Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1)
\(ĐKXĐ:x\ne-1\)
\(\dfrac{x^2+2x+1}{x+1}\\ =\dfrac{\left(x+1\right)^2}{x+1}\\ =x+1\)
2)
ĐKXĐ x khác 0 và x khác 3
\(\dfrac{x^2-6x+9}{x\left(x-3\right)}\\ =\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\\ =\dfrac{x-3}{x}\)
3)
ĐKXĐ: x khác 0 và x khác -2
\(\dfrac{x^2-4}{2x\left(x+2\right)}\\ =\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)}\\ =\dfrac{x-2}{2x}\)
4)
DKXĐ: x khác 0 và x khác 2
\(\dfrac{x^2-2x}{5x^2-10x}\\ =\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}\\ =\dfrac{1}{5}\)
`1)` Biểu thức xác định `<=>x+1 \ne 0<=>x \ne -1`
`[x^2+2x+1]/[x+1]=[(x+1)^2]/[x+1]=x+1`
`2)` Bth xác định `<=>x(x-3) \ne 0<=>{(x \ne 0),(x \ne 3):}`
`[x^2-6x+9]/[x(x-3)]=[(x-3)^]/[x(x-3)]=[x-3]/x`
`3)` Bth xác định `<=>2x(x+2) \ne 0<=>{(x \ne 0),(x \ne -2):}`
`[x^2-4]/[2x(x+2)]=[(x-2)(x+2)]/[2x(x+2)]=[x-2]/[2x]`
`4)` Bth xác định `<=>5x^2-10x \ne 0<=>5x(x-2) \ne 0<=>{(x \ne 0),(x \ne 2):}`
`[x^2-2x]/[5x^2-10x]=[x(x-2)]/[5x(x-2)]=1/5`
1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
\(=4\)
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=\left(2x^3-5x^3\right)+\left(x^2-x^2\right)-3x\)
\(=-3x^3-3x\)
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)
\(=-11x+24\)
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=36x^2+12x+1+36x^2-12x+1-72x^2+2\)
\(=4\)
c) \(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^3-x^2+x^2\)
\(=-3x^3-3x\)
d) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)
a) (6x+1)2 + (6x-1)2 - 2(1+6x)(6x-1)
= (6x+1+6x-1)2
=144x2
b) x(2x2 -3) - x2(5x+1) +x2
=2x3 - 3x - 5x3 -x2+x2
=-3x3-3x
=-3x(x2+1)
c) 3(22+1)(24+1)(28+1)(216+1)
= (22-1)(22+1)(24+1)(28+1)(216+1)
= (24-1)(24+1)(28+1)(216+1)
= (28-1)(28+1)(216+1)
= (216-1)(216+1)
= 232 -1
d) 3x(x-2) - 5x(1-x) - 8(x2 -3)
= 3x2-6x - 5x + 5x2 - 8x2 +24
= -11x +24
(6x+1)2+(6x-1)2-2(1+6x)(6x-1)=(6x+1+1-6x)2=4
x(2x2-3)-x2(5x+1)+x2=2x3-3x-5x3-x2= -3x3-x2-3x
3x(x-2)-5x(1-x)-8(x2-3)=3x2-6x-5x+5x2-8x2+24= -11x+24
chưa chắc là đúng đâu nhé
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
2: ĐKXĐ: \(x\ne2\)
\(\dfrac{3x^2+6x+12}{x^3-8}=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)