đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

ĐKXĐ: x>=0

\(\sqrt{7+\sqrt{2x}}=3+\sqrt5\)

=>\(7+\sqrt{2x}=\left(3+\sqrt5\right)^2=14+6\sqrt5\)

=>\(\sqrt{2x}=7+6\sqrt5\)

=>\(2x=\left(7+6\sqrt5\right)^2=49+180+84\sqrt5=229+84\sqrt5\)

=>\(x=\frac{229}{2}+42\sqrt5\) (nhận)

d: \(\frac{4}{\sqrt7-\sqrt3}+\frac{6}{3+\sqrt3}+\frac{\sqrt7-7}{\sqrt7-1}\)

\(=\frac{4\left(\sqrt7+\sqrt3\right)}{7-3}+\frac{6\left(3-\sqrt3\right)}{\left(3+\sqrt3\right)\left(3-\sqrt3\right)}-\frac{\sqrt7\left(\sqrt7-1\right)}{\sqrt7-1}\)

\(=\sqrt7+\sqrt3+3-\sqrt3-\sqrt7=3\)

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt5}\right)\left(4-\sqrt{10+2\sqrt5}\right)}\)

=>\(A^2=8+2\cdot\sqrt{16-10-2\sqrt5}=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

\(\sqrt{\frac{\left(-5\right)^2}{7}}=\frac{\sqrt{\left(-5\right)^2}}{\sqrt{7}}=\frac{|5|}{\sqrt{7}}=\frac{5\sqrt{7}}{7}\)

\(\frac{-\sqrt{\left(-5\right)^2}}{-\sqrt{49}}=\frac{\sqrt{\left(-5\right)^2}}{\sqrt{49}}=\frac{|5|}{|7|}=\frac{5}{7}\)

\(\frac{5\sqrt{7}}{7}>\frac{5}{7}\leftrightarrow\sqrt{\frac{\left(-5\right)^2}{7}}>\frac{-\sqrt{\left(-5\right)^2}}{-\sqrt{49}}\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)

\(=3\sqrt{2}\)

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

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

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}\)

=21

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

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

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

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

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

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

Trần Đức Thắng lm nốt đi

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)