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a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)
\(\Leftrightarrow x^3+9x+2=x^3+8\)
\(\Leftrightarrow x^3+9x=x^3+8-2\)
\(\Leftrightarrow x^3+9x=x^3+6\)
\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)
\(\Leftrightarrow\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^4-4=8x-16+16\)
\(\Leftrightarrow x^2+12=8x\)
\(\Leftrightarrow x^2+12=8x-8x\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)
b: \(\Leftrightarrow2\left(x^2-2x+1\right)-3x^2+5x-1=0\)
\(\Leftrightarrow2x^2-4x+2-3x^2+5x-1=0\)
\(\Leftrightarrow-x^2+x+1=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-1\right)=5\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{5}}{2}\\x_2=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
c: \(\Leftrightarrow x^2+6x+9-1-\left(x^2+8x-4x-32\right)=0\)
\(\Leftrightarrow x^2+6x+8-x^2-4x+32=0\)
=>2x+40=0
hay x=-20
d: \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow7x^2+8x+13-7x^2+63=36\)
=>8x+76=36
hay x=-5
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a, \(x^2-25-\left(x+5\right)=0\)
\(\Rightarrow x^2-5^2-\left(x+5\right)=0\)
\(\Rightarrow\left(x-5\right)\times\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-5-1\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-6\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\x-6=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-5=\left(-5\right)\\x=0+6=6\end{cases}}\)
b, \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(\left(2x\right)^2-1^2\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(2x-1\right)\times\left(2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-\left(2x+1\right)\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(-2\right)=0\)\(\Rightarrow\left(-4x\right)+2=0\)
\(\Rightarrow\left(-4x\right)=0-2=-2\)
\(\Rightarrow x=\frac{-2}{-4}=\frac{1}{2}\)
c, \(x^2\times\left(x^2+4\right)-x^2-4=0\)
\(\Rightarrow x^2\times\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Rightarrow\left(x^2-1\right)\times\left(x^2+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2=1\\x^2=\left(-4\right)\end{cases}}\)
\(\Rightarrow x=1\)
a) (2x - 1)(x^2 - 1 + 1) = 2x^3 - 3x^2 + 2
(2x - 1).x^2 = 2x^3 - 3x^2 + 2
2x^3 - x^2 = 2x^3 - 3x^2 + 2
-x^2 = -3x^2 + 2
2x^2 = 2
x^2 = 1
=> x = 1; -1
b) (x + 2)(x + 2) - (x - 2)(x - 2) = 8x
(x + 2)^2 - (x - 2)^2 = 8x
x^2 + 4x + 4 - x^2 + 4x - 4 = 8x
8x = 8x
=> x thuộc N*
c) (x + 1)(x + 2)(x + 5) - x^3 - 8x^2 = 27
x^3 + 5x^2 + 2x^3 + 10x + x^2 + 5x + 2x + 10x - x^3 - x^2 = 27
17x + 10 = 27
17x = 27 - 10
17x = 17
=> x = 1
d) (x + 1)(x^2 + 2x + 4) - x^3 - 3x^2 + 16 = 0
x^3 + 2x^2 + 4x + x^2 + 2x + 4 - x^3 - 3x^2 + 16 = 0
6x + 20 = 0
6x = -20
x = -20/6
=> x = -10/3
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)