a. ĐK \(x\ge0\)và \(x\ne1\)

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

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

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

\(=\frac{x+1}{4\sqrt{x}}\)

b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)

c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)

Vậy A >1/2

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

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

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

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

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

\(A=\frac{3x+3}{-2\sqrt{x}.\left(\sqrt{x}+1\right)}\)
P/s: hình như đề sai hay sao á, thường thì người ta không cho mẫu là 2 số trừ được như ( x - 3x ) đâu

\(DK:x\ge0\)

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

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

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

\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)

\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)

\(\Leftrightarrow x=1\)

Vay nghiem cua PT la \(x=1\)

Ta có

\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)

\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)

Thế vào ta được

\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)

\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)

1.

đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)

có \(a^2+b^2=4\)

pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)

\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)

\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)

vì a,b>o nên \(a-b=\sqrt{2}\)

\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)

Bình phương 2 vế:

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

\(\Leftrightarrow\sqrt{4-x}=1\)

\(\Rightarrow x=3\)

Nếu đúng thì tích giùm mình cái nha!!!!!!!!!!!

2.ĐKXĐ D=R
Đặt \(a=\sqrt[3]{7-x},b=\sqrt[3]{x-5}\)
ta có: \(\hept{\begin{cases}a^3+b^3=2\\a^3-b^3=12-2x=2\left(6-x\right)\end{cases}}\)
Vậy ta có:

\(\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\Leftrightarrow\left(a-b\right)\left(2-\left(a+b\right)\left(a^2+ab+b^2\right)\right)=0\)
Th1: \(a-b=0\Leftrightarrow\sqrt[3]{7-x}=\sqrt[3]{x-5}\Leftrightarrow x=6\)
Th2: \(\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\a^3+b^3=12\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)\left(a^2+ab+b^2\right)=2\\\left(a+b\right)\left(a^2-ab+b^2\right)12\end{cases}}\)
Từ đó suy ra: 

\(\frac{a^2-ab+b^2}{a^2+ab+b^2}=6\Leftrightarrow5a^2-7ab+6b^2=0\)
nếu \(b=0\Leftrightarrow\sqrt[3]{x-5}=0\Leftrightarrow x=5\)thay vào phương trình ta thấy không thỏa mãn.
nếu \(b\ne0\Rightarrow5a^2-7ab+5b^2=0\Leftrightarrow5\left(\frac{a}{b}\right)^2-7\frac{a}{b}+5=0\)(1)
phương trình (1) vô nghiệm với ẩn \(\frac{a}{b}\). nên trường hợp này không xảy ra.
vậy phương trình có duy nhất nghiệm x = 6.

ĐK: \(x^4-4x^3+14x-11\ge0\) (*)

\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)

e/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x+3-\sqrt{x-1}=4\)

\(\Leftrightarrow\sqrt{x-1}=x-1\)

\(\Leftrightarrow x-1=x^2-2x+1\)

\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)

a/ ĐKXĐ: ...

\(\Leftrightarrow9x+3\sqrt{x^2-x-1}=7x+7\)

\(\Leftrightarrow3\sqrt{x^2-x-1}=7-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\9\left(x^2-x-1\right)=\left(7-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{7}{2}\\5x^2+19x-58=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\frac{29}{5}\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ne1\)

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

\(\Leftrightarrow\frac{1}{\left|x-1\right|}=\frac{1}{x-1}\)

\(\Rightarrow x-1>0\Rightarrow x>1\)