Giup em với ạ em đg cần rất gấp :((
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x và y tỉ lệ thuận
nên \(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}\)
Áp dụng tính chất của DTSBN, ta được:
\(\dfrac{x_1}{y_1}=\dfrac{x_2}{y_2}=\dfrac{2x_1-3x_2}{2y_1-3y_2}=\dfrac{42.5}{-8.5}=-5\)
=>x=-5y
1 sickness - disability
2 opposition
3 pride
4 professional
5 unemployed
6 Poverty
7 deafened
8
9 photographing
10 education
1 sickness - disability
2 opposition
3 pride
4 professional
5 unemployed
6 Poverty
7 deafened
8
9 photographing
10 education
Cho xin một like đi các dân chơi à.
5. There is a boy behind that tree
6. There are some girls in front of the house
7. There isn't a telephone in her office
8. There aren't any chairs downstairs
II
1. Minh lives in a house near a lake
2. There is a big yard in front of our school
3. Are there many flowers to the right of the museum?
4. What is there next to the photocopy store?
5. My father works in a hospital in the city
6. How many people are there in Linh's family
7. My friend in Hanoi doesn't live with his family
8. Hoa gets up at 6 o'clock and brushes her teeth
9. Our classroom is on the first floor
10. There are six rooms in Minh's house
a: \(AO=\dfrac{1}{2}AC\)(O là trung điểm của AC)
nên AO=AD
hay ΔAOD cân tại A
\(11,\\ a,M=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\left(\dfrac{1}{\sqrt{a}}>0\right)\)
\(9,\\ a,=\left|2-\sqrt{7}\right|=\sqrt{7}-2\\ b,=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\\ c,=3-4+2=1\\ d,=6\sqrt{3a}-4\sqrt{3a}=2\sqrt{3a}\\ 10,\)
a, Áp dụng HTL: \(x=\sqrt{9\cdot25}=15\)
b, Áp dụng HTL: \(\left\{{}\begin{matrix}8^2=10x\\y^2=x\left(x+10\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6,4\\y=\sqrt{6,4\cdot16,4}\approx10,245\end{matrix}\right.\)
Bài 1:
a) \(\left(2x+y\right)^2-\left(y-2x\right)^2=\left(2x+y-y+2x\right)\left(2x+y+y-2x\right)=8xy\)
b) \(\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\\ =x^2-2xy+y^2+2x^2-2y^2+x^2+2xy+y^2\\ =4x^2\)
c) \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
d) \(\left(y-\dfrac{1}{2}\right)\left(y+\dfrac{1}{2}\right)=y^2-\dfrac{1}{4}\)
Bài 1:
a: \(\left(2x+y\right)^2-\left(y-2x\right)^2\)
\(=4x^2+4xy+y^2-y^2+4xy-4x^2\)
=8xy
b: Ta có: \(\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
c: \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
d: \(\left(y-\dfrac{1}{2}\right)\left(y+\dfrac{1}{2}\right)=y^2-\dfrac{1}{4}\)