a,A.√2= √(4+2√3)-√(4-2√3)

= √(1+√3)² -√( √3 -1)²

= 1+√3-√3+1= 2 

=> A= 2/√2=√2

B²= (4+√15)².(4-√15).(√10-√6)²

= (4+√15).1.(16-4√15)

= (4+√15).(4-√15).4

= 4

=> B = √4 = 2

a) \(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(A^2=\left(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right)^2\)

\(A^2=3+\sqrt{5}+3-\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(A^2=6+2\sqrt{3^2-5}\)

\(A^2=6+4\)

\(A^2=10\)

\(\Rightarrow\orbr{\begin{cases}A=10\\A=-10\end{cases}}\)

Mà \(A>0\Rightarrow A=10\)

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

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

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

\(B^2=8-2\sqrt{4^2-7}\)

\(B^2=8-6\)

\(B^2=2\)

\(\Rightarrow\orbr{\begin{cases}B=2\\B=-2\end{cases}}\)

Mà \(B< 0\Rightarrow B=-2\)

Cách khác :

b) \(4-\sqrt{7}=\frac{8-2\sqrt{7}}{2}=\frac{7-2\sqrt{7}+1}{2}=\left(\frac{\sqrt{7}-1}{\sqrt{2}}\right)^2\)

\(4+\sqrt{7}=\frac{8+2\sqrt{7}}{2}=\frac{7+2\sqrt{7}+1}{2}=\left(\frac{\sqrt{7}+1}{\sqrt{2}}\right)^2\)

do đó : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\left(\frac{\sqrt{7}-1}{\sqrt{2}}\right)^2}-\sqrt{\left(\frac{\sqrt{7}+1}{\sqrt{2}}\right)^2}=\frac{\sqrt{7}-1}{\sqrt{2}}-\frac{\sqrt{7}+1}{\sqrt{2}}=-\sqrt{2}\)

tương tự câu a.

ĐK: \(x\ge1\)

Bình phương 2 vế ta có:

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

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

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

\(\Leftrightarrow0=0\)(đúng) 

Vậy x>=1

dk :x>=1

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

xet 2 th => tim x

a) ĐKXĐ : \(\orbr{\begin{cases}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{cases}}\)

\(\sqrt{x^2-3}=x^2-3\)

\(\Leftrightarrow\sqrt{x^2-3}=\sqrt{x^2-3}\cdot\sqrt{x^2-3}\)

\(\Leftrightarrow\sqrt{x^2-3}-\sqrt{x^2-3}\cdot\sqrt{x^2-3}=0\)

\(\Leftrightarrow\sqrt{x^2-3}\left(1-\sqrt{x^2-3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2-3}=0\\\sqrt{x^2-3}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-3=0\\x^2-3=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\in\left\{\pm\sqrt{3}\right\}\\x\in\left\{\pm2\right\}\end{cases}}\)( thỏa mãn )

b) ĐKXĐ : \(x\le6\)

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

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)

\(\Leftrightarrow\left|x-3\right|=6-x\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=6-x\\x-3=x-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=9\\0x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x\in\varnothing\end{cases}}\)( thỏa mãn )

a)\(ĐKXĐ:x\ge\frac{-1}{2}\)

 \(\sqrt{x^2+4x+4}=2x+1\)

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

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

\(\Leftrightarrow-x=-1\)

\(\Leftrightarrow x=1\)

Vậy nghiệm duy nhất của phương trình là 1.

b)\(ĐKXĐ:x\ge3\)

 \(\sqrt{4x^2-12x+9}=x-3\)

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

\(\Leftrightarrow2x-3=x-3\)

\(\Leftrightarrow2x=x\)

\(\Leftrightarrow x=0\)(không t/m đkxđ)

Vậy phương trình vô nghiệm

\(a,\sqrt{2x-1}=x-2\Leftrightarrow2x-1=\left(x-2\right)^2\)

ĐK \(x\ge2\)

\(\Leftrightarrow2x-1=x^2-4x+4\)

\(\Leftrightarrow x^2-6x+5=0\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}}\)

Vì x=1 (KTM)=> x=5 thì t/m đề bài

\(C=\sqrt{2x^2-6x+5}=\sqrt{2\left(x^2-3x+\frac{9}{4}\right)+\frac{1}{2}}\)

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

Vậy GTNN của C là \(\frac{1}{2}\) \(\Leftrightarrow x=\frac{3}{2}\)

mình nhầm. thay GTNN \(\frac{1}{2}\)thành \(\sqrt{\frac{1}{2}}\)

\(B=\sqrt{2x^2-4x+10}=\sqrt{2\left(x^2-2x+1\right)+8}\)

\(B=\sqrt{2\left(x-1\right)^2+8}\ge8\)

Vậy GTNN của B là 8 \(\Leftrightarrow x=1\)

\(B=\sqrt{2x^2-4x+10}=\sqrt{\left(2x^2-4x+2\right)+8}=\sqrt{2\left(x^2-2x+1\right)+8}=\sqrt{2\left(x-1\right)^2+8}\)

Ta có \(2\left(x-1\right)^2\ge0\)

để \(2\left(x-1\right)^2\)nhỏ nhất thì \(x=1\)

Vậy tại \(x=1\)thì \(GTNN_B=\sqrt{2\left(1-1\right)^2+8}=\sqrt{0+8}=\sqrt{8}\)

\(a,=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)

\(\sqrt{29^2-20^2}=\sqrt{\left(29-20\right)\left(29+20\right)}=\sqrt{3^2.7^2}=21\)

\(\text{Đặt: }\)\(\hept{\begin{cases}\sqrt{4-\sqrt{15}}=a\\\sqrt{4+\sqrt{15}}=b\end{cases}}\)\(\text{cần tính: a-b}\)

\(\hept{\begin{cases}ab=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=1\\a^2+b^2=8\end{cases}}\Rightarrow\left(a-b\right)^2=6\Rightarrow a-b=-\sqrt{6}\left(vì:a< b\right)\)

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

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

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

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

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