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`a)M=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/(x^4+4x^2+3)`
`=(x^4+2)/(x^6+1)+(x^2-1)/(x^4-x^2+1)-(x^2+3)/((x^2+1)(x^2+3))`
`=(x^4+2)/(x^6+1)+((x^2-1)(x^2+1))/(x^6+1)-1/(x^2+1)`
`=(x^4+2+x^4-1-x^4+x^2-1)/(x^2+1)`
`=(x^4+x^2)/(x^2+1)`
`=(x^2(x^2+1))/(x^2+1)`
`=x^2`
`b)` tìm gtnn chứ?
`M=x^2>=0`
Dấu '=" `<=>x=0`
\(M=\dfrac{x}{\left(x-2\right)\cdot\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\)
\(=\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-6}{x^2-4}\)
a)
\(ĐKXĐ:\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x^2-4\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b)
\(\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x^2+4x}{x^2-4}\)
\(=\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x\left(x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x+2}{x-2}\)
c)
\(\dfrac{x+2}{x-2}=\dfrac{x-2+4}{x-2}=\dfrac{x-2}{x-2}+\dfrac{4}{x-2}=1+\dfrac{4}{x-2}\)
vậy M nhận giá trị nguyên thì 4⋮x-2
=> x-2 thuộc ước của 4
\(Ư\left(4\right)\in\left\{-1;1;-2;2;;4;-4\right\}\)
ta có bảng sau
| x-2 | -1 | 1 | -2 | 2 | 4 | -4 |
| x | 1(tm) | 3(tm) | 0(tm) | 4(tm) | 6(tm | -2(loại) |
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
ĐKXĐ: \(x\ne\pm1\)
\(A=\left(\dfrac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)\left(1+x\right)}+\dfrac{4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(=\left(\dfrac{x^2+2x+1-\left(x^2-2x+1\right)+4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)
\(=\left(\dfrac{4x^2+4x}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)
\(=\dfrac{4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)}{4\left(x+1\right)}=-\dfrac{x}{x+1}\)
a: ĐKXĐ: x<>2; x<>0
b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
c: M>=-3
=>(x+1+6x)/2x>=0
=>(7x+1)/x>=0
=>x>0 hoặc x<=-1/7
1,
\(A=\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x^2+x-2-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x^2-4}{\left(x-2\right)\left(x+2\right)}\)
\(x=4\Rightarrow A=\dfrac{4.x^2-4}{\left(4-2\right)\left(4+2\right)}=...\)
2.
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3-5x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)+3-5x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
3.
Đề lỗi, thiếu dấu trước \(\dfrac{6+5x}{4-x^2}\)
4.
\(A=\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{2x-5\left(x+5\right)-\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4}{x-5}\)
\(x=\dfrac{4}{5}\Rightarrow A=\dfrac{-4}{\dfrac{4}{5}-5}=\dfrac{20}{21}\)
5.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
\(A=\left(\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{x^2+x-2+x^2-x+2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x^2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)
\(=\dfrac{2x\left(x^2-2\right)\left(x-3\right)}{2x\left(x-2\right)\cdot\left(x+2\right)^2}=\dfrac{\left(x^2-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)^2}\)
a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)
\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)
\(=\dfrac{x+1}{2x}\)
b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)
\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(8-x\right)}{x+2}\)
\(=\dfrac{x-8}{x+2}\)
c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)
\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)
\(=-2\)
d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
ĐKXĐ: \(x\neq0;x\neq-1\)
\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)
$---$
ĐKXĐ: \(x\neq y\)
\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)
MTC = (x - 2)(x + 2)
\(M=\dfrac{1}{x-2}-\dfrac{1}{x+2}+\dfrac{x^2+4x}{x^2-4}\)
\(=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2-x+2+x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x+2}{x-2}\)