Bài 5A:

a: Ta có: \(M=\left(2x+y\right)^2-\left(y-2x\right)^2\)

\(=\left(2x+y-y+2x\right)\left(2x+y+y-2x\right)\)

\(=4x\cdot2y=8xy\)

b: Ta có: \(N=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)

\(=\left(3x+2+1-2y\right)^2\)

\(=\left(3x-2y+3\right)^2\)

c: Ta có: \(P=\left(x^2+2xy\right)^2+2\left(x^2+2xy\right)y^2+y^4\)

\(=\left(x^2+2xy+y^2\right)^2\)

\(=\left(x+y\right)^4\)

\(\frac{2}{7}+\frac{7}{12}=\frac{24}{84}+\frac{49}{84}=\frac{73}{84}\)

\(\frac{25}{48}+\frac{22}{48}=\frac{47}{48}\)

\(\frac{5}{7}+\frac{3}{8}=\frac{40}{56}+\frac{21}{56}=\frac{61}{56}\)

\(\frac{15}{24}+\frac{15}{16}=15.\left(\frac{1}{24}+\frac{1}{16}\right)=\frac{25}{16}\)

\(\frac{5}{6}+\frac{3}{4}=\frac{20+18}{24}=\frac{19}{12}\)

\(\frac{9}{24}+\frac{14}{24}=\frac{23}{24}\)

học tót tích mình hem

2/7 + 7/12 = 73/84.

25/48 + 11/24 = 47/48.

5/7 + 3/8 = 61/56.

15/24 + 15/16 = 25/16.

5/6 + 3/4 = 19/12.

3/8 + 7/12 = 23/24.

hok tốt!

\(\frac{2}{7}+\frac{7}{12}=\frac{24}{84}+\frac{49}{84}=\frac{73}{84}\)

\(\frac{25}{48}+\frac{11}{24}=\frac{25}{48}+\frac{22}{48}=\frac{47}{48}\)

\(\frac{5}{7}+\frac{3}{8}=\frac{40}{56}+\frac{21}{56}=\frac{61}{56}\)

\(\frac{5}{6}+\frac{3}{4}=\frac{20}{24}+\frac{18}{24}=\frac{38}{24}=\frac{19}{12}\)

\(\frac{3}{8}+\frac{7}{12}=\frac{36}{96}+\frac{56}{96}=\frac{92}{96}=\frac{23}{24}\)

Học tốt

1 distance

2 interesting

3 unusual

4 traditional

5 regularly

6 rainy

7 impossible

8 convinced

1. distance 

2. interesting

3. unusual

4. traditional

5. regularly 

6. rainy 

7. impossible

8. convinced

1. distance

2. interesting

3. unusual

4. traditional

5. regularly

6. rainy

7. impossible

8. convinced

Bài 3: 

a) Ta có: \(A=\left(x+y\right)^2-\left(x-y\right)^2\)

\(=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\)

\(=x^2+2xy+y^2-x^2+2xy-y^2\)

=4xy

b) Ta có: \(B=\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

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

\(=\left(2y\right)^2=4y^2\)

c) Ta có: \(C=\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-\left(x^3-3x^2y+3xy^2-y^3\right)-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

Bài 1 : 

a,

Trên hình vẽ ta thấy : \(i=90^o\)

\(i=i'\Leftrightarrow i'=90^o\)

b,

\(i=30^o\)

\(i=i'\Leftrightarrow i'=30^o\)

c,

\(i=130^o-90^o=40^o\)

\(i=i'\Leftrightarrow i'=40^o\)

d,

\(i'=90^o-45^o=45^o\)

\(i=i'\Leftrightarrow i'=45^o\)

Bài 2 :

a)

 

b,

c,