Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ta có : \(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)\(=\frac{x+2}{x+3}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{1}{2-x}\)
\(=\frac{\left(x+2\right)\left(2-x\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{1\left(x+3\right)}{\left(x+3\right)\left(2-x\right)}\)
\(=\frac{-x^2+4+5+x+3}{\left(x+3\right)\left(2-x\right)}\) \(=\frac{-x^2+x+12}{\left(x+3\right)\left(2-x\right)}\)
\(=\frac{\left(4-x\right)\left(x+3\right)}{\left(x+3\right)\left(2-x\right)}\) = \(\frac{4-x}{2-x}\)
xin lỗi mình viết nhầm cho gửi lại câu hỏi!
CHO \(A=\left(\left(\frac{x+7}{x+9}+\frac{x+7}{x^2+81-18x}+\frac{x+5}{x^2-81}\right)\left(\frac{x-9}{x+3}\right)^2\right)^{ }:\left(\frac{x+7}{x+3}\right)\)
a) Rút gọn A
b) Tìm số nguyên x để A nguyên
a, 8/x-8 + 11/x-11 = 9/x-9 + 10/ x-10
b, x/x-3 - x/x-5 = x/x-4 - x/x-6
c, 4/x^2-3x+2 - 3/2x^2-6x+1 +1 = 0
d, 1/x-1 + 2/ x-2 + 3/x-3 = 6/x-6
e, 2/2x+1 - 3/2x-1 = 4/4x^2-1
f, 2x/x+1 + 18/x^2+2x-3 = 2x-5 /x+3
g, 1/x-1 + 2x^2 -5/x^3 -1 = 4/ x^2 +x+1
a)
\(DKXD:\left[{}\begin{matrix}x^2+x\ne0\\x\ne0\\x+1\ne0\end{matrix}\right.< =>\left[{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b)
\(\left(\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)
\(=\left(\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}\right)\cdot\dfrac{x+1}{3}\)
\(=\dfrac{x^2+3x}{x\left(x+1\right)}\cdot\dfrac{x+1}{3}\\ =\dfrac{x\left(x+3\right)\cdot\left(x+1\right)}{x\left(x+1\right)\cdot3}\\ =\dfrac{x+3}{3}\)
Ghi hẳn hoi đi
What ?? Đề bài j kì z bn ??