\(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

\(=\dfrac{5+4\sqrt{5}+4-8\sqrt{5}}{2\sqrt{5}-4}\)

\(=\dfrac{9-4\sqrt{5}}{2\sqrt{5}-4}\)

\(=\dfrac{\left(9-4\sqrt{5}\right)\left(2\sqrt{5}+4\right)}{4}\)

\(=\dfrac{\left(9-4\sqrt{5}\right)\cdot2\left(\sqrt{5}+2\right)}{4}\)

\(=\dfrac{\left(9-4\sqrt{5}\right)\left(\sqrt{5}+2\right)}{2}\)

\(=\dfrac{9\sqrt{5}+18-20-8\sqrt{5}}{2}\)

\(=\dfrac{\sqrt{5}-2}{2}\)

sáng mai nha

Gọi N = 16a=25b=30c

N chiahetcho 16,25,30

Do a,b,c nhỏ nhất nên N=BCNN(16;25;30)=1200

=> a=1200/16=75

b=48

c=40

4000 + 500 + 60 + 7 = 4567

tik nha!!!!

\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

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

\(=\dfrac{10\sqrt{5}-10\sqrt{2}+2\sqrt{50}-2\sqrt{20}}{3}+\left[-2\left(1+\sqrt{5}\right)\right]\)

\(=\dfrac{10\sqrt{5}-10\sqrt{2}+10\sqrt{2}-2\sqrt{20}}{3}+\left[-2-2\sqrt{5}\right]\)

\(=\dfrac{10\sqrt{5}-4\sqrt{5}}{3}-2-2\sqrt{5}\)

\(=\dfrac{6\sqrt{5}}{3}-2-2\sqrt{5}\)

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

\(=-2\)

1)

a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow2x=\dfrac{7}{2}-\dfrac{5}{2}\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)

\(\Leftrightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=-\dfrac{1}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48}{5}\\x=\dfrac{52}{5}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{48}{5};x_2=\dfrac{52}{5}\)

\(\left(-113+39\right)+\left(239-400\right)-100-287\)

\(=\left(-74\right)+\left(-161\right)-100-287\)

\(=-74-161-100-287\)

\(=-622\)

Nhị thức Niu - Tơn | Toán Học Việt Nam - Kho tài liệu toán học chất lượng | toan-hoc-viet-nam-kho-tai-lieu-toan-hoc-chat-luong

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