ai giúp mình vs tới 4h chiều nay hết hạn r. cảm ơn nhìu ạ
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PL
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2021
Tam giác ABC vuông tại A có AM là trung tuyến ứng với cạnh huyền
\(\Rightarrow AM=\dfrac{1}{2}BC\Rightarrow BC=2AM=50\left(m\right)\)
a. Áp dụng định lý Pitago:
\(AB=\sqrt{BC^2-AC^2}=30\left(m\right)\)
b. Kẻ \(MH\perp AC\Rightarrow MH||AB\) (cùng vuông góc AC)
Mà M là trung điểm BC \(\Rightarrow MH\) là đường trung bình tam giác ABC
\(\Rightarrow MH=\dfrac{1}{2}AB=15\left(m\right)\)
\(\Rightarrow S_{AMC}=\dfrac{1}{2}MH.AC=\dfrac{1}{2}.15.40=300\left(m^2\right)\)
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NC
2
9 tháng 11 2021
Bài 1:
\(a,A=6\sqrt{2}-6\sqrt{2}+2\sqrt{5}=2\sqrt{5}\\ b,B=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{3}+\sqrt{2}\\ c,=2\sqrt{3}-6\sqrt{3}+15\sqrt{3}-4\sqrt{3}=7\sqrt{3}\\ d,=1+6\sqrt{3}-\sqrt{3}-1=5\sqrt{3}\\ e,=4\sqrt{2}+\sqrt{2}-6\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Bài 2:
\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}=5\Leftrightarrow2x-3=25\Leftrightarrow x=14\\ b,PT\Leftrightarrow x^2=\sqrt{\dfrac{98}{2}}=\sqrt{49}=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\\ c,ĐK:x\ge3\\ PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+1\right)=0\\ \Leftrightarrow\sqrt{x-3}=0\left(\sqrt{x+3}+1>0\right)\\ \Leftrightarrow x=3\\ d,ĐK:x\ge1\\ PT\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\\ \Leftrightarrow\sqrt{x-1}=1\Leftrightarrow x=2\left(tm\right)\\ e,PT\Leftrightarrow2x-1=16\Leftrightarrow x=\dfrac{17}{2}\\ f,PT\Leftrightarrow\left|2x-1\right|=\sqrt{3}-1\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{3}-1\\2x-1=1-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\\x=\dfrac{2-\sqrt{3}}{2}\end{matrix}\right.\)
9 tháng 11 2021
Bài 3:
\(a,Q=\dfrac{1+5}{3-1}=3\\ b,P=\dfrac{x+\sqrt{x}-6+x-2\sqrt{x}-3-x+4\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ c,M=\dfrac{\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{3-\sqrt{x}}{\sqrt{x}+5}=\dfrac{-\sqrt{x}}{\sqrt{x}+5}\)
Vì \(-\sqrt{x}\le0;\sqrt{x}+5>0\) nên \(M< 0\)
Do đó \(\left|M\right|>\dfrac{1}{2}\Leftrightarrow M< -\dfrac{1}{2}\Leftrightarrow-\dfrac{\sqrt{x}}{\sqrt{x}+5}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}-\sqrt{x}-5}{2\left(\sqrt{x}+5\right)}< 0\Leftrightarrow\sqrt{x}-5< 0\left(\sqrt{x}+5>0\right)\\ \Leftrightarrow0\le x< 25\)
Bài 4:
\(a,A=\dfrac{16+2\cdot4+5}{4-3}=29\\ b,B=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ c,P=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x+2\sqrt{x}+5}{\sqrt{x}+1}\\ P=\dfrac{\left(\sqrt{x}+1\right)^2+4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\\ P\ge2\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=2\sqrt{4}=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow x=1\left(tm\right)\)
PL
1
2 tháng 1 2022
1, The homework is too difficult for him to do
2, The homework isn't easy enough for him to do
3, The homework is so difficult that he can't do it
4, It is such difficult homework that he can't do it
PL
2
DT
7 tháng 1 2022
1. My sister says she went to school by bus this morning.
My sister said she had gone to school by bus that morning.
2. Nga says she have done her homework.
Nga said she had done her homework.
7 tháng 1 2022
1, My sister says that she goes to school by bus this morning
My sister said that she had gone to school by bus this morning
2, Nga says that she have done her homework
Nga said that she had done her homework
PL
1
DT
24 tháng 12 2021
1.........................................
=>the computer hasn't been fixed by them yet
2.......................................
=>Last night I was helped by a stranger.
23 tháng 10 2023
3: \(\left(\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}\right):\sqrt{3-2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{2-1}:\left(\sqrt{2}-1\right)\)
\(=\dfrac{2}{\sqrt{2}-1}=2\left(\sqrt{2}+1\right)=2\sqrt{2}+2\)
5:
\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
6:
\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2-\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{6-2\sqrt{5}-6-2\sqrt{5}}{4}=\dfrac{-4\sqrt{5}}{4}=-\sqrt{5}\)
4:
\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}-\sqrt{3}+3\right)}\)
\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{-6\left(\sqrt{2}-\sqrt{3}-3\right)}{4+2\sqrt{6}}\)
\(=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)}{\sqrt{6}+2}=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{6}-2\right)}{2}\)
\(=\dfrac{-3\left(2\sqrt{3}-2\sqrt{2}-3\sqrt{2}+2\sqrt{3}-6\sqrt{3}+6\right)}{2}\)
\(=\dfrac{-3\left(-2\sqrt{3}-5\sqrt{2}+6\right)}{2}\)
PL
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