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Bài 1:
a, Ta có:
\(\dfrac{x.\dfrac{xy}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{y.\dfrac{xy}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\dfrac{x^2y}{x-y}}{x+\dfrac{xy}{x-y}}-\dfrac{\dfrac{xy^2}{x-y}}{y-\dfrac{xy}{x-y}}\)
\(=\dfrac{\left(\dfrac{x^2y}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)-\left(\dfrac{xy^2}{x-y}\right)\left(x+\dfrac{xy}{x-y}\right)}{\left(x+\dfrac{xy}{x-y}\right)\left(y-\dfrac{xy}{x-y}\right)}\)
\(=\dfrac{\dfrac{x^2y^2}{x-y}-\dfrac{x^3y^2}{\left(x-y\right)^2}-\dfrac{x^2y^2}{x-y}-\dfrac{x^2y^3}{\left(x-y\right)^2}}{xy-\dfrac{x^2y}{x-y}+\dfrac{xy^2}{x-y}-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{-\left(\dfrac{x^3y^2+x^2y^3}{\left(x-y\right)^2}\right)}{xy-\left(\dfrac{x^2y-xy^2}{x-y}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=-\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{xy-\left(\dfrac{xy\left(x-y\right)}{\left(x-y\right)}\right)-\dfrac{x^2y^2}{\left(x-y\right)^2}}\)
\(=\dfrac{\dfrac{x^2y^2\left(x+y\right)}{\left(x-y\right)^2}}{\dfrac{x^2y^2}{\left(x-y\right)^2}}=x+y\)
Chúc bạn học tốt!! Làm một câu mà toát cả mồ hôi!
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
a, Ta có : \(A=\frac{1}{x+2}-\frac{2x}{4-x^2}+\frac{3}{x-2}\)
\(=\frac{1}{x+2}-\frac{2x}{\left(2-x\right)\left(x+2\right)}+\frac{3}{x-2}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra : \(M=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3x+2}\)
\(=\frac{2\left(3x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}=\frac{2}{x-2}\)
a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\) (ĐK : \(\forall x\in R\))
\(=\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}\)
* Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)
*Nếu x<2 => M=2-x-x-2=-2x
b,Để M=2\(\ne-4\)
=>M=-2x
=>-2x=-4
=>x=2
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P=\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
* Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
* Nếu x<2 =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
VẬY.......
Tk nha!
a) Biểu thức M xác định <=> \(\hept{\begin{cases}2-2x\ne0\\2-2x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}2x\ne2\\2x^2\ne2\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x^2\ne1\end{cases}}\) <=> \(\hept{\begin{cases}x\ne1\\x\ne\pm1\end{cases}}\)
Vậy đk xác định biểu thức M <=> x \(\ne\)\(\pm\)1
b) Ta có:
M = \(\frac{x}{2-2x}-\frac{x^2+1}{2-2x^2}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x^2\right)}\)
M = \(\frac{x}{2\left(1-x\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x\left(x+1\right)}{2\left(1-x\right)\left(x+1\right)}-\frac{x^2+1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x^2+x-x^2-1}{2\left(1-x\right)\left(x+1\right)}\)
M = \(\frac{x-1}{-2\left(x-1\right)\left(x+1\right)}\)
M = \(-\frac{1}{2\left(x+1\right)}\) (đk : x + 1 \(\ne\)0 => x \(\ne\)-1)
a, \(I=s\left(s^2-t\right)+\left(t^2+s\right)=s^3-st+t^2+s\)
Thay t = -1 và s = 1 vào biểu thức trên ta được :
\(1+1+1+1=4\)
b, \(N=u^2\left(u-v\right)-v\left(v^2-u^2\right)=u^2\left(u-v\right)+v\left(u+v\right)\left(u-v\right)\)
\(=\left(u-v\right)\left(u^2+v\left(u+v\right)\right)\)
Thay \(u=0,5=\frac{1}{2};v=-\frac{1}{2}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right).\frac{1}{4}=\frac{1}{4}\)
a) Thay phân thức P vào biểu thức A rồi rút gọn chúng ta thu được A = u + v với điều kiện các biểu thức có nghĩa.
b) Tương tự a) ta có B = 1.