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

Theo BĐT Cosi ta được 

\(P\ge\left|1-2x+2x+1\right|=2\)

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}2x-1\le0\\2x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{2}\)

khó quá rồi

a) Ta có : \(AB^2+AC^2=6^2+8^2=100\)

               \(BC^2=10^2=100\)

\(\Rightarrow AB^2+AC^2=BC^2\)

\(\Rightarrow\bigtriangleup ABC\) vuông tại \(A\)  (đpcm)

b) Từ \(AB\cdot AC=AH\cdot BC\)

\(\Rightarrow6\cdot8=AH\cdot10\)

\(\Rightarrow AH=4,8\)

c) Từ \(AB^2=BC\cdot BH\)

\(\Rightarrow6^2=10\cdot HB\)

\(\Rightarrow HB=3,6\)

Từ \(HB+HC=BC\)

\(\Rightarrow3,6+HC=10\)

\(\Rightarrow HC=6,4\)

\(S_{\bigtriangleup ABC}=\dfrac{1}{2}AB\cdot AC\)  .

Xét tam giác ABC vuông tại A, đường cao AH 

Áp dụng hệ thức \(AB^2=HB.BC=HB\left(HC+HB\right)=HB\left(16+HB\right)\Leftrightarrow225=16HB+HB^2\)

\(\Leftrightarrow HB^2+16BH-225=0\Leftrightarrow HB=9cm\)

BC = HC + HB = 9 + 16 = 25 cm

Áp dụng hệ thức \(AH^2=HB.HC=144\Leftrightarrow AH=12cm\)

.

\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+2}\)

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

\(\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{6}+\sqrt{8}+2}{\sqrt{2}+\sqrt{3}+2}\)

\(\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

`=`\(\sqrt{2}+1\)

\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

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

\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)

\(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(\sqrt{3+2})^2}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\)

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

\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)

\(=\sqrt{4+\sqrt{25}}\)

\(=\sqrt{4+5}=3\)

a, đk x >= 0 

\(\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)(tmđk) 

b, đk x >= 0 

\(\Leftrightarrow x< 2\)Kết hợp với đk vậy 0 =< x < 2 

c, đk x >= 0  \(\Leftrightarrow2x< 16\Leftrightarrow x< 8\)

Kết hợp đk vậy 0 =< x < 8 

a) \(2\sqrt{x}=14\)

Vì \(x\ge0\) nên bình phương hai vế ta được :

\(x=7^2\Leftrightarrow x=49\)

Vậy \(x=49\)

b) \(\sqrt{x}< \sqrt{2}\Leftrightarrow(\sqrt{x})^2< (\sqrt{2})^2\Leftrightarrow x< 4\)

c) \(\sqrt{2x}< 4\Leftrightarrow(\sqrt{2x})^2< 4^2\Leftrightarrow2x< 16\Leftrightarrow x< 8\)

95%

\(\sqrt{\dfrac{2}{5}}\) + \(\sqrt{\dfrac{5}{2}}\) = \(\dfrac{\sqrt{2}\times\sqrt{2}}{\sqrt{10}}\)+ \(\dfrac{\sqrt{5}\times\sqrt{5}}{\sqrt{10}}\) = \(\dfrac{7}{\sqrt{10}}\)= \(\dfrac{7\sqrt{10}}{10}\)

Lời giải:

\(\sqrt{\frac{2}{5}}+\sqrt{\frac{5}{2}}=\frac{\sqrt{2}.\sqrt{2}+\sqrt{5}.\sqrt{5}}{\sqrt{5}.\sqrt{2}}=\frac{7}{\sqrt{10}}=\frac{7\sqrt{10}}{10}\)

a) \((2+\sqrt{3})\sqrt{7-4\sqrt{3}}\)

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

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

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

b) \(\sqrt{(1-\sqrt{2023})^2}\cdot\sqrt{2024+2\sqrt{2023}}\)

\(=|1-\sqrt{2023}|\sqrt{2023+2\sqrt{2023}+1}\)

\(=(\sqrt{2023}-1)\sqrt{(\sqrt{2023}+1)^2}\)

\(=(\sqrt{2023}-1)(\sqrt{2023}+1)\)

\(=\sqrt{2023^2}-1^2=2023-1=2022\)