MX
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
DT
1
7 tháng 12 2016
\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x^2-4\right)}{x\left(x+2\right)}=\frac{A}{x}\)
\(\Leftrightarrow\frac{4\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}=\frac{A}{x}\)\(\Leftrightarrow\frac{4\left(x-2\right)}{x}=\frac{A}{x}\)
\(\Leftrightarrow4\left(x-2\right)=A\Leftrightarrow A=4x-8\)
30 tháng 1 2019
a) \(\frac{x^2-2x+2}{x^2+x+1}-\frac{x^2}{x^2+x+1}=\frac{3}{\left(x^4+x^2+1\right)x}\)
\(\Leftrightarrow\frac{x^2-2x+2}{x^2-x+1}.x\left(x^2-x+1\right)\left(x^2+x+1\right)-\frac{x^2}{x^2+x+1}.x\left(x^2-x+1\right)\left(x^2+x+1\right)\)\(=\frac{3}{\left(x^4+x^2+1\right)x}.x\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x^2-2x+2\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)-x^3\left(x^2-x+1\right)\left(x^4+x^2+1\right)\)\(=3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x=\frac{3}{2}\)
b) làm tương tự nhé
11 tháng 12 2016
\(\frac{a^2x^3-a^2}{ax^2+ax+a}=\frac{a^2\left(x^3-1\right)}{a\left(x^2+x+1\right)}=\frac{a^2\left(x-1\right)\left(x^2+x+1\right)}{a\left(x^2+x+1\right)}=a\left(x-1\right)=ax-a\)
\(\frac{12x^2-26x-16}{4x^2+4x+1}=\frac{\left(6x-16\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{6x-16}{2x+1}\)
\(\frac{\left(x+a\right)^2-x^2}{2x+a}=\frac{\left(x+a+x\right)\left(x+a-x\right)}{2x+a}=\frac{\left(2x+a\right)a}{2x+a}=a\)
11 tháng 12 2018
a) Để \(\frac{2x+3}{4x-5}=0\)
=> 2x + 3 = 0
x = -3/2
b) Để \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-3\right).\left(x-1\right)}=\frac{x+2}{x-3}=0\)
=> x + 2 = 0=> x = -2
c) để \(\frac{x^2-1}{x^2-2x+1}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-1\right)^2}=\frac{x+2}{x-1}=0\)
=> x + 2 = 0 => x = -2
d) để \(\frac{x^2-4}{x^2+3x-10}=\frac{\left(x+2\right).\left(x-2\right)}{\left(x-2\right).\left(x+5\right)}=\frac{x+2}{x+5}=0\)
=> ...
e) để \(\frac{x^3-16x}{x^3-3x^2-4x}=\frac{x.\left(x-4\right).\left(x+4\right)}{x.\left(x-4\right).\left(x+1\right)}=\frac{x+4}{x+1}=0\)
=> ....
14 tháng 6 2020
con lựa a;b còn đâu bác lm nhé ... ko c;d dài dòng lắm bác )):
a, \(\frac{2x-3}{x+2}-\frac{x+2}{x-2}=\frac{2}{x^2-4}ĐKXĐ:x\ne\pm2\)
\(\frac{\left(2x-3\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\frac{2}{\left(x+2\right)\left(x-2\right)}\)
Khử mẫu ta đc : \(\left(2x-3\right)\left(x-2\right)-\left(x+2\right)^2=2\)
\(x^2-11x+2=2\)
\(x^2-11x=0\Leftrightarrow x\left(x-11\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=11\end{cases}}\)
28 tháng 3 2020
ĐKXĐ bạn tự tìm nha : )
k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)
\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)
i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)
\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)
h, = k,
f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)
\(\Rightarrow A.\left(x^2+2x\right)=\left(4x^2-16\right).x\)
\(\Rightarrow A=\frac{\left[\left(2x\right)^2-4^2\right].x}{x^2+2x}\)
\(A=\frac{\left(2x-4\right)\left(2x+4\right).x}{x\left(x+2\right)}\)
\(A=\frac{2.2.\left(x-2\right)\left(x+2\right).x}{x\left(x+2\right)}\)
\(A=4\left(x-2\right)\)\(\left(x\ne0;x+2\ne0\right)\)
\(A=4x-8\)