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\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
em c.ơn trước
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a, <=>x2 +6x+9-4x-12=0
<=> x2 +2x -3=0
<=> x2 +3x -x-3=0
<=> x.(x+3) - (x+3) =0
<=> (x-1)(x+3)=0
<=> x=1 hoặc x=-3
b, <=> x(x2 -25) - (x-3)(x+3)2 -7=0
<=> x3 -25x + (9-x2) (x+3) -7=0
<=> x3 -25x+ 9x+27-x3 -3x2 -7=0
<=> -3x2 -16x +20=0
<=>(3x-10)(x-2) =0 (đoạn này tự phân tích nha ^ ^)
<=> x= 10/3 hoặc x=2
Chúc bạn học tốt nha!
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a) Ta có: \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2+4\left(x^2-5x\right)+6\left(x^2-5x\right)+24=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x+4\right)+6\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x^2-5x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x-3x+6\right)\left(x^2-x-4x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)-3\left(x-2\right)\right]\left[x\left(x-1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy: S={1;2;3;4}
b) Ta có: \(\left(2x+1\right)^2-2x-1=2\)
\(\Leftrightarrow\left(2x+1\right)^2-\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)+\left(2x+1\right)-2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-2\right)+\left(2x+1-2\right)=0\)
\(\Leftrightarrow\left(2x+1+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{2}\right\}\)
c) Ta có: \(x\left(x-1\right)\left(x^2-x+1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-x^2+x-x^2+x-1\right)-6=0\)
\(\Leftrightarrow x\left(x^3-2x^2+2x-1\right)-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-x-6=0\)
\(\Leftrightarrow x^4-2x^3+2x^2-4x+3x-6=0\)
\(\Leftrightarrow x^3\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x^2-1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(x-1\right)\left(x+1\right)+3\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x^2-x+3\right)=0\)
mà \(x^2-x+3>0\forall x\)
nên (x-2)(x+1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: S={2;-1}
d) Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+1+2x\right)+x\left(x^2+1+2x\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+1=0
hay x=-1
Vậy: S={-1}
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
a) \(\left(x-2\right)^2-\left(x^2-3x\right)=9\)
\(\Rightarrow x^2-4x+4-x^2+3x-9=0\)
\(\Rightarrow-x-5=0\)
=> x = -5
b) \(\left(5x-2\right)^2=\left(4-x\right)^2\)
\(\Rightarrow25x^2-10x+4-16+8x-x^2=0\)
\(\Rightarrow24x^2-2x-12=0\)
\(\Rightarrow12x^2-x-6=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
c) \(x^2-4x-5=0\)
=> (x - 5).(x + 1) = 0
=> x = 5 hoặc x = -1
a) 2x2 - 5x3 = 0
⇔ x2( 2 - 5x ) = 0
⇔ \(\orbr{\begin{cases}x^2=0\\2-5x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{5}\end{cases}}\)
b) ( x + 1 )( 2 - x ) - ( 3x + 5 )( x + 2 ) = -4x2 + 2
⇔ -x2 + x + 2 - ( 3x2 + 11x + 10 ) + 4x2 - 2 = 0
⇔ 3x2 + x - 3x2 - 11x - 10 = 0
⇔ -10x - 10 = 0
⇔ -10x = 10
⇔ x = -1
c) ( x + 3 )( x2 - 3x + 9 ) - x( x - 2 )2 = 27
⇔ x3 + 27 - x( x2 - 4x + 4 ) - 27 = 0
⇔ x3 - x3 + 4x2 - 4x = 0
⇔ 4x( x - 1 ) = 0
⇔ \(\orbr{\begin{cases}4x=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
d) ( x - 1 )( x - 5 ) + 3 = 0
⇔ x2 - 6x + 5 + 3 = 0
⇔ x2 - 6x + 8 = 0
⇔ x2 - 2x - 4x + 8 = 0
⇔ x( x - 2 ) - 4( x - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ \(\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)