\(ĐKXĐ:a\ge0\)

\(A=\left(\frac{2\sqrt{a}}{a\sqrt{a}+a+\sqrt{a}+1}+\frac{1}{\sqrt{a}+1}\right):\left(1+\frac{\sqrt{a}}{a+1}\right)\)

\(\Leftrightarrow A=\left(\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}+1\right)}+\frac{1}{\sqrt{a}+1}\right):\frac{a+\sqrt{a}+1}{a+1}\)

\(\Leftrightarrow A=\frac{2\sqrt{a}+a+1}{\left(a+1\right)\left(\sqrt{a}+1\right)}\cdot\frac{a+1}{a+\sqrt{a}+1}\)

\(\Leftrightarrow A=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(a+\sqrt{a}+1\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{a}+1}{a+\sqrt{a}+1}\)

ai giúp mình với ạ

a . i ) Vì CM,CA là tiếp tuyến của (O) 

\(\Rightarrow CM\perp OM,CA\perp OA\Rightarrow CMOA\) nội tiếp đường tròn đường kính CO 

Tương tự : = > DMOB nội tiếp 

ii ) Vì CM,CA là tiếp tuyến của (O) \(\Rightarrow OC\) là phân giác của \(\widehat{AOM}\)

Tương tự OD là phân giác \(\widehat{BOM}\)

Mà \(\widehat{AOM}+\widehat{MOB}=180^0\Rightarrow OC\perp OD\)

Ta có : CMOA , OBDM nội tiếp 

\(\Rightarrow\widehat{AOC}=\widehat{AMC}=\widehat{ABM}=\widehat{OBM}=\widehat{ODM}\) vì CM là tiếp tuyến của (O) 

b ) Ta có : \(\widehat{MAB}=60^0\Rightarrow\widehat{DMB}=\widehat{MAB}=60^0\) vì DM là tiếp tuyến của (O) 

Mà \(DM=DB\Rightarrow\Delta DMB\) đều 

Lại có : \(\widehat{MOB}=2\widehat{MAB}=120^0\)

\(\Rightarrow\frac{S_{MB}}{S_O}=\frac{120^0}{360^0}=\frac{1}{3}\)

\(\Rightarrow S_{MB}=\frac{1}{3}S_O=\frac{1}{3}.\pi.R^2\)

\(ĐKXĐ:a>1\)

\(P=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow P=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow P=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\cdot\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow P=\frac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(\Leftrightarrow P=\frac{2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(\Leftrightarrow P=\frac{2}{a-1}\)

\(ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)

Ta có :

 \(P=\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\frac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\left(\frac{\left(a+\sqrt{a}-2\right)-\left(a-\sqrt{a}-2\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\frac{2\sqrt{a}}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{2}{a-1}\)

Vậy \(P=\frac{2}{a-1}\left(a>0;a\ne1\right)\)

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Ta có : A = \(\left(\frac{x+2}{x.\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

                 = \(\frac{x+2+x+\sqrt{x}-2-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)

                = \(\frac{x-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=1\)

Vậy A = 1

Bạn xem lại đề chỗ : \(y=f\left(x\right)=\frac{2}{3}\)

TỰ TỬ ĐÊ CU AI THẤY HAY CHO

\(\hept{\begin{cases}\frac{x}{x-1}+\frac{2y}{y+1}=3\\\frac{x}{x+1}+\frac{y}{y-1}=2\end{cases}}\)\(ĐKXĐ:x,y\ne\pm1\)

\(< =>\hept{\begin{cases}\frac{x\left(y+1\right)}{\left(x-1\right)\left(y+1\right)}+\frac{\left(x-1\right)2y}{\left(x-1\right)\left(y+1\right)}=3\\\frac{x\left(y-1\right)}{\left(x+1\right)\left(y-1\right)}+\frac{y\left(x+1\right)}{\left(x+1\right)\left(y-1\right)}=2\end{cases}}\)

\(< =>\hept{\begin{cases}\frac{xy+x}{xy+x-y-1}+\frac{2xy-2y}{xy+x-y-1}=3\\\frac{xy-x}{xy-x+y-1}+\frac{xy+y}{xy-x+y-1}=2\end{cases}}\)

\(< =>\hept{\begin{cases}\frac{xy+x+2xy-2y}{xy+x-y-1}=3\\\frac{xy-x+xy+y}{xy-x+y-1}=2\end{cases}}\)

\(< =>\hept{\begin{cases}\frac{3xy+x-2y}{xy+x-y-1}=3\\\frac{2xy-x+y}{xy-x+y-1}=2\end{cases}}\)

\(< =>\hept{\begin{cases}3xy+x-2y=3xy+3x-3y-3\\2xy-x+y=2xy-2x+2y-2\end{cases}}\)

\(< =>\hept{\begin{cases}3xy+x-2y-3xy-3x+3y+3=0\\2xy-x+y-2xy+2x-2y+2=0\end{cases}}\)

\(< =>\hept{\begin{cases}-2x+y+3=0\\x-y+2=0\end{cases}}\)

\(< =>\hept{\begin{cases}-2\left(-2+y\right)+y+3=0\left(1\right)\\x=-2+y\left(2\right)\end{cases}}\)

\(\left(1\right)< =>4-2y+y+3=0\)

\(< =>7-y=0< =>y=7\left(tmđk\right)\)

\(\left(2\right)< =>x=-2+7=5\left(tmđk\right)\)

Vậy nghiệm của hệ phương trình trên là \(\left\{5;7\right\}\)

bài này mình thấy chỉ cần thế này là xong

\(x^2+y^2=1\Leftrightarrow x+y=\sqrt{1}=1\)( có đúng ko nhỉ )

=>\(x+y=0+1=1+0\)

\(=>\left\{x,y\right\}\in\left(0,1\right);\left(1,0\right)\)

P/s : làm thử , e ms lớp 8 .

\(\hept{\begin{cases}x^2+y^2=1\\x^2-x=y^2-y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1-y^2\\x^2-y^2=x-y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1-y^2\\\left(x-y\right)\left(x+y\right)=x-y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1-y^2\\x+y=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1-y^2\left(1\right)\\x=1-y\left(2\right)\end{cases}}\)

Thay ( 2 ) vào ( 1 ) ta có :

\(\left(1-y\right)^2=1-y^2\)

\(\Leftrightarrow1-2y+y^2=1-y^2\)

\(\Leftrightarrow1+y^2-1+y^2=2y\)

\(\Leftrightarrow2y^2=2y\)

\(\Leftrightarrow2y^2-2y=0\)

\(\Leftrightarrow2y\left(y-1\right)=0\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

T/h 1 : y = 0 

=> x = 1 - 0 = 1

T/h 2 : y = 1

=> x = 1 - 1 = 0

Vậy ...................