Mình làm một câu ví dụ thui nha

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

\(\frac{5x}{50}=2\Rightarrow x=20\)

\(\frac{y}{6}=2\Rightarrow y=12\)

\(\frac{2z}{42}=2\Rightarrow x=42\)

mấy câu khác thì tương tự

tíc mình nha bạn

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

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

C=\(\left(2a+b\right)^2-\left(2a-b\right)^2=\left(2a+b-2a+b\right)\left(2a+b+2a-b\right)=\left(2b\right).\left(4a\right)\)

D=\(\left(2x-1\right)^2-2\left(2x-3\right)^2+4=4x^2-4x+1-4x+6+4=4x^2-8x+11\)

E=\(\left(x+3y\right)^2-\left(x-3y\right)^2=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)=\left(6y\right).\left(2x\right)\)

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

G=\(\left(x-2y\right)^2+4\left(x-2y\right)y+4y^2=x^2-4xy+4y^2+4xy-8y^2+4y^2=x^2\)

H=\(\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y^{ }\right)^2=x^2-2xy+y^2-4\left(x^2+2xy-xy-2y^2\right)+4x+8y=x^2-2xy+y^2-4x^2-8xy+4xy+8y^2+4x+8y=3x^2+12xy-9y^2+4x+8y\)

Ta có:

a) A= (x-y)^2 + (x+y)^2

A= x^2 -2xy + y^2 + x^2 + 2xy + y^2

A= 2x^2+ 2y^2

b) B= (x+y)^2 -( x-y)^2

B= (x+y-x+y)(x+y+x-y)

B= 2y.2x= 4xy

c) C= (2a+b)^2 -( 2a-b)^2

C= (2a+b-2a+b)(2a+b+2a-b)

C= 2b.4a

C= 8ab

d) D= (2x-1)^2 -2(2x-3)^2+4

D= 4x^2 -4x+1 -2( 4x^2 -12x + 9) +4

D= 4x^2 -4x+1 -8x^2 + 24x -18 +4

D= -4x^2 + 20x-13

e) E= (x+3y)^2-(x-3y)^2

E= (x+3y-x+3y)(x+3y+x-3y)

E= 6y.2x= 12xy

f) F= (2x+y)^2-(2x-y)^2

F=(2x+y-2x+y)(2x+y+2x-y)

F= 2y.4x= 8xy

g) G= (x-2y)^2 + 4(x-2y)y + 4y^2

G= (x-2y)^2 + 2(x-2y)2y + (2y)^2

G= (x-2y+2y)^2

G= x^2

h) H= (x-y)^2 -4(x-y)(x+2y)+ 4(x+2y)^2

H= (x-y)^2 - 2(x-y)2(x+2y) + [2(x+2y)]^2

H= (x-y- 2x-4y)^2

H= (-x-5y)^2

Lưu ý (-A-B)^2 = ( A+ B)^2

=> H= (x+5y)^2

5

a)\(\left(x+y\right)^2:\left(x+y\right)=x+y\)

b)\(\left(x-y\right)^5:\left(y-x\right)^4=\left(x-y\right)^5:\left(x-y\right)^4=x-y\)

c)\(\left(5x^4-3x^3+x^2\right):3x^2=\frac{5}{3}x^2-x+\frac{1}{3}^{ }\)

d)\(\left(x^3y^3-\frac{1}{2}x^2y^3+x^3y^2\right):\frac{1}{2}x^2y^2=2xy-y+x\)

a: \(=2\left[\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\right]-3\left(x^4+y^4\right)\)

\(=2x^4-2x^2y^2+2y^4-3x^4-3y^4\)

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

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

b: Bạn xem lại đề chỗ 3y^ nhé bạn