Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
\(\dfrac{x^2-9y^2}{x^2-6xy+9y^2}\) tại x = 1 , y = -\(\dfrac{2}{3}\)
= \(\dfrac{x^2-\left(3y\right)^2}{\left(x-3y\right)^2}\)
= \(\dfrac{\left(x-3y\right)\left(x+3y\right)}{\left(x-3y\right)}\)
= (x + 3y)
Thay x = 1 , y = -\(\dfrac{2}{3}\) vào
x + 3y
= 1 +3 . -\(\dfrac{2}{3}\)
= -1
Chúc bạn học tốt
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.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
\(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\) (1)
a) ĐKXĐ: \(x\ne\pm3\)
b) \(\left(1\right)=\left[\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{4}{5x+15}\)
\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5x+15}\)
\(=\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{4}{5\left(x-3\right)}\)
c) Thay \(x=19\) vào \(A=\dfrac{4}{5\left(x-3\right)}\) ta có:
\(A=\dfrac{4}{5.\left(19-3\right)}=\dfrac{4}{80}=\dfrac{1}{20}\)
Vậy \(x=19\) thì \(A=\dfrac{1}{20}\)
a) ĐK: \(x\)≠\(+-3\)
b) \(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\)
\(=\dfrac{2x^2+3\left(x+3\right)-x\left(x-3\right)}{x^2-9}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x+3\right)\left(x-3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4\left(x+3\right)^2}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4}{5\left(x-3\right)}=\dfrac{4}{5x-15}\)
c) Tại x=19
⇒ \(A=\dfrac{4}{5.19-15}=\dfrac{4}{80}=\dfrac{1}{20}\)
Vậy ...
a) \(A=4x^2-4x+1+9-4x^2=-4x+10\)
\(=-4.\dfrac{1}{4}+10=9\)
b) \(B=x^3+xy-x^3-8y^3=y\left(x-8y^2\right)\)
\(=\left(-2\right).\left(32-32\right)=0\)
a: Ta có: \(A=\left(2x-1\right)^2+\left(3-2x\right)\left(3+2x\right)\)
\(=4x^2-4x+1+9-4x^2\)
\(=-4x+10\)
\(=-4\cdot\dfrac{1}{4}+10=-1+10=9\)
\(A=\left(\dfrac{4}{x-2}-\dfrac{3}{x+2}\right):\dfrac{x+14}{x^2}\\ =\left(\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x^2}{x+14}\\ =\dfrac{4x+8-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\ =\dfrac{x+14}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\=\dfrac{x^2}{x^2-4}\)
Tại `x=-3` Ta có :
\(\dfrac{x^2}{x^2-4}\\ =\dfrac{-3^2}{-3^2-4}\\ =\dfrac{9}{9-4}\\ =\dfrac{9}{5}\)