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a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)
b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)
1) a) \(\frac{x}{x+1}+\frac{x^3-2x^2}{x^3+1}=\frac{x}{x+1}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3-x^2+x+x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{2x^3-3x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b) \(\frac{x+1}{2x-2}+\frac{3}{x^2-1}+\frac{x+3}{2x+2}=\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{x+3}{2\left(x+1\right)}\)
\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)
\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{2x^2+4x+2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)
2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)
Thay x = 14 ; y = -15 vào biểu thức ta được
\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)
\(=\frac{16+x}{x^2-2x}-\frac{18}{x^2-2x}\)
\(=\frac{16+x-18}{x\left(x-2\right)}\)
\(=\frac{-2+x}{x\left(x-2\right)}\)
a) \(\frac{16+x}{x^2-2x}+\frac{18}{2x-x^2}=\frac{16+x-18}{x^2-2x}=\frac{x-2}{x\left(x-2\right)}=\frac{1}{x}\)
b) \(\frac{2y}{2x^2-xy}+\frac{4x}{xy-2x^2}=\frac{2y-4x}{2x^2-xy}=\frac{-2\left(2x-y\right)}{x\left(2x-y\right)}=\frac{-2}{x}\)
c) \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}=\frac{4-x^2+2x^2-2x+5-4x}{x-3}=\frac{x^2-6x+9}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)
a) = \(12a^2b\left(a^2-b^2\right)\)
= \(12a^4b-12a^2b^3\)
b)nhân ra :
= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)
= \(2x^4-19x^3+33x^2-46x+10\)
Tìm x:
a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)
= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)
=\(2x=-14=>x=-7\)
b) \(x^3+27-x\left(x^2-1\right)=27\)
= \(x^3+27-x^3+x=27\)
= \(27+x=27=>x=0\)
a, đkxđ:x# 2 , x# -2
b,
A = \(\frac{x+1}{x-2}\)=0
<=> x + 1 = 0
<=> x = -1
c,B=\(\frac{x2}{x^2-4}\)
Mà x= \(-\frac{1}{2}\)
<=> \(\frac{1}{4}:\left(\frac{1}{4}-4\right)\)
<=>\(\frac{1}{4}:\frac{-15}{4}\)
<=>\(\frac{1}{4}.\frac{4}{-15}\)
<=>\(\frac{-1}{15}\)
d, \(A-B=\frac{x+1}{x-2}-\frac{x^2}{x^2-4}\)
\(=\frac{\left(x+1\right)\left(x+2\right)-x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+3x+2-x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\frac{a}{x-2}+\frac{b}{\left(x+1\right)^2}=\frac{a\left(x+1\right)^2+b\left(x-2\right)}{\left(x-2\right)\left(x+1\right)^2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)
Đồng nhất hệ số, ta có :
\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)
Chọn đáp án A