theo mk là 39

a) đk: \(a\ge0;a\ne4\)

Ta có:

\(A=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}}\)

\(A=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\frac{1}{\sqrt{a}-2}\)

\(A=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(A=\frac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(A=\frac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(A=\frac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-4\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(A=\frac{\sqrt{a}-4}{\sqrt{a}-2}\)

b) Ta có: \(a=7-4\sqrt{3}\)

\(\Leftrightarrow a=4-4\sqrt{3}+3\)

\(\Leftrightarrow a=\left(2-\sqrt{3}\right)^2\)

\(\Rightarrow\sqrt{a}=\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

Thay vào ta tính:

\(A=\frac{2-\sqrt{3}-4}{2-\sqrt{3}-2}=\frac{3+2\sqrt{3}}{3}\)

giúp mình với mình cần gấp lắm huhuu

a. ĐKXĐ : \(\hept{\begin{cases}x\ge0\\y\ge0\\y-x\ne0\end{cases}}\)<=> \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)

b. \(R=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\left(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{y-x}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(\Leftrightarrow R=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(\Leftrightarrow R=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{x-\sqrt{xy}+y}\)

\(\Leftrightarrow R=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

c. Với \(\hept{\begin{cases}x\ge0\\y\ge0\\x\ne y\end{cases}}\)thì \(\sqrt{xy}\ge0\)  ( 1 )

Ta có : \(x-\sqrt{xy}+y=\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}\)

Mà \(\orbr{\begin{cases}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(1\right)\end{cases}}\)=> \(x-\sqrt{xy}+y\ge0\)( 2 )

Từ ( 1 ) và ( 2 ) => \(R\ge0\) ( Đpcm )

a. ĐKXĐ : \(\orbr{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\end{cases}}\)<=> \(\orbr{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

b. \(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow P=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)

là bằng 2 phần 3 phải ko

Ta có: \(X=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)

<=> \(X^2=6-3\sqrt{2+\sqrt{3}}+2+\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{4-\left(2+\sqrt{3}\right)}\)

<= \(X^2=8-2\sqrt{2+\sqrt{3}}-2\sqrt{3}.\sqrt{2-\sqrt{3}}\)

<=> \(X^2=8-\sqrt{2}\left(\sqrt{3}+1\right)-\sqrt{6}\left(\sqrt{3}-1\right)\)

<=> \(X^2=8-4\sqrt{2}\)

<=> \(X^2-8=-4\sqrt{2}\)

=> \(X^4-16X+64=32\)

<=> \(X^4-16X^2+32=0\)

Vậy X là nghiệm phương trình \(X^4-16X^2+32=0\)

\(G=\sqrt{8\sqrt{3}-4\sqrt{6}-4\sqrt{2}+18}-\sqrt{14-4\sqrt{6}}\)

\(G=\sqrt{12+4+2-4\sqrt{6}+8\sqrt{3}-4\sqrt{2}}-\sqrt{12+4-4\sqrt{6}}\)

\(G=\left(2\sqrt{3}+2-\sqrt{2}\right)-\left(2\sqrt{3}-2\right)=4-\sqrt{2}\)

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

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

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

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

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

\(=\frac{2}{x+\sqrt{x}+1}\)

b) \(x+\sqrt{x}+1>0\Rightarrow G>0\)

\(x+\sqrt{x}+1>0+0+1=1\)

\(\Rightarrow\frac{2}{x+\sqrt{x}+1}< \frac{2}{1}=2\Rightarrow G< 2\)

\(\Rightarrow O< G< 2\)