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\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
a) A \(=\)\(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)\(=\)\(\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\)\(\frac{2\left(x-2\right)}{x+2}\)\(=\)\(\frac{2x-4}{x+2}\)
Tại x = \(\frac{1}{2}\)thì:
A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
Lời giải:
a)
\(A=\frac{x^2y(y-x)-xy^2(x-y)}{3y^2-2x^2}=\frac{x^2y(y-x)+xy^2(y-x)}{3y^2-2x^2}=\frac{(xy^2+x^2y)(y-x)}{3y^2-2x^2}\)
\(=\frac{xy(x+y)(y-x)}{3y^2-2x^2}=\frac{xy(y^2-x^2)}{3y^2-2x^2}\)
Với $x=-3; y=\frac{1}{2}$ thì:
$xy=\frac{-3}{2}; x^2=9; y^2=\frac{1}{4}$
Do đó $A=\frac{-35}{46}$
b)
\(B=\frac{(8x^3-y^3)(4x^2-y^2)}{(2x+y)(4x^2-4xy+y^2)}=\frac{(2x-y)(4x^2+2xy+y^2)(2x-y)(2x+y)}{(2x+y)(2x-y)^2}\)
\(=4x^2+2xy+y^2=4.2^2+2.2.\frac{-1}{2}+(\frac{-1}{2})^2=\frac{57}{4}\)
\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2+y+2}\)
\(P=\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right):\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)
\(P=\left(\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(x+y\right)\left(2y-x\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\right):\frac{2x^2+y+2}{x+1}\)
\(P=\left(\frac{2x^2+y-2}{2y-x}.\frac{x+1}{2x^2+y-2}\right).\frac{1}{x+1}\)
\(P=\frac{1}{2y-x}\)
Tại \(x=-1,76\) và \(y=\frac{3}{25}\) thì giá trị của \(Q=\frac{1}{2}\)
a) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\)
\(=\frac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)
c) \(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}\)
\(=\frac{2x\left(x-2\right)^2}{x\left(x+2\right)\left(x-2\right)}\)
\(=\frac{2x\left(x-2\right)}{x\left(x+2\right)}\)
\(=\frac{2x^2-4x}{x^2+2x}\)
d) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)
\(=\frac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x}{x+y}\)