Áp dụng CT căn phức tạp : \(\sqrt{A\pm\sqrt{B}}=\sqrt{\frac{A+\sqrt{A^2-B}}{2}}\pm\sqrt{\frac{A-\sqrt{A^2-B}}{2}}\)

ĐKXĐ : \(-1\le x\le1\)

Áp dụng CT căn phức tạp , ta được : \(\sqrt{1+\sqrt{1-x^2}}=\sqrt{\frac{1+\sqrt{1-1+x^2}}{2}}+\sqrt{\frac{1-\sqrt{1-1+x^2}}{2}}\)

\(=\sqrt{\frac{1+\left|x\right|}{2}}+\sqrt{\frac{1-\left|x\right|}{2}}=\hept{\begin{cases}\frac{1}{\sqrt{2}}\left(\sqrt{1+x}+\sqrt{1-x}\right)\text{ nếu x }\ge0\\\frac{1}{\sqrt{2}}\left(\sqrt{1-x}+\sqrt{1+x}\right)\text{ nếu x }< 0\end{cases}}\)( kết quả như nhau )

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

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

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

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

Ket qua ra 3 nha, ban tinh tung ve roi tim hieu nhe

bạn có thể giải vế đầu giúp mình không

1) ĐK:x\(\ge\frac{1}{2}\)

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

\(\Leftrightarrow\begin{cases}x\ge0\\2x-1=x^2\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge0\\x=1\end{cases}\)

\(\Leftrightarrow x=1\)   (thỏa mãn)

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

\(A=\frac{18+10}{4}\)

\(A=7\)

\(B=\sqrt{9-3\times2\sqrt{3}+3}+\sqrt{12-2\times3\times2\sqrt{3}+9}\)

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

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

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

\(B=\sqrt{3}\)

Dat \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\)

Bien doi PT thanh \(a^2+4b^2=5ab\)

\(\Leftrightarrow a^2-5ab+4b^2=0\)

\(\Leftrightarrow\left(a^2-ab\right)-\left(4ab-4b^2\right)=0\)

\(\Leftrightarrow a\left(a-b\right)-4b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a=4b\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{65+x}=\sqrt[3]{65-x}\)

\(\Leftrightarrow65+x=65-x\)

\(\Leftrightarrow x=0\left(n\right)\)

\(\left(2\right)\Leftrightarrow\sqrt[3]{65+x}=4\sqrt[3]{65-x}\)

\(\Leftrightarrow65+x=64.65-64x\)

\(\Leftrightarrow65x=64.65-65\)

\(\Leftrightarrow x=63\left(n\right)\)

Vay nghiem cua PT la \(x=0,x=63\)

Bài 1 :

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

b, ĐKXĐ : \(-x^2+10x-25\ge0\)

=> \(x^2-10x+25\le0\)

=> \(\left(x-5\right)^2\le0\)

=> \(x-5\le0\)

=> \(x\le5\)

Bài 2 :

a, Ta có : \(A=\sqrt{\left(2\sqrt{2}-5\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}\)

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

b, Ta có : \(B=\sqrt{9+4\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

=> \(B=\sqrt{4+2.2\sqrt{5}+5}-\sqrt{1-2\sqrt{5}+5}\)

=> \(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)

=> \(B=2+\sqrt{5}-\sqrt{5}+1=3\)

c, Ta có : \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

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

=> \(C=\frac{1+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

\(\sqrt{4\left(a-3\right)^2}\)

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

\(=\sqrt{4a^2-24a+36}\)

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

\(=\left|2a-6\right|\)

\(=2a-6\)

Lời giải:

1)

Để biểu thức có nghĩa thì:

\(2x^2-5x+3\geq 0\)

\(\Leftrightarrow 2x(x-1)-3(x-1)\geq 0\)

\(\Leftrightarrow (2x-3)(x-1)\geq 0\)

\(\Leftrightarrow \left[\begin{matrix} x\geq \frac{3}{2}\\ x\leq 1\end{matrix}\right.\)

2)

\(\sqrt{6.5+\sqrt{12}}+\sqrt{6.5-\sqrt{12}}+2\sqrt{6}\)

\(=\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2+2\sqrt{6}.\frac{1}{\sqrt{2}}}+\sqrt{(\sqrt{6})^2+(\frac{1}{\sqrt{2}})^2-2\sqrt{6}.\frac{1}{\sqrt{2}}}+2\sqrt{6}\)

\(=\sqrt{(\sqrt{6}+\frac{1}{\sqrt{2}})^2}+\sqrt{(\sqrt{6}-\frac{1}{\sqrt{2}})^2}+2\sqrt{6}\)

\(=\sqrt{6}+\frac{1}{\sqrt{2}}+\sqrt{6}-\frac{1}{\sqrt{2}}+2\sqrt{6}=4\sqrt{6}\)