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\(\dfrac{-6x^4+7x^3+5x+2}{3x+1}\)
\(=\dfrac{-6x^4-2x^3+9x^3+3x^2-3x^2-x+6x+2}{3x+1}\)
\(=\dfrac{-2x^3\left(3x+1\right)+3x^2\left(3x+1\right)-x\left(3x+1\right)+2\left(3x+1\right)}{3x+1}\)
\(=-2x^3+3x^2-x+2\)
3x.(x-2)-x2+2x=0
⇔3x2-6x-x2+2x=0
⇔2x2-4x=0
⇔2x(x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
vậy x=0 và x=2
3x(x-2)-x^2+2x=0
<=>3x(x-2)-x(x-2)=0
<=>(3x-x)(x-2)=0
<=>2x(x-2)=0
<=>2x=0 hoặc x-2=0
<=>x=0 hoặc x=2
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
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
\(A=\left(x+y\right).\left(x^2-xy+y^2\right)-\left(x-y\right).\left(x^2+xy+y^2\right)=\left(x^3+y^3\right)-\left(x^3-y^3\right)=2y^3\)
=> Biểu thức A phụ thuộc vào giá trị của y
\(\left(x-1\right)^3+3x.\left(x-4\right)+1=0\Leftrightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)
\(\Leftrightarrow x^3-9x=0\Leftrightarrow x.\left(x^2-9\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)
1)
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)
2) Bạn xem lại đề!
a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)
\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Leftrightarrow-10x^2-10x=0\)
\(\Leftrightarrow-10x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c: Ta có: \(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
hay x=-4
\(\left(x+3\right)\left(x^2-3x+9\right)=7x^3+21x\\ \Leftrightarrow x^3+27=7x^3+21x\\ \Leftrightarrow6x^3+21x-27=0\\ \Leftrightarrow\left(6x^3-6x^2\right)+\left(6x^2-6x\right)+\left(27x-27\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x^2+6x+27=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{51}{2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\6\left(x+\dfrac{1}{2}\right)^2+\dfrac{51}{2}=0\left(vô.lí\right)\end{matrix}\right.\)
Vậy \(x=1\)
\(\Leftrightarrow x^3+27-7x^3-21x=0\)
\(\Leftrightarrow-6x^3-21x+27=0\)
\(\Leftrightarrow-6x^3+6x-27x+27=0\)
\(\Leftrightarrow-6x\left(x-1\right)\left(x+1\right)-27\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^2+6x+27\right)=0\)
hay x=1
a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
a ) ( 2x + 1 )2 - 4 ( x + 2 )2 = 9
4x2 + 4x + 1 - 4 ( x2 +4x + 4 ) = 9
4x2 + 4x + 1 - 4x2 -16x -16 = 9
-12x - 15 = 9
-12x = 24
x = -2
b) 3 ( x - 1 )2 - 3x ( x - 5 ) = 1
3 ( x2 - 2x + 1 ) - 3x2 + 15x = 1
3x2 - 6x + 3 - 3x2 + 15x = 1
9x + 3 = 1
9x = -2
x = \(\frac{-2}{9}\)
lồn nhé
ờ , bạn thì tự nhìn lại bản thân mình đi đã nhé