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The building that was destroyed in the fire has now been rebuilt.
"Make a sandwich," Keiro said to me.
"Stay back," the firefighter ordered us.
"I will get up early tomorrow," Mary said.
"We are meeting our friends today," he said.
"I sent them an email yesterday," Jack said.
"You should be more careful in the future," Nick told Tina.
"I like football," Chris told me.
"Where are we going this evening?" Lan wondered.
"Where is my brother?" Ron wanted to know.
The global warming controversy concerns the public debate over whether global warming is occurring, how much has occurred in modern times, what has caused it, what its effects will be, whether any action can or should be taken to curb it, and if so what that action should be. In the scientific literature, there is a very strong consensus that global surface temperatures have increased in recent decades and that the trend is caused by human-induced emissions of Greenhouse glasses No scientific body of national or international standing disagrees with this view, though a few organizations with members in extractive industries hold non-committal positions, and some have tried to persuade the public that climate change is not happening, or if the climate is changing it is not because of human influence, attempting to sow doubt in the scientific consensus.
Ta có: \(\frac{x\sqrt{x}-1}{\sqrt{x}-1}-\frac{x\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)=2\sqrt{x}\)
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}=\frac{2x+2}{x-1}\)
Ta có: \(\frac{x\sqrt{x}-1}{\sqrt{x}-1}-\frac{x\sqrt{x}+1}{\sqrt{x}+1}+\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\cdot\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(=2\sqrt{x}+\frac{2x+2}{x-1}\cdot\frac{x-1}{\sqrt{x}}=2\sqrt{x}+\frac{2x+2}{\sqrt{x}}=\frac{2x+2x+2}{\sqrt{x}}=\frac{4x+2}{\sqrt{x}}\)
a) Hệ phương trình có nghiệm duy nhất là
\(\left\{{}\begin{matrix}2x-y=3\\x+4y=6\end{matrix}\right.\)
b) Hệ phương trình có vô số nghiệm là
\(\left\{{}\begin{matrix}2x-y=3\\4x-2y=6\end{matrix}\right.\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}+\sqrt{5}}{2}\)
\(=\dfrac{3\sqrt{5}}{2}\)
A B C D O E F G H x y I
1/
Ta có
sđ cung AC = sđ cung BC (1)
\(sđ\widehat{CFG}=\dfrac{1}{2}\left(sđcungBC+sđcungAE\right)\) (góc có đỉnh ở trong hình tròn) (2)
\(sđ\widehat{CHE}=\dfrac{1}{2}sđcungCAE=\dfrac{1}{2}\left(sđcungAC+sđcungAE\right)\) (góc nội tiếp) (3)
Từ (1) (2) (3) \(\Rightarrow\widehat{CFG}=\widehat{CHE}\)
Ta có
\(\widehat{CFG}+\widehat{EFG}=\widehat{EFC}=180^o\)
\(\Rightarrow\widehat{CHE}+\widehat{EFG}=180^o\)
=> EFGH là tứ giác nội tiếp (Tứ giác có hai góc đối bù nhau là tứ giác nội tiếp)
2/
sđ cung AC = sđ cung BC (4)
\(sđ\widehat{AGC}=\dfrac{1}{2}\left(sđcungAC+sđcungBH\right)\) (5) (góc có đỉnh ở trong hình tròn)
\(sđ\widehat{CHy}=\dfrac{1}{2}sđcungCBH=\dfrac{1}{2}\left(sđcungBC+sđcungBH\right)\) (6) (Góc giữa tiếp tuyến và dây cung)
Từ (4) (5) (6) \(\Rightarrow\widehat{AGC}=\widehat{CHy}\)
Mà AC = AG (gt) => tgACG cân tại A \(\Rightarrow\widehat{AGC}=\widehat{ACG}\)
\(\Rightarrow\widehat{ACG}=\widehat{CHy}\) mà 2 góc trên ở vị trí so le trong => xy//AC
Có VT = \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{zx}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xyz}\left(x+y+z\right)}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|=VP\) (Vì x + y + z = 0)