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=(\(\dfrac{99}{2}+1+\dfrac{98}{3}+1+...+\dfrac{1}{100}+1\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2
=(\(\dfrac{101}{2}+\dfrac{101}{3}+...\dfrac{101}{100}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)) -2
=101(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\)):(\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}\))-2
=101 -2 =99
-_-
a: \(=2\sqrt{x-3}+3\sqrt{x-3}-4\sqrt{x-3}+3-x\)
\(=\sqrt{x-3}+3-x\)
c: \(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=18\)
=>2 căn x-2=18
=>x-2=81
=>x=83
\(20\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=20\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-2\sqrt{20\sqrt{3}}\)
\(=80\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=72\sqrt{5\sqrt{3}}\)
\(\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.3+9}-2\sqrt{3}\)
\(\sqrt{\left(2\sqrt{3}-3\right)^2}-2\sqrt{3}\)
\(2\sqrt{3}-3-2\sqrt{3}\)
= -3
\(\frac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}\)
\(=\frac{5\left(\sqrt{6}-1\right)^2}{5}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{1}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(\sqrt{6}-1\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{2}-1\right)\)
\(=6-2\sqrt{6}+1-2+2\sqrt{6}-3+\sqrt{2}-1=\sqrt{2}\)
1) \(x+5\sqrt{x}+6=x+2\sqrt{x}+3\sqrt{x}+6\)
\(=\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)\)
\(=\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\)
2) \(=\left(4\sqrt{3}+3\sqrt{15}-12\sqrt{15}\right)\sqrt{3}\)
\(=\left(4\sqrt{3}-8\sqrt{15}\right)\sqrt{3}=12-24\sqrt{5}\)
Bài làm:
\(A=\left(3\sqrt{32}-2\sqrt{18}-\sqrt{50}\right)\div\sqrt{2}\)
\(A=\left(12\sqrt{2}-6\sqrt{2}-5\sqrt{2}\right)\div\sqrt{2}\)
\(A=\sqrt{2}\div\sqrt{2}\)
\(A=1\)
a: \(B=3\sqrt{7}-6\sqrt{12}+\sqrt{84}-24\)
b: \(B=3\cdot\dfrac{\sqrt{2}-\sqrt{5}}{\left(\sqrt{2}-\sqrt{5}\right)^2}=\dfrac{3}{\sqrt{2}-\sqrt{5}}=-\sqrt{5}-\sqrt{2}\)
ko heu