a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)

\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)

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

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

1/ \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

\(x=\left(1+\frac{\sqrt{10}\left(\sqrt{10}+1\right)}{1+\sqrt{10}}\right)\left(\frac{\sqrt{10}\left(\sqrt{10}-1\right)}{\sqrt{10}-1}-1\right)\)

\(x=\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)\)

\(x=10-1=9\)

Thay \(x=9\) vào A:

\(A=\frac{2\sqrt{9}+1}{9+\sqrt{9}}=\frac{7}{12}\)

Vậy với \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\Leftrightarrow A=\frac{7}{12}\)

2/ \(B=\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)

\(\Leftrightarrow B=\frac{9x-1-2\sqrt{x}\left(3\sqrt{x}-1\right)+\sqrt{x}+1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\cdot\frac{3\sqrt{x}+1}{3}\)

\(\Leftrightarrow B=\frac{9x-1-6x+2\sqrt{x}+\sqrt{x}+1}{3\left(3\sqrt{x}-1\right)}\)

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

3/ \(P=A.B=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\cdot\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{2\sqrt{x}+1}{3\sqrt{x}-1}\)

Để \(P\in Z\Leftrightarrow2\sqrt{x}+1⋮3\sqrt{x}-1\)

\(\Leftrightarrow6\sqrt{x}+2⋮3\sqrt{x}-1\)

\(\Leftrightarrow2\left(3\sqrt{x}-1\right)+4⋮3\sqrt{x}-1\)

\(\Leftrightarrow4⋮3\sqrt{x}-1\)

\(\Leftrightarrow3\sqrt{x}-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow3\sqrt{x}\in\left\{0;2;-1;3;-3;5\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;\frac{2}{3};-\frac{1}{3};1;-1;\frac{5}{3}\right\}\)

\(\Leftrightarrow x\in\left\{0;\frac{4}{9};\frac{1}{9};1;\frac{25}{9}\right\}\)

Loại bỏ những giá trị x < 0 , x \(x\notin Z\)và x không thỏa mãn ĐKXĐ

Vậy để \(P\in Z\Leftrightarrow x\in\left\{1\right\}\)

1: Ta có: \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

\(=\left(\frac{1+\sqrt{10}+10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}-\left(\sqrt{10}-1\right)}{\sqrt{10}-1}\right)\)

\(=\frac{1+2\sqrt{10}\cdot1+\left(\sqrt{10}\right)^2}{1+\sqrt{10}}\cdot\frac{\left(\sqrt{10}\right)^2-2\cdot\sqrt{10}\cdot1+1}{\sqrt{10}-1}\)

\(=\left(1+\sqrt{10}\right)\cdot\left(\sqrt{10}-1\right)\)

\(=10-1=9\)

Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\), ta được:

\(A=\frac{2\cdot\sqrt{9}+1}{9+\sqrt{9}}=\frac{2\cdot3+1}{9+3}=\frac{7}{12}\)

Vậy: \(\frac{7}{12}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) tại \(x=\left(1+\frac{10+\sqrt{10}}{1+\sqrt{10}}\right)\cdot\left(\frac{10-\sqrt{10}}{\sqrt{10}-1}-1\right)\)

2: Ta có: \(B=\left(1-\frac{2\sqrt{x}}{3\sqrt{x}+1}+\frac{\sqrt{x}+1}{9x-1}\right):\frac{3}{3\sqrt{x}+1}\)

\(=\left(\frac{9x-1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}-\frac{2\sqrt{x}\left(3\sqrt{x}-1\right)}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right)\cdot\frac{3\sqrt{x}+1}{3}\)

\(=\frac{9x-1-6x+2\sqrt{x}+\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\cdot\frac{3\sqrt{x}+1}{3}\)

\(=\frac{3x+3\sqrt{x}+2}{9\sqrt{x}-3}\)

1/Em không chắc nha, nhất là câu c ý, nó sai sai hay là em làm sai nhỉ?

a) ĐK \(x\ge0\). Bình phương hai vế:

\(x+5=x+2\sqrt{x}+1\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\) (TMĐK)

b)ĐK \(0\le x\le1\) . Bình phương hai vế:

\(2\sqrt{x\left(1-x\right)}=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(TMĐK\right)\)

c) ĐK: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\Leftrightarrow5\le x\le3\) (vô lí))

Vậy không tồn tại x thỏa mãn đề bài.

ĐKXĐ: \(x\ge-1\)

\(\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-6\sqrt{x+1}+9}=2\sqrt{x+1-2\sqrt{x+1}+1}\)

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

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

Ta có:

\(\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|\ge\left|\sqrt{x+1}+1+\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\)

Dấu "=" xảy ra khi và chỉ khi:

\(\sqrt{x+1}-3\ge0\Rightarrow x\ge8\)

Vậy nghiệm của pt là \(x\ge8\)

ĐKXĐ: \(x\ge-1\)

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

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

Áp dụng BĐT trị tuyệt đối:

\(\left|\sqrt{x+1}+1\right|+\left|\sqrt{x+1}-3\right|\ge\left|\sqrt{x+1}+1+\sqrt{x+1}-3\right|=\left|2\sqrt{x+1}-2\right|\)

Dấu "=" xảy ra khi và chỉ khi \(\left(\sqrt{x+1}+1\right)\left(\sqrt{x+1}-3\right)\ge0\)

\(\Leftrightarrow\sqrt{x+1}-3\ge0\)

\(\Leftrightarrow x+1\ge9\)

\(\Leftrightarrow x\ge8\)

1.

Xét riêng 2 căn lớn đầu tiên

Bình phương, thu gọn được căn(12-8 căn 2)

Giờ kết hợp kết quả này với căn lớn còn lại

Tiếp tục bình phương, thu gọn là xong