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\(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\) ( ĐKXĐ : \(x\ne0,x\ne2\) )
\(\Leftrightarrow4x^2-4x^3+x^4+2\left(x^3-2x^2\right)=0\)
\(\Leftrightarrow x^4-2x^3=0\)
\(\Leftrightarrow x^3\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) ( Loại do không thỏa mãn ĐKXĐ )
Vậy : không có giá trị của x thỏa mãn đề.
a: Để A là số nguyên thì
x^3-2x^2+4 chia hết cho x-2
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
b: Để B là số nguyên thì
\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)
=>\(3x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(\dfrac{4x-4}{2x^2-2}\)
\(=\dfrac{4\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2}{x+1}\)
Để phân thức có giá trị bằng -2 thì \(\dfrac{2}{x+1}=-2\)
\(\Leftrightarrow x+1=-1\)
hay x=-2(thỏa ĐK)
a) Ta có: \(A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\)
\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right):\dfrac{2x-1}{4x-8}\)
\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}\)
\(=\dfrac{x\left(2x-1\right)}{x+2}\cdot\dfrac{2}{2x-1}\)
\(=\dfrac{2x}{x+2}\)
\(A=\frac{x^2+4x+7}{x-3}=\frac{x\left(x-3\right)+3x+4x+7}{x-3}=\frac{x\left(x-3\right)+7\left(x-3\right)+21+7}{x-3}\)\(=\frac{\left(x-3\right)\left(x+7\right)+28}{x-3}=x+7+\frac{28}{x-3}\)
(x-3) phải thuộc ước của 28=[+-1,+-2,+,4,+-7,+-14,+-28}
x={-25,-11,-4,1,2,4,5,7,10,17,31} nhiêu quá
2:
a: \(=\left(2x^2-xy\right)+\left(2xz-yz\right)\)
\(=x\left(2x-y\right)+z\left(x-2y\right)=\left(x-2y\right)\left(x+z\right)\)
b: \(=\left(x^2-4y^2\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y-1\right)\)
c: \(=\left(y^2+10y+25\right)-9z^2\)
\(=\left(y+5\right)^2-\left(3z\right)^2\)
\(=\left(y+5+3z\right)\left(y+5-3z\right)\)
d: \(=\left(x+2y\right)^3-\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x+2y\right)\left[\left(x+2y\right)^2-\left(x-2y\right)\right]\)
\(=\left(x+2y\right)\left(x^2+4xy+4y^2-x+2y\right)\)
1:
a: \(x\left(3-4x\right)+5\left(3-4x\right)=\left(3-4x\right)\left(x+5\right)\)
b: \(2y\left(5y-6\right)-4\left(6-5y\right)\)
\(=2y\left(5y-6\right)+4\left(5y-6\right)\)
\(=2\left(5y-6\right)\left(y+2\right)\)
c: \(=27\left(x-2\right)^3-3x\left(x-2\right)^2\)
\(=3\left(x-2\right)^2\cdot\left[9\left(x-2\right)-x\right]\)
\(=3\left(x-2\right)^2\left(8x-18\right)=6\left(x-2\right)^2\cdot\left(4x-9\right)\)
d: \(=6y\left(x-y\right)\left(x+y\right)-8y\left(x+y\right)^2\)
\(=2y\left(x+y\right)\left[3\left(x-y\right)-4\left(x+y\right)\right]\)
\(=2y\left(x+y\right)\left(3x-3y-4x-4y\right)\)
\(=2y\left(x+y\right)\left(-x-7y\right)\)
Bài 1
a) x(3 - 4x) + 5(3 - 4x)
= (3 - 4x)(x + 5)
b) 2y(5y - 6) - 4(6- 5y)
= 2y(5y - 6) + 4(5y - 6)
= (5y - 6)(2y + 4)
= 2(5y - 6)(y + 2)
c) 27(x - 2)³ - 3x(2 - x)²
= 27(x - 2)³ - 3x(x - 2)²
= 3(x - 2)²[9(x - 2) - x]
= 3(x - 2)²(9x - 18 - x)
= 3(x - 2)²(8x - 18)
= 6(x - 2)²(4x - 9)
d) 6y(x² - y²) - 8y(x + y)²
= 6y(x - y)(x + y) - 8y(x + y)²
= 2y(x + y)[3(x - y) - 4(x + y)]
= 2y(x + y)(3x - 3y - 4x - 4y)
= 2y(x + y)(-x - 7y)
= -2y(x + y)(x + 7y)
\(ĐKXĐ:x\ne0;x\ne2\)
\(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)
\(\Leftrightarrow4x^2-4x^3+x^4=-2\left(x^3-2x^2\right)\)
\(\Leftrightarrow4x^2-4x^3+x^4=-2x^3+4x^2\)
\(\Leftrightarrow x^4-2x^3=0\Leftrightarrow x^3\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\left(ktm\right)\)
Vậy không có x để phân thức bằng -2
Ta có : \(\frac{4x^2-4x^3+x^4}{x^3-2x^2}=-2\)
( ĐKXĐ : \(x\ne0,x\ne\pm\sqrt{2}\) )
\(\Leftrightarrow\frac{4x^2-4x^3+x^4}{x^3-2x^2}+2=0\)
\(\Leftrightarrow4x^2-4x^3+x^4+2\left(x^3-2x^2\right)=0\)
\(\Leftrightarrow-2x^3+x^4=0\)
\(\Leftrightarrow x^3\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\) ( Loại \(x=0\) không thỏa mãn ĐKXĐ )
Vậy : \(x=2\) thỏa mãn đề.