giải hpt sau:
a, 2(x+y) + 3(x-y) =4 b,(x+1)(y-1) =xy-1
(x+y) +2(x-y) =5 (x-3)(y+3)= xy-3
mong mn giúp mình ạ:(
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ΔABC=ΔDEF
=>\(\widehat{A}=\widehat{D}\)
=>\(\widehat{D}=55^0\)
ΔABC=ΔDEF
=>\(\widehat{B}=\widehat{E}\)
=>\(\widehat{B}=75^0\)
Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=>\(\widehat{C}=180^0-55^0-75^0=50^0\)
=>\(\widehat{F}=\widehat{C}=50^0\)
1. protect
2. skillful
3. services
4. collected
5. community
6. development
7. voluntary
8. elderly
9. organization
10. donation
1 protect
2 skillful
3 services
4 collected
5 community
6 development
7 voluntary
8 elderly
9 organization
10 donation
\(\dfrac{11}{3}+\left|x\right|=\dfrac{9}{4}\)
=>\(\left|x\right|=\dfrac{9}{4}-\dfrac{11}{3}=\dfrac{27}{12}-\dfrac{44}{12}=-\dfrac{17}{12}\)
mà \(\left|x\right|>=0\forall x\)
nên \(x\in\varnothing\)
66.2
3 hat => a hat
4 job => a job
5 OK
6 apple => an apple
7 party => a party
8 wonderful thing => a wonderful thing
9 island => an island
10 key => a key
11 OK
12 good idea => a good idea
13 car => a car
14 cup => a cup
15 OK
16 umbrella => an umbrella
66.3
2 a piece of wood
3 a glass of water
4 a bar of chocolate
5 a cup of tea
6 a piece of paper
7 a bowl of soup
8 a loaf of bread
9 a jar of honey
66.2
3 a hat
4 a job
5 ok
6 an apple
7 a party
8 a wonderful thing
9 an island
10 a key
11 ok
12 a good idea
13 a car
14 a cup
14 ok
16 an umbrella
\(\left(\dfrac{1}{2}\right)^{x+2}=16^{4-2x}\)
=>\(2^{-x-2}=2^{4\left(4-2x\right)}\)
=>-x-2=4*(4-2x)
=>-x-2=16-8x
=>-x+8x=16+2
=>7x=18
=>\(x=\dfrac{18}{7}\)
a: \(\dfrac{x}{y}+\dfrac{y}{x}>=2\cdot\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)
b: \(\dfrac{1}{x}+\dfrac{1}{y}>=\dfrac{4}{x+y}\)
=>\(\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)
=>\(\left(x+y\right)^2>=4xy\)
=>\(x^2+2xy+y^2-4xy>=0\)
=>\(x^2-2xy+y^2>=0\)
=>\(\left(x-y\right)^2>=0\)(luôn đúng)
Bài 2:
a) \(\dfrac{-7}{-13}=\dfrac{7}{13}\) là số hưu tỉ dương
b) \(\dfrac{2}{-17}=-\dfrac{2}{17}\) là số hưu tỉ âm
c) \(-\dfrac{-6}{5}=\dfrac{6}{5}\) là số hưu tỉ dương
Bài 3:
a) \(-2\dfrac{1}{4}=-\left(2+\dfrac{1}{4}\right)=-\dfrac{9}{4}\)
b) \(6\dfrac{2}{3}=6+\dfrac{2}{3}=\dfrac{20}{3}\)
c) \(-3\dfrac{1}{4}=-\left(3+\dfrac{1}{4}\right)=-\dfrac{13}{4}\)
a: \(x^2+y^2>=2xy\)
=>\(x^2-2xy+y^2>=0\)
=>\(\left(x-y\right)^2>=0\)(luôn đúng)
b: \(x^2+4xy>=-4y^2\)
=>\(x^2+4xy+4y^2>=0\)
=>\(\left(x+2y\right)^2>=0\)(luôn đúng)
c: \(2\left(x^2+y^2\right)>=\left(x+y\right)^2\)
=>\(2x^2+2y^2-x^2-2xy-y^2>=0\)
=>\(x^2-2xy+y^2>=0\)
=>\(\left(x-y\right)^2>=0\)(luôn đúng)
a: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\x+y+2\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-y-3x+y=4-5\\3x-y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=-1\\y=3x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3\cdot\dfrac{-1}{2}-5=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y-x+y=0-2\\x-y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}0y=-2\\x-y=0\end{matrix}\right.\Leftrightarrow\left(x;y\right)\in\varnothing\)