b, Ta có : \(M=\frac{4\sqrt{x}}{x+2\sqrt{x}+1}=\frac{8}{9}\Rightarrow36\sqrt{x}=8x+16\sqrt{x}+8\)

\(\Leftrightarrow8x-20\sqrt{x}+8=0\Leftrightarrow2x-5\sqrt{x}+2=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\Leftrightarrow x=4;x=\frac{1}{4}\)

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

\(=2-x\cdot\frac{x-1}{x-2}=\frac{2x-4}{x-2}-\frac{x^2-x}{x-2}=\frac{-x^2+3x-4}{x-2}\)

\(B=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+\frac{3\sqrt{5}x^2}{x}=\frac{2\sqrt{5}x}{x-2}\cdot\left|x-2\right|+3\sqrt{5}x\)

Với 0 < x < 2 \(B=-2\sqrt{5}x+3\sqrt{5}x=\sqrt{5}x\)

Với x > 2 \(B=2\sqrt{5}x+3\sqrt{5}x=5\sqrt{5}x\)

\(C=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}\left(\sqrt{x}+5\right)}+\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-5\right)^2}}=\frac{\sqrt{x}-5}{\sqrt{x}}+\left|\frac{\sqrt{x}-1}{\sqrt{x}-5}\right|\)

Với 0 < x < 1 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với 1 < x < 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}-\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}-\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{-9\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

Với x > 5 \(C=\frac{\sqrt{x}-5}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-5}=\frac{x-10\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}+\frac{x-\sqrt{x}}{x\left(\sqrt{x}-5\right)}=\frac{2x-11\sqrt{x}+25}{x\left(\sqrt{x}-5\right)}\)

tích hộ mình nha

đề là  \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)đúng ko bạn

\(\left(\sqrt{2}+1-\sqrt{2}+1\right)\left[\left(\sqrt{2}+1\right)^2+\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\left(\sqrt{2}-1\right)^2\right]\)

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

\(=2\left(3+\sqrt{1}+3\right)\)

\(=2\left(6+1\right)\)

\(=14\)

\(b,\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)

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

\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)

\(=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(3-\sqrt{3}\right)+\left(2\sqrt{2}-6\right)\left(3+\sqrt{3}\right)}{9-3}\)

\(=\frac{6\sqrt{2}+3\sqrt{6}-2\sqrt{6}-3\sqrt{2}+6\sqrt{2}-3\sqrt{6}+2\sqrt{6}-3\sqrt{2}}{6}\)

\(=\frac{6\sqrt{2}}{6}=\sqrt{2}\)

\(\hept{\begin{cases}\left(x+y\right)\left(x+y+z\right)=45\left(1\right)\\\left(y+z\right)\left(x+y+z\right)=63\left(2\right)\\\left(z+x\right)\left(x+y+z\right)=54\left(3\right)\end{cases}}\)

lấy (1)+(2)+(3) 

\(\left(x+y+z\right)\left(x+y+y+z+z+x\right)=162\)

\(\left(x+y+z\right)\left(2x+2y+2z\right)=162\)

\(\left(x+y+z\right)^2=81\)

\(x+y+z=9\)

thế vào pt (1)(2)(3) ta đc

\(\hept{\begin{cases}9\left(x+y\right)=45\\9\left(y+z\right)=63\\9\left(z+x\right)=54\end{cases}\hept{\begin{cases}x+y=5\\y+z=7\\z+x=6\end{cases}\hept{\begin{cases}x=5-y\left(4\right)\\z=7-y\left(5\right)\\z+x=6\left(6\right)\end{cases}}}}\)

thế(4);(5) vào (6)

\(5-y+7-y=6\)

\(y=3\)

\(x=5-3=2\)

\(z=7-3=4\)

thử lại lấy x+y+z \(3+2+4=9\left(TM\right)\)

góp ý nhỏ cho bài của bạn Như Quỳnh

Đoạn \(\left(x+y+z\right)^2=81\) ta phải có hai trường hợp là 

-9 và 9, trong cả hai trường hợp ta đều giải ra nghiệm thỏa mãn.

b. rõ ràng (0,0,0) là nghiệm của hệ

Xét x khác 0 dễ dàng ta chỉ ra được y và z khác 0

khi đó hệ \(\Leftrightarrow\hept{\begin{cases}6\left(\frac{1}{x}+\frac{1}{y}\right)=5\\12\left(\frac{1}{y}+\frac{1}{z}\right)=7\\4\left(\frac{1}{x}+\frac{1}{z}\right)=3\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=\frac{1}{2}\\\frac{1}{y}=\frac{1}{3}\\\frac{1}{z}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\\z=4\end{cases}}\) vậy hệ có hia nghiệm (0,0,0) và (2,3,4)

\(c.\frac{2x^2}{1+x^2}.\frac{2y^2}{1+y^2}.\frac{2z^2}{1+z^2}=xyz\Leftrightarrow\orbr{\begin{cases}xyz=0\\\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}=1\end{cases}}\)

\(xyz=0\Rightarrow x=y=z=0\) còn \(\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}=1\)

ta có \(1+x^2\ge2x\Leftrightarrow\frac{2x}{1+x^2}\le1\)

tương tự ta có : \(\frac{2x}{1+x^2}.\frac{2y}{1+y^2}.\frac{2z}{1+z^2}\le1\)

Dấu = xảy ra khi x=y=z =1 

Vậy hệ có hai nghiệm (0,0,0) và (1,1,1)

a. ta có : \(P=\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)\(=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

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

b.\(x=\frac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{3}-1\)

Vậy \(P=\frac{\left(\sqrt{3}-1+1\right)^2}{\sqrt{3}-1}=\frac{3}{\sqrt{3}-1}=\frac{3\left(\sqrt{3}+1\right)}{2}\)

c. ta có \(P=\frac{\left(1+\sqrt{x}\right)^2}{\sqrt{x}}\ge\frac{4\sqrt{x}}{\sqrt{x}}=4>2\)

d.\(P\sqrt{x}=\left(1+\sqrt{x}\right)^2=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow x-4\sqrt{x}+4+\sqrt{x-4}=0\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x}-2=0\\\sqrt{x-4}=0\end{cases}\Leftrightarrow x=4}\)

\(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}+\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)

\(=\frac{\sqrt{3}+\sqrt{3}\sqrt{\sqrt{3}+1}+\sqrt{3}-\sqrt{3}\sqrt{\sqrt{3}+1}}{1-\left(\sqrt{3}+1\right)}\)

\(=\frac{2\sqrt{3}}{1-\sqrt{3}-1}=\frac{2\sqrt{3}}{-\sqrt{3}}=-2\)

Độ dài cạnh huyền là: \(5.2=10\left(cm\right)\)

Gọi độ dài hai cạnh góc vuông lần lượt là \(a,b\left(cm\right);a,b>0\).

Ta có: \(ab=4.10=40\)

\(a^2+b^2=10^2=100\Rightarrow\left(a+b\right)^2=100+2ab=180\Rightarrow a+b=6\sqrt{5}\)

Do đó \(a,b\)là hai nghiệm của phương trình \(X^2-6\sqrt{5}X+40=0\)

\(\Leftrightarrow\left(X-4\sqrt{5}\right)\left(X-2\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}X=4\sqrt{5}\\X=2\sqrt{5}\end{cases}}\)

Do đó độ dài hai cạnh góc vuông lần lượt là \(4\sqrt{5}\left(cm\right),2\sqrt{5}\left(cm\right)\).

\(\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{2^2+4\sqrt{2}+\sqrt{2}^2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{2^2-4\sqrt{2}+\sqrt{2}^2}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)

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

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

\(=\sqrt{2}+1+\sqrt{2}-1\)

\(=2\sqrt{2}\)