a) \(\sqrt{\frac{1}{20}}=\frac{\sqrt{1}}{\sqrt{20}}=\frac{1}{\sqrt{2^2\cdot5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\sqrt{5}\cdot\sqrt{5}}=\frac{\sqrt{5}}{10}\)

b) \(\sqrt{\frac{1}{60}}=\frac{\sqrt{1}}{\sqrt{60}}=\frac{1}{\sqrt{2^2\cdot15}}=\frac{1}{2\sqrt{15}}=\frac{\sqrt{15}}{2\sqrt{15}\cdot\sqrt{15}}=\frac{\sqrt{15}}{30}\)

b) \(\sqrt{\frac{3}{98}}=\frac{\sqrt{3}}{\sqrt{98}}=\frac{\sqrt{3}}{\sqrt{7^2\cdot2}}=\frac{\sqrt{3}}{7\sqrt{2}}=\frac{\sqrt{3}\cdot\sqrt{2}}{7\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{6}}{14}\)

Chờ từ trưa không idol nào đụng thì thôi em xin vậy :))

BT1:

Ta có: \(A\cdot B=\sqrt{4+\sqrt{10+2\sqrt{5}}}\cdot\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(=\sqrt{16-10-2\sqrt{5}}\)

\(=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

Từ đó thay vào: \(\left(A-B\right)^2\)

\(=A^2-2AB+B^2\)

\(=4+\sqrt{10+2\sqrt{5}}-2\left(\sqrt{5}-1\right)+4-\sqrt{10+2\sqrt{5}}\)

\(=10-2\sqrt{5}\)

\(\Rightarrow A-B=\sqrt{10-2\sqrt{5}}\)

BT2:

Đặt \(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(\Leftrightarrow B^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)

\(=8-2\sqrt{16-7}=8-2\cdot3=2\)

\(\Rightarrow B=\sqrt{2}\)

\(\Rightarrow A=B-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)

BT3:

đk: \(\orbr{\begin{cases}x\ge2\\x< -2\end{cases}}\)

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

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

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

\(C=\frac{2x^2+8x+8+2x^2-8}{4x+8}\)

\(C=\frac{4x^2+8x}{4x+8}=x\)

Vậy C = x

ta có vế trái không bao giờ chia hết cho 3 nên vô nghiệm

x nguyên dương nên x có 3 dạng 3k; 3k + 1; 3k + 2

*) Xét x = 3k thì \(x^2+x-1=\left(3k\right)^2+3k-1\) không chia hết cho 3

*) Xét x = 3k + 1 thì \(x^2+x-1=\left(3k+1\right)^2+3k+1-1=BS3+1\)không chia hết cho 3

*) Xét x = 3k + 2 thì \(x^2+x-1=\left(3k+2\right)^2+3k+2-1=BS3+2\)không chia hết cho 3

Vậy \(x^2+x-1\)luôn không chia hết cho 3 (1)

Xét vế phải: vì y nguyên dương nên 2y + 1 > 0 suy ra \(3^{2y+1}⋮3\)(2)

Từ (1) và (2) suy ra phương trình vô nghiệm

a) Để hàm xác định thì \(\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

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

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

và \(f\left(a^2\right)=\frac{\sqrt{a^2}+1}{\sqrt{a^2}-1}=\frac{\left|a\right|+1}{\left|a\right|-1}\)(với \(a\ne\pm1\))

* Nếu \(a\ge0;a\ne1\)thì \(f\left(a^2\right)=\frac{a+1}{a-1}\)

* Nếu \(a< 0;a\ne-1\)thì \(f\left(a^2\right)=\frac{a-1}{a+1}\)

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

Để f(x) nguyên thì \(\frac{2}{\sqrt{x}-1}\)nguyên hay \(2⋮\sqrt{x}-1\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Mà \(\sqrt{x}-1\ge-1\)nên ta xét ba trường hợp:

+) \(\sqrt{x}-1=-1\Rightarrow x=0\left(tmđk\right)\)

+) \(\sqrt{x}-1=1\Rightarrow x=4\left(tmđk\right)\)

+) \(\sqrt{x}-1=2\Rightarrow x=9\left(tmđk\right)\)

Vậy \(x\in\left\{0;4;9\right\}\)thì f(x) có giá trị nguyên 

d) \(f\left(x\right)=\frac{\sqrt{x}+1}{\sqrt{x}-1}\); \(f\left(2x\right)=\frac{\sqrt{2x}+1}{\sqrt{2x}-1}\)

f(x) = f(2x) khi \(\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{2x}+1}{\sqrt{2x}-1}\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{2x}+1\right)\)\(\Leftrightarrow\sqrt{2}x+\sqrt{2x}-\sqrt{x}-1=\sqrt{2}x-\sqrt{2x}+\sqrt{x}-1\)\(\Leftrightarrow\sqrt{2x}-\sqrt{x}=-\sqrt{2x}+\sqrt{x}\Leftrightarrow2\sqrt{2x}=2\sqrt{x}\Leftrightarrow\sqrt{2x}=\sqrt{x}\Leftrightarrow x=0\)(tmđk)

Vậy x = 0 thì f(x) = f(2x)

Ta có: \(x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}\)

\(=\left(x^4+2x^2y^2+y^4\right)-2x^2y^2+\frac{x^4y^4}{\left(x^2+y^2\right)}\)

\(=\left(x^2+y^2\right)^2-2x^2y^2+\left(\frac{x^2y^2}{x^2+y^2}\right)^2\)

\(=\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2\)

Thay vào ta tính được:

\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\left(x^2+y^2-\frac{x^2y^2}{x^2+y^2}\right)^2}}\)

Mà \(x^2+y^2-\frac{x^2y^2}{x^2+y^2}=\frac{\left(x^2+y^2\right)^2-x^2y^2}{x^2+y^2}=\frac{x^4+x^2y^2+y^4}{x^2+y^2}>0\left(\forall x,y\right)\)

Khi đó:

\(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2-\frac{x^2y^2}{x^2+y^2}}\)

\(P=\sqrt{x^2+y^2+\frac{x^2y^2}{\left(x+y\right)^2}}\)

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

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

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

\(P=\left|x+y-\frac{xy}{x+y}\right|=\left|\frac{x^2+xy+y^2}{x+y}\right|=\frac{x^2+xy+y^2}{x+y}\)

Vậy \(P=\frac{x^2+xy+y^2}{x+y}\)

sử dụng \((t+1/t)^2 = t^2 + 1/t^2 +2\)

\(\sqrt{x +4} = \frac{-2x}{5+x} -2 = \frac{-4x+5}{5+x}\\ \)

Bình phương cả 2 vế sẽ cho phương trình bậc 3....

\(ĐKXĐ:x\ne0\)

\(A=\sqrt{\frac{\left(x^2-3\right)^2}{x^2}+12}+\sqrt{\left(x+2\right)^2-8x}\)

\(=\sqrt{\frac{x^4-6x^2+9}{x^2}+\frac{12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)

\(=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2-4x+4}\)

\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{\left(x-2\right)^2}\)

\(=\sqrt{\frac{\left(x^2+9\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)

\(=\frac{x^2+9}{\left|x\right|}+\left|x-2\right|=\frac{x^2}{\left|x\right|}+\frac{9}{\left|x\right|}+\left|x-2\right|\)

Vì \(x\inℤ\)\(\Rightarrow\frac{x^2}{\left|x\right|}\inℕ\), \(\left|x-2\right|\inℕ^∗\)

\(\Rightarrow\)Để A có giá trị nguyên thì \(\frac{9}{\left|x\right|}\inℕ^∗\)

\(\Rightarrow\left|x\right|\in\left\{1;3;9\right\}\)\(\Rightarrow x\in\left\{\pm1;\pm3;\pm9\right\}\)

Vậy \(x\in\left\{\pm1;\pm3;\pm9\right\}\)

Bình phương 2 số đó rồi so sánh