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. 

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

\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\\left(x+y\right)-\sqrt{x+1}=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}+\left(x+y\right)-\sqrt{x+1}=4-5=-1\\\left(x+y\right)-\sqrt{x+1}=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3\left(x+y\right)=-1\\\sqrt{x+1}=-\dfrac{1}{3}+5=\dfrac{14}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x+y=-\dfrac{1}{3}\\x+1=\dfrac{196}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{187}{9}\\y=-\dfrac{1}{3}-\dfrac{187}{9}=-\dfrac{190}{9}\end{matrix}\right.\left(nhận\right)\)

Nhiều quá em, em chỉ nên đăng những câu nào cảm thấy khó khăn khi giải quyết thôi

17325 nhé bạn

17325 bn nha

17325

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 " , " 

có 2 vế câu
vế câu 1 nối với vế câu 2 bằng dấu phẩy

Trong câu ghép sau có mấy vế câu? Các vế câu được nối với nhau bằng cách nào?

  “Đôi mắt khuyên vòng trắng long lanh, đôi chân thoăn thoắt.”

- CN: Đôi mắt.

- VN: khuyên vòng trắng long lanh.

− CN: Đôi chân.

− VN: thoăn thoắt.

  ⇒  Có 2 vế câu.

  ⇒  Các vế câu được nối với nhau bằng dấu phẩy, nối trực tiếp.

#zinc

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

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

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 \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

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)

\(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\\ =\sqrt{2+2\sqrt{2}+1}-\sqrt{4-4\sqrt{2}+2}\\ =\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\\ =\sqrt{2}+1-2+\sqrt{2}=2\sqrt{2}-1\)

uk

lm wen

ok bạn ơi!!

Luôn luôn có mk bên cạnh bạn. ^_^

ukm:)