Rút gọn biểu thức M
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
\(\left(-m+n-p\right)-\left(-m-n-p\right)\)
\(=-m+n-p+m+n+p\)
\(=\left(-m+m\right)+\left(-p+p\right)+\left(n+n\right)\)
\(=2n\)
Vậy \(\left(-m+n-p\right)-\left(-m-n-p\right)=2n\)
\(\left(-m+n-p\right)-\left(-m-n-p\right)\)
\(=-m+n-p+m+n+p\)
\(=\left(-m+m\right)+\left(-p-p\right)+\left(n+n\right)\)
\(=0+0+\left(n+n\right)\)
\(=0+\left(n+n\right)\)
\(=n+n\)
\(=2n\)
Vậy biểu thức (-m+n-p)-(-m-n-p) =2n
1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5
This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured
M = 2 2 + 3 8 − 18 = 2 2 + 3 4.2 − 9.2 = 2 2 + 6 2 − 3 2 = 5 2
a: Ta có: \(M=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1\)
\(=a-\sqrt{a}\)
a: \(M=\dfrac{18+5x+15+3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{8x+24}{\left(x+3\right)\left(x-3\right)}=\dfrac{8}{x-3}\)
b: Thay x=11 vào M, ta được:
\(M=\dfrac{8}{11-3}=1\)
a) \(M=\dfrac{18}{x^2-9}+\dfrac{5}{x-3}+\dfrac{3}{x+3}.\left(x\ne\pm3\right).\)
\(M=\dfrac{18}{\left(x-3\right)\left(x+3\right)}+\dfrac{5}{x-3}+\dfrac{3}{x+3}=\dfrac{18+5\left(x+3\right)+3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{18+5x+15+3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{24+8x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{8\left(3+x\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{8}{x-3}.\)
b) Thay \(x=11\left(TM\right)\) vào biểu thức M:
\(\dfrac{8}{11-3}=\dfrac{8}{8}=1.\)
ĐKXĐ: \(x\notin\left\{0;-2;1;-1\right\}\)
\(M=\dfrac{1}{x^2+2x}-\dfrac{1}{x^2-1}\)
\(=\dfrac{x^2-1-x^2-2x}{\left(x^2-1\right)\left(x^2+2x\right)}\)
\(=\dfrac{-2x-1}{\left(x-1\right)\left(x+1\right)\cdot x\left(x+2\right)}\)