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17 tháng 12 2020
Nowadays, our life is very different from our life in the past. It was not as convenient as now. There were no modern equipments, electricity and the children didn't have to go to school, they used to stay at home and do chores. Almost people did on the farm. Besides, there weren't any motorbikes, cars or planes so the people used to travel on foot. The air was extremely fresh because there were any big factories.
13 tháng 1
Bài 5:
Thay x=1 và y=2 vào hệ phương trình, ta được:
\(\left\{{}\begin{matrix}-m\cdot1+2=-2m\\1+m^2\cdot2=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2m=-m+2\\2m^2=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2=4\\-m=2\end{matrix}\right.\)
=>m=-2
Bài 6:
a: ĐKXĐ: x>=1 và y>=-2
\(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}-3\sqrt{y+2}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y+2}=1\\\sqrt{x-1}=2+3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\x-1=25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=26\\y=-1\end{matrix}\right.\left(nhận\right)\)
b: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{8}{12}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{y}=\dfrac{-1}{3}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=21\\\dfrac{1}{x}=\dfrac{1}{12}-\dfrac{1}{21}=\dfrac{7-4}{84}=\dfrac{3}{84}=\dfrac{1}{28}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=28\\y=21\end{matrix}\right.\left(nhận\right)\)
c: ĐKXĐ: x<>0 và y<>2
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y-2}=4\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{6}{y-2}=8\\\dfrac{4}{x}-\dfrac{1}{y-2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y-2}=7\\\dfrac{2}{x}+\dfrac{3}{y-2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y-2=1\\\dfrac{2}{x}=4-\dfrac{3}{1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\left(nhận\right)\)
d: ĐKXĐ: x<>-2y và x<>-y/2
\(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{2x+y}=3\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x+2y}+\dfrac{3}{2x+y}=9\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{10}{x+2y}=10\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=1\\\dfrac{3}{2x+y}=4-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=1\\2x+y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x+4y=2\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\left(nhận\right)\)
e: ĐKXĐ: x>4 và y<>-2
\(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x-4}}+\dfrac{4}{y+2}=7\\\dfrac{5}{\sqrt{x-4}}-\dfrac{1}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x-4}}+\dfrac{4}{y+2}=7\\\dfrac{20}{\sqrt{x-4}}-\dfrac{4}{y+2}=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{23}{\sqrt{x-4}}=23\\\dfrac{5}{\sqrt{x-4}}-\dfrac{1}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x-4}=1\\\dfrac{1}{y+2}=5-4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-4=1\\y+2=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\left(nhận\right)\)
f: ĐKXĐ: x>=-1
\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\\left(x+y\right)-\sqrt{x+1}=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}+\left(x+y\right)-\sqrt{x+1}=4-5=-1\\\left(x+y\right)-\sqrt{x+1}=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\left(x+y\right)=-1\\\sqrt{x+1}=-\dfrac{1}{3}+5=\dfrac{14}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=-\dfrac{1}{3}\\x+1=\dfrac{196}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{187}{9}\\y=-\dfrac{1}{3}-\dfrac{187}{9}=-\dfrac{190}{9}\end{matrix}\right.\left(nhận\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
Nhiều quá em, em chỉ nên đăng những câu nào cảm thấy khó khăn khi giải quyết thôi
CT
27 tháng 3 2022
Trong câu ghép : " Đôi mắt khuyên vòng trắng long lanh, đôi chân thanh thoát "
→ Có 2 vế câu
→ Các vế câu được nối bằng dấu " , "
17 tháng 7 2018
a,<=> x2-4x+22+y2-8y+42-14
<=> (x2-2x2+22)+(y2-2x4+42)-14
<=> (x-2)2+(y-4)2-14
Vì (x-2)2+(y-4)2>= 0
=> F >= -14 => MIn F = -14 <=> x=2, y=4
b, <=> (x2+52+(2y)2-4xy+10x-20y) +(y2-2y+1)+2
<=> (x+5-2y )2+(y-1)2+2
Vì (x+5-2y) 2+(y-1)2 >= 0
=> G >= 2 => Min =2 <=> y=1, x= -3
17 tháng 7 2018
\(F=x^2-4x+y^2-8y+6\)
\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)
\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y-4\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)
\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)
17 tháng 9 2021
a)
\(\sqrt{9+4\sqrt{5}}\cdot\sqrt{6-2\sqrt{5}}\\ =\sqrt{4+4\sqrt{5}+5}\cdot\sqrt{1-2\sqrt{5}+5}\\ =\sqrt{\left(2+\sqrt{5}\right)^2}\cdot\sqrt{\left(1-\sqrt{5}\right)^2}\\ =\left(2+\sqrt{5}\right)\left(1-\sqrt{5}\right)\)
b)
\(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\\ =\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\\ =\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\\ =\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)
ờ dg r
Bn để thẳng như z ko có đọc đc nha