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

\(=1-\sqrt{3}-\sqrt{3}-2\)

\(=-2\sqrt{3}-1\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)^2}\)

\(=2-\sqrt{3}+4-2\sqrt{3}\)

\(=6-3\sqrt{3}\)

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

\(A=\sqrt{3}-1-\sqrt{3}-2\)

\(A=-3\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(4-2\sqrt{3}\right)}\)

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

\(B=1\)

Ta có: \(B=21\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)

\(=21\cdot\left[2+\sqrt{3}+3-\sqrt{5}+2\sqrt{\left(2+\sqrt{3}\right)\left(3-\sqrt{5}\right)}\right]-6\cdot\left[2-\sqrt{3}+3+\sqrt{5}+2\cdot\sqrt{\left(2-\sqrt{3}\right)\left(3+\sqrt{5}\right)}\right]-15\sqrt{15}\)

\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{\left(4+2\sqrt{3}\right)\left(6-2\sqrt{5}\right)}\right)-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\sqrt{\left(4-2\sqrt{3}\right)\left(6+2\sqrt{5}\right)}\right]-15\sqrt{15}\)

\(=21\cdot\left[5+\sqrt{3}-\sqrt{5}+\left(\sqrt{3}+1\right)\left(\sqrt{5}-1\right)\right]-6\cdot\left[5-\sqrt{3}+\sqrt{5}+\left(\sqrt{3}-1\right)\left(\sqrt{5}+1\right)\right]-15\sqrt{15}\)

\(=21\cdot\left(5+\sqrt{3}-\sqrt{5}+\sqrt{15}-\sqrt{3}+\sqrt{5}-1\right)-6\cdot\left(5-\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt{3}-\sqrt{5}-1\right)-15\sqrt{15}\)

\(=21\cdot\left(4+\sqrt{15}\right)-6\left(4+\sqrt{15}\right)-15\sqrt{15}\)

\(=84+21\sqrt{15}-24-6\sqrt{15}-15\sqrt{15}\)

\(=60\)

Giúp e câu a nữa ạ

a: \(\sqrt{6-4\sqrt2}+\sqrt{22-12\sqrt2}\)

\(=\sqrt{4-2\cdot2\cdot\sqrt2+2}+\sqrt{18-2\cdot3\sqrt2\cdot2+4}\)

\(=\sqrt{\left(2-\sqrt2\right)^2}+\sqrt{\left(3\sqrt2-2\right)^2}\)

\(=2-\sqrt2+3\sqrt2-2=2\sqrt2\)

b: \(\sqrt{\left(\sqrt3-\sqrt2\right)^2}+\sqrt2=\sqrt3-\sqrt2+\sqrt2=\sqrt3\)

c: \(3\sqrt5-\sqrt{\left(1-\sqrt5\right)^2}\)

\(=3\sqrt5-\left|1-\sqrt5\right|\)

\(=3\sqrt5-\left(\sqrt5-1\right)=2\sqrt5+1\)

d:Sửa đề: \(\sqrt{17-12\sqrt2}+\sqrt{6+4\sqrt2}\)

\(=\sqrt{9-2\cdot3\cdot2\sqrt2+8}+\sqrt{4+2\cdot2\cdot\sqrt2+2}\)

\(=\sqrt{\left(3-2\sqrt2\right)^2}+\sqrt{\left(2+\sqrt2\right)^2}=3-2\sqrt2+2+\sqrt2=5-\sqrt2\)

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

\(\Rightarrow\)\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

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

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

                       \(=2\sqrt{3}\)

\(\Rightarrow\)\(A=\sqrt{6}\)   (đpcm)

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

\(VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

\(=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\)

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

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

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

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

\(=\frac{2\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{12}}{\sqrt{2}}=\sqrt{6}\)

\(=VP\)

Vậy đẳng thức được chứng minh .

1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)

\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)

\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)

\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)

\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)

\(=\frac{-29\sqrt{2}}{2}\)

2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)

\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

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

\(=10-4\sqrt{3}+8+4\sqrt{3}\)

\(=18\)

3) ĐK để phương trình có nghiệm là: x≥0

Ta có: \(\sqrt{x^2-6x+9}=2x\)

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

\(\Leftrightarrow\left|x-3\right|=2x\)

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

Vậy: S={1}

4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{4x^2+1}=2x-1\)

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

\(\Leftrightarrow4x^2+1=4x^2-4x+1\)

\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)

\(\Leftrightarrow4x=0\)

hay x=0(loại)

Vậy: S=∅

\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)

5.

\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)

Đặt \(\sqrt{x^2+7}=t>0\)

\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)

\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)

Câu 6 bạn coi lại đề

4.

ĐKXĐ: ...

Đặt \(\sqrt{x+3}=a\ge0\)

\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)

\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)

\(\Rightarrow2x^2-ax-a^2=0\)

\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)

\(a,\sqrt{x-2\sqrt{x}-1}-\sqrt{x-1}=1.\)

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

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

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

\(\Rightarrow x-1=x^2\Rightarrow x^2-x+1=0\) ( vô nghiệm vì nó luôn lớn hơn 0 )

\(đkxđ\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\)

\(c,\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}.\)

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

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

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

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

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

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

\(\Rightarrow2x-1=1\Rightarrow2x=2\Leftrightarrow x=1\)\(\left(tm\right)\)

d tương tự nha , nhân thêm 2 vế với \(\sqrt{6}\)là ra

1)

ĐK: \(x\geq 5\)

PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)

\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)

2)

ĐK: \(x\geq -1\)

\(\sqrt{x+1}+\sqrt{x+6}=5\)

\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)

\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)

\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)

Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$

\(\Rightarrow x=3\) (thỏa mãn)

Vậy .............

Làm hơi tắt xíu, có gì ko hiểu cmt nha :>

\(a.\sqrt{x-1}=3\left(ĐK:x\ge1\right)\Leftrightarrow x-1=9\Leftrightarrow x=10\)

\(b.\sqrt{x^2-4x+4}=2\\ \Leftrightarrow\sqrt{\left(x-2\right)^2}=2\\ \Leftrightarrow\left|x-2\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-2=2\left(x\ge2\right)\\2-x=2\left(x< 2\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

\(c.\sqrt{25x^2-10x+1}=4x-9\\ \Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x-9\\ \Leftrightarrow\left|5x-1\right|=4x-9\\\Leftrightarrow \left[{}\begin{matrix}5x-1=4x-9\left(x\ge\frac{1}{5}\right)\\1-5x=4x-9\left(x< \frac{1}{5}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-8\left(ktm\right)\\x=\frac{10}{9}\left(ktm\right)\end{matrix}\right.\)

\(d.\sqrt{x^2+2x+1}=\sqrt{x+1}\left(ĐK:x\ge-1\right)\\ \Leftrightarrow x^2+2x+1=x+1\\ \Leftrightarrow x^2+x=0\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

e. ĐK: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\\ \Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\\ \Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\)

Câu cuối chưa nghĩ ra, sorry :<