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a: \(P=\dfrac{x^2+x-x^2+x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x-1}\)
a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(1-3xy\right).\frac{-x-1}{1-3xy}+\frac{x^2}{x+1}\)
\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`
\(=\frac{-\left(x+1\right)^2+x^2}{x+1}\)
\(=\frac{-x^2-2x-1+x^2}{x+1}\)
\(=\frac{-2x-1}{x+1}\)(1)
b) Thay \(x=-3,y=2014\)vào (1) ta được:
\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)
Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014
1: \(C=\left(x-\dfrac{4xy}{x+y}+y\right):\left(\dfrac{x}{x+y}+\dfrac{y}{y-x}+\dfrac{2xy}{x^2-y^2}\right)\)
\(=\dfrac{\left(x+y\right)^2-4xy}{x+y}:\left(\dfrac{x}{x+y}-\dfrac{y}{x-y}+\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\right)\)
\(=\dfrac{x^2+2xy+y^2-4xy}{x+y}:\dfrac{x\left(x-y\right)-y\left(x+y\right)+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{x^2-2xy+y^2}{x+y}:\dfrac{x^2-xy-xy-y^2+2xy}{\left(x+y\right)\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)^2}{x+y}\cdot\dfrac{x^2-y^2}{x^2-y^2}=\dfrac{\left(x-y\right)^2}{x+y}\)
2: \(\left(x^2-y^2\right)\cdot C=-8\)
=>\(\left(x-y\right)\left(x+y\right)\cdot\dfrac{\left(x-y\right)^2}{x+y}=-8\)
=>\(\left(x-y\right)^3=-8\)
=>x-y=-2
=>x=y-2
\(M=x^2\left(x+1\right)-y^2\left(y-1\right)-3xy\left(x-y+1\right)+xy\)
\(=\left(y-2\right)^2\left(y-2+1\right)-y^2\left(y-1\right)-3xy\left(-2+1\right)+xy\)
\(=\left(y-1\right)\left[\left(y-2\right)^2-y^2\right]+3xy+xy\)
\(=\left(y-1\right)\left(-4y+4\right)+4xy\)
\(=-4\left(y-1\right)^2+4y\left(y-2\right)\)
\(=-4y^2+8y-4+4y^2-8y\)
=-4
a) \(A=\left(x-y\right).\left(x^2+x+y\right)-x.\left(2x^2+2y^3\right)\)
\(=x^3+x^2+xy-x^2y-xy-y^2-2x^3-2xy^3\)
\(=-x^3-y^2-2xy^3\)
b) Ta thay \(x=-1;y=-5\)
\(-x^3-y^2-2xy^3\)
\(=-\left(-1\right)^3-\left(-5\right)^2-2.\left(-1\right).\left(-5\right)^3\)
\(=1-25-250\)
\(=-274\)
a) ĐKXĐ:
\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)
\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)
\(A=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(A=\dfrac{x-1}{x+1}\)
c) Thay x = 3 vào A ta có:
\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
a) ĐKXĐ:
\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)
\(\Leftrightarrow3x\ne\pm y\)
b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)
\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)
\(B=\dfrac{2\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}\)
\(B=\dfrac{2}{3x+y}\)
Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:
\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)
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.
\(M=\dfrac{x^2}{x\left(x+2\right)}+\dfrac{2x}{x\left(x+2\right)}+\dfrac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\dfrac{x^2+2x+2\left(x+2\right)}{x\left(x+2\right)}=\dfrac{x^2+4x+4}{x\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x+2}{x}\)
\(x=-\dfrac{3}{2}\Rightarrow M=\dfrac{-\dfrac{3}{2}+2}{-\dfrac{3}{2}}=-\dfrac{1}{3}\)
đkxđ:\(x\ne5,x\ne-5\)
\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\)
\(\dfrac{2x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5x+25}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}=\dfrac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=-\dfrac{4}{x-5}\)
thay x=1 vào bt A, ta được:
\(-\dfrac{4}{1-5}=1\)
Câu 1:
\(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left(5x-5y\right)^2-\left(4x+4y\right)^2\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
Bài 2:
a: ĐKXĐ: \(x\notin\left\{1;-\dfrac{1}{2}\right\}\)
b: \(P=\left(\dfrac{1}{x-1}-\dfrac{x}{1-x^3}\cdot\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{x^2+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{x^2+1}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x^2+1}{2x+1}\)
\(=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2+1}{2x+1}=\dfrac{x^2+1}{x^2-1}\)
c: Thay x=1/2 vào P, ta được:
\(P=\dfrac{\left(\dfrac{1}{2}\right)^2+1}{\left(\dfrac{1}{2}\right)^2-1}=\dfrac{5}{4}:\dfrac{-3}{4}=\dfrac{5}{4}\cdot\dfrac{-4}{3}=-\dfrac{5}{3}\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm1\end{matrix}\right.\)
Ta có : \(A=\left(\dfrac{x}{x-1}-\dfrac{1}{x^2-x}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{x^2-1}\right)\)
\(=\left(\dfrac{x}{x-1}-\dfrac{1}{x\left(x-1\right)}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\left(\dfrac{x^2-1}{x\left(x-1\right)}\right):\left(\dfrac{x-1+2}{\left(x-1\right)\left(x+1\right)}\right)\)\(=\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}:\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{x}:\dfrac{1}{x-1}=\dfrac{x+1}{x}.\left(x-1\right)=\dfrac{x^2-1}{x}\)
b, Thay x = 0,5 vào A ta được : A = -3/2
Vậy ...
a) Ta có: \(A=\left(\dfrac{x}{x-1}-\dfrac{1}{x^2-x}\right):\left(\dfrac{1}{x+1}+\dfrac{2}{x^2-1}\right)\)
\(=\left(\dfrac{x^2}{x\left(x-1\right)}-\dfrac{1}{x\left(x-1\right)}\right):\left(\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}+\dfrac{2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(=\dfrac{x^2-1}{x\left(x-1\right)}:\dfrac{x-1+2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2-1}{x}\)
b) Thay \(x=\dfrac{1}{2}\) vào A, ta được:
\(A=\left(\dfrac{1}{4}-1\right):\dfrac{1}{2}=\dfrac{-3}{4}\cdot2=-\dfrac{3}{2}\)
Vậy: Khi \(x=\dfrac{1}{2}\) thì \(A=-\dfrac{3}{2}\)