x2 - 7x + 6 = 0
<=> x2 - 6x - x + 6 = 0
<=> x( x - 6 ) - ( x - 6 ) = 0
<=> ( x - 6 )( x - 1 ) = 0
<=> \(\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy tập nghiệm ptrình S = { 6; 1 }
g. x4 + x3 + x + 1 = 0
<=> x3( x + 1 ) + ( x + 1 ) = 0
<=> ( x3 + 1 )( x + 1 ) = 0
<=> \(\left[{}\begin{matrix}x^3+1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm ptrình S = { 1; -1 }
# Học tốt #
