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\(B=\frac{2}{x^2-y^2}\cdot\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\sqrt{\frac{9\left(x+y\right)^2}{4}}\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{3\left(x+y\right)}{2}\)(vì x > -y <=> x + y > 0)
\(=\frac{3}{x-y}\)
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\frac{a}{2}\)(vì a > = 0)
\(D=\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}=\frac{1}{a-b}\cdot a^2\left(a-b\right)=a^2\)(a > b > 0)
câu cuối điều kiện là a>b
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}=\frac{a^2\left|a-b\right|}{a-b}=\frac{a^2\left(a-b\right)}{a-b}=a^2\) (vì a>b)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(A=\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{x-2\sqrt{2x-4}}}{2}=\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(2x-4\right)-4\sqrt{2x-4}+4}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}=\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
Đến đây có hai trường hợp :
- Với \(2\le x< 4\)\(\Rightarrow\left|\sqrt{2x-4}-2\right|=2-\sqrt{2x-4}\Rightarrow A=\frac{2-\sqrt{2x-4}}{2}\)
- Với \(x\ge4\Rightarrow\left|\sqrt{2x-4}-2\right|=\sqrt{2x-4}-2\Rightarrow A=\frac{\sqrt{2x-4}-2}{2}\)
b) \(B=\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}+a+1=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}+a+1=a-\sqrt{a}-a-\sqrt{a}+a+1=a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2\)
a) ta thấy x-4=(canx-2)(cãnx+2)
2-canx=-(cãnx - 2)
tự học mới giỏi
b)rut gọn roi giai cho
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
\(P=\frac{\frac{1}{4}\left(4a+4\sqrt{a}+1\right)}{2\sqrt{a}+1}.\frac{4\left(2\sqrt{b}-1\right)}{\left(2\sqrt{b}\right)^2-1}\)
\(P=\frac{\left(2\sqrt{a}+1\right)^2}{\left(2\sqrt{a}+1\right)}.\frac{\left(2\sqrt{b}-1\right)}{\left(2\sqrt{b}-1\right)\left(2\sqrt{b}+1\right)}\)
\(P=\frac{2\sqrt{a}+1}{2\sqrt{b}-1}\)