Thay x=4 vào \(y=f\left(x\right)=\sqrt{x}\), ta được

\(f\left(4\right)=\sqrt{4}=2\)

=>A(4;2) thuộc đồ thị hàm số \(y=f\left(x\right)=\sqrt{x}\)

Thay \(x=2\) vào \(y=f\left(x\right)=\sqrt{x}\), ta được;

\(f\left(2\right)=\sqrt{2}>1\)

=>B(2;1) không thuộc đồ thị hàm số \(y=f\left(x\right)=\sqrt{x}\)

Thay \(x=8\) vào \(y=\sqrt{x}\), ta được:

\(y=\sqrt{8}=2\sqrt{2}\)

=>\(C\left(8;2\sqrt{2}\right)\) thuộc đồ thị hàm số \(y=\sqrt{x}\)

Thay \(x=4-2\sqrt{3}\) vào \(y=\sqrt{x}\), ta được:

\(y=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1< >1-\sqrt{3}\)

=>\(D\left(4-2\sqrt{3};1-\sqrt{3}\right)\) không thuộc đồ thị hàm số \(y=f\left(x\right)=\sqrt{x}\)

Thay \(x=6+2\sqrt{5}\) vào \(y=f\left(x\right)=\sqrt{x}\), ta được:

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

\(=\left|\sqrt{5}+1\right|=\sqrt{5}+1\)

vậy: \(E\left(6+2\sqrt{5};1+\sqrt{5}\right)\) thuộc đồ thị hàm số \(y=f\left(x\right)=\sqrt{x}\)

a: Để A<0 thì 2*căn x-4<0

=>căn x<2

=>0<=x<4

=>\(x\in\left\{0;1;2;3\right\}\)

b: \(A-2=\dfrac{2\sqrt{x}-4-2\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+1}< 0\)

=>A<2

c: A<1

=>A-1<0

=>\(\dfrac{2\sqrt{x}-4-\sqrt{x}-1}{\sqrt{x}+1}< 0\)

=>căn x-5<0

=>0<=x<25

d: A>-1

=>A+1>0

=>\(\dfrac{2\sqrt{x}-4+\sqrt{x}+1}{\sqrt{x}+1}>0\)

=>3*căn x-3>0

=>x>1

e: A<=(-x+6căn x-8)/(căn x+1)

=>2*căn x-4<=-x+6căn x-8

=>x-4căn x+4<=0

=>x=4

a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)

\(A=5\sqrt{6}-10\)

\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)

\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)

Đến đây ko rút gọn được nữa, nhưng nếu đề là:

\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)

c/

\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)

\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)

\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)

\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))

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

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

\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)

\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)

\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)

\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)

\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)

\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)

\(\Rightarrow G=2\)

a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)

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

b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)

\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)

Th1: x-3 < 0

\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)

Th2: x-3 > 0

\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)

c)

Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)

\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)

\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)

Th1: \(x-2\ge0\Leftrightarrow x\ge2\)

\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)

Th2: \(x-2\le0\Leftrightarrow x\le2\)

kết hợp với đk, ta được: 1 \< x \< 2

\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)

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

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Bài 2:

\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}}\)

\(=\sqrt{8-4\sqrt{3}}\cdot\sqrt{\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{6-2}}\)

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

\(=\dfrac{6-2}{2}=\dfrac{4}{2}=2\)

a: TXĐ: \(D=R\backslash\left\{-\dfrac{1}{2}\right\}\)

b: TXĐ: \(D=R\backslash\left\{-3;1\right\}\)

c: TXĐ: \(D=\left[-\dfrac{1}{2};3\right]\)