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3.clothes were washed in the past by my mother
1. Energy consumption will be reduced as much as possible.
2. A lot of energy next century will be offered by wind power.
3. In the past, clothes were washed with my mother's hand. ;-;
4. Their house is decorated with beautiful Christmas trees.
5. Ann will be invited to the party by Bill.
Ta có: \(\sqrt{\dfrac{3\sqrt{5}-1}{2\sqrt{5}+3}}-\sqrt{\dfrac{\sqrt{5}+11}{7-2\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{11}}-\sqrt{\dfrac{\left(11+\sqrt{5}\right)\left(7+2\sqrt{5}\right)}{29}}\)
\(=\sqrt{\dfrac{11\cdot\left(30-9\sqrt{5}-2\sqrt{5}+3\right)}{121}}-\sqrt{29\cdot\left(\dfrac{77+22\sqrt{5}+7\sqrt{5}+20}{841}\right)}\)
\(=\dfrac{\sqrt{363-121\sqrt{5}}}{11}+\dfrac{\sqrt{2813+841\sqrt{5}}}{29}\)
\(=\dfrac{\sqrt{726-242\sqrt{5}}}{11\sqrt{2}}+\dfrac{\sqrt{5626+1682\sqrt{5}}}{29\sqrt{2}}\)
\(x=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left|\sqrt{5}+2\right|-\left|\sqrt{5}-2\right|\)
\(=\sqrt{5}+2-\sqrt{5}+2=4\)
\(y=\sqrt{3+2\sqrt{5}}-\sqrt{3-2\sqrt{5}}\)
Xem lại đề, \(\sqrt{3-2\sqrt{5}}\) không xác định.
a: ĐKXĐ x>0; x<>1
\(A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{x-1}{\sqrt{x}}\)
b: A<0
=>x-1<0
=>0<x<1
she is interested in reading comic
Tomorrow in this time I wil be lieing on the beach
My car isn't as expensive as his
She has started live in Paris since 2005
A new dam will build near the rive next year
Bài 9.a) Quãng đường vật rơi trong 3s đầu tiên:
\(S_3=\dfrac{1}{2}gt^2=\dfrac{1}{2}\cdot9,8\cdot3^2=44,1\left(m\right)\)
Quãng đường vật đi trong giây thứ 3 là: \(S'=\dfrac{1}{2}gt^2-\dfrac{1}{2}g\left(t-1\right)^2=\dfrac{1}{2}\cdot9,8\cdot3^2-\dfrac{1}{2}\cdot9,8\cdot2^2=24,5\left(m\right)\)
Bài 10.
a) Thời gian vật rơi khi chạm đất: \(t=\sqrt{\dfrac{2S}{g}}=\sqrt{\dfrac{2\cdot45}{10}}=3\left(s\right)\)
Vận tốc vật khi chạm đất: \(v=g\cdot t=10\cdot3=30\)m/s
b) Quãng đường vật rơi trong giây cuối cùng:
\(S=\dfrac{1}{2}gt^2-\dfrac{1}{2}g\left(t-1\right)^2=\dfrac{1}{2}\cdot10\cdot3^2-\dfrac{1}{2}\cdot10\cdot2^2=25\left(m\right)\)
Bài 11.
a) Quãng đường vật rơi trong 2s đầu tiên:
\(S=\dfrac{1}{2}gt^2=\dfrac{1}{2}\cdot10\cdot2^2=20\left(m\right)\)