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Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
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) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`
`<=> 4 + 3 + (-5x) + (-2)=0`
`<=> -5x+5=0`
`<=>-5x=-5`
`<=>x=1`
`2,(25x^2-10x):5x +3(x-2)=4`
`<=> 5x - 2 + 3x-6=4`
`<=> 8x -8=4`
`<=> 8x=12`
`<=>x=12/8`
`<=>x=3/2`
`3,(3x+1)^2-(2x+1/2)^2=0`
`<=> [(3x+1)-(2x+1/2)][(3x+1)+(2x+1/2)]=0`
`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`
`<=>( x+1/2) (5x+3/2)=0`
`@ TH1`
`x+1/2=0`
`<=>x=0-1/2`
`<=>x=-1/2`
` @TH2`
`5x+3/2=0`
`<=> 5x=-3/2`
`<=>x=-3/2 : 5`
`<=>x=-15/2`
`4, x^2+8x+16=0`
`<=>(x+4)^2=0`
`<=>x+4=0`
`<=>x=-4`
`5, 25-10x+x^2=0`
`<=> (5-x)^2=0`
`<=>5-x=0`
`<=>x=5`
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
x² + 2x + 3x + 6 = 0
(x² + 2x) + (3x + 6) = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x + 2 = 0 hoặc x + 3 = 0
*) x + 2 = 0
x = -2
*) x + 3 = 0
x = -3
Vậy x = -3; x = -2
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
a) x(4x2-1)=0
=>x(2x-1)(2x+1)=0
=>\(\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
vậy x\(\in\) {\(\dfrac{-1}{2}\) ;0;\(\dfrac{1}{2}\) }
c)x3-x2-x+1=0
=>(x3-x2)-(x-1)=0
=>x2(x-1)-(x-1)=0
=>(x-1)(x2-1)=0
=>\(\left[{}\begin{matrix}x-1=0\\x^2-1=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)
Bổ sung thêm \(x^2=1\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\).
\(a)x^3-0,25=0\\ \Leftrightarrow x^3-\dfrac{1}{4}=0\\ \Leftrightarrow x^3=\dfrac{1}{4}\\ x=\dfrac{\sqrt[3]{2}}{2}\)
Vậy...
\(b)x^2-10=-25\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x-5=0\\ \Leftrightarrow x=5\)
Vậy...
\(c)x^2-2x-3=0\\ \Leftrightarrow x^2+x-3x-3=0\\ \Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy...
\(d)2x^2+5x-3=0\\ \Leftrightarrow2x^2+6x-x-3=0\\ \Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy...