a/A\(=\frac{x+2}{x-\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+1}-\frac{1-\sqrt{x}}{\sqrt{x}-2}\)
\(=\frac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2-2x+4\sqrt{x}-1+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
Thay x=16 vào A ta có: A\(=\frac{3}{2}\)
b/ B= \(1-\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
\(\frac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}=\frac{1}{\sqrt{x}-2}\)
=>C=\(\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{1}{\sqrt{x}-2}\)=\(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\)
c/Để C thuộc Z thì \(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\) thuộc Z
C\(=\text{​​}\frac{4\sqrt{x}-1}{\sqrt{x}+1}=\frac{4\sqrt{x}+4}{\sqrt{x}+1}-\frac{5}{\sqrt{x}+1}=4-\frac{5}{\sqrt{x}+1}\)
=> \(5⋮\left(\sqrt{x}+1\right) \Leftrightarrow\sqrt{x}+1\in\left\{-5;-1;1;5\right\}\)
Nhận xét: \(\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x}+1\in\left\{1;5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;4\right\} \Leftrightarrow x\in\left\{0;16\right\}\)
Vậy \(x\in\left\{0;16\right\}\) thì C thuộc Z
Chúc bạn học tốt!

1b) Ta có : \(49x^2-81=0\)
\(\Leftrightarrow x^2=\frac{81}{49}\)
\(\Leftrightarrow x=\pm\frac{9}{7}\)

Bài 1a có thiếu đề ko?

1) Đặt \(x^2\)+2x=t
Ta có \(\left(x^2+2x-2\right)\left(x^2+2x+3\right)-6\)=(t-2)(t+3)-6=\(t^2+t-6-6\)\(=t^2+t-12\)\(\left(t-3\right)\left(t+4\right)\)\(=\left(x^2+2x-3\right)\left(x^2+2x+4\right)=\left(x+3\right)\left(x-1\right)\left(x^2+2x+4\right)\)

2) Đặt \(x^2-4x+7=t\)
Ta có : (\(\left(x^2-4x+6\right)\left(x^2-4x+8\right)\)\(-8\)\(=\left(t-1\right)\left(t+1\right)-8=t^2-1-8=t^2-9=\left(t-3\right)\left(t+3\right)\)\(=\left(x^2-4x+4\right)\left(x^2-4x+10\right)\)\(=\left(x-2\right)^2\left(x^2-4x+10\right)\)

Áp dụng bđt bu-nhi-a cho VT ta có:
\(\left(\sqrt{x^2+x-1}\right)^2+\left(\sqrt{-x^2+x+1}\right)^2\ge\frac{\left(\sqrt{x^2+x-1}+\sqrt{-x^2+x+1}\right)^2}{2}\)
\(\Leftrightarrow\)\(x^2+x-1-x^2+x+1\ge\frac{VT^2}{2}\)
=>VT^2\(\le\)4x
=>VT\(\le\)\(2\sqrt{x}\)\(\le\)x+1
Lại có:VP=x^2-x+2\(\ge\)x+1
Mà VT=VP => VT=VT=x+1
Dấu "=" xảy ra <=>x=1

b) Ta có : \(\sqrt{5}\)-\(\sqrt{3-\sqrt{29-6\sqrt{20}}}\)= \(\sqrt{5}-\sqrt{3-\sqrt{29-2\sqrt{180}}}\)=\(\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)^2}\)=\(\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)=\(\sqrt{5}-\sqrt{6-2\sqrt{5}}\)=\(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)=\(\sqrt{5}-\sqrt{5}+1\)=1

a) Ta có :\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\) =\(\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Tương tự : \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\) = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
=>\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)+\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)= \(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)=\(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{3-2}\)=10