giúp em vs

T_T

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.

1. sickness/ disabilities

2. opposition

3. pride

4. professional

5. unemployed

6. poverty

7. deafened

8. severity

9. photography

10. education

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

bài này nghĩa là như nèo??ko có đề bài 1 à

\(\dfrac{12x4x72}{36x2x9}=\dfrac{48x72}{72x9}=\dfrac{48}{9}=\dfrac{16}{3}\)

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: \(\left(y-\dfrac{1}{2}\right)\left(y+\dfrac{1}{2}\right)=y^2-\dfrac{1}{4}\)