a,\(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{a+b}{2}\right)^2-1^2=\left(\frac{a+b}{2}-1\right)\left(\frac{a+b}{2}+1\right)\)

b,\(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(x-y\right)^2\left(9-4\right)=5\left(x-y\right)^2\)

c,\(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)\)

\(=\left(-p-4q\right)3p\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{p^2}{6}\right)^2=\left(5pm-\frac{p^2}{6}\right)\left(5pm+\frac{p^2}{6}\right)\)

a, \(\frac{1}{4}\left(a+b\right)^2-1=\left(\frac{1}{2}a+\frac{1}{2}b\right)^2-1=\left(\frac{a}{2}+\frac{b}{2}-1\right)\left(\frac{a}{2}+\frac{b}{2}+1\right)\)

b, \(9\left(x-y\right)^2-4\left(x-y\right)^2=\left(3x-3y\right)^2-\left(2x-2y\right)^2\)

\(=\left(3x-3y-2x+2y\right)\left(3x-3y+2x-2y\right)=5\left(x-y\right)^2\)

c, \(\left(p-2q\right)^2-4\left(p+q\right)^2=\left(p-2q\right)^2-\left(2p+2q\right)^2\)

\(=\left(p-2q-2p-2q\right)\left(p-2q+2p+2q\right)^2=9p^2\left(-p-4q\right)\)

d, \(25p^2m^4-\frac{1}{36}p^4=\left(5pm^2\right)^2-\left(\frac{1}{6}p^2\right)^2=\left(5pm^2-\frac{1}{6}p^2\right)\left(5pm^2+\frac{1}{6}p^2\right)\)

\(=p^2\left(5m^2-\frac{1}{6}p\right)\left(5m^2+\frac{1}{6}p\right)\)

271​x3−32​x2+4x−8

 x  -  x  +  4x  -  8

=  4x  -  8

=  4( x  -  8)

b, \(\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-18x^2+81\right)=\left(x^4-81\right)-x^4+18x^2-81\)=\(18x^2-162\)

a, \(5x^2\)\(-3x-x^3+x^2+x^3-6x^2\)\(-10+3x\)=\(-10\)

b,\(x^3+x^2+x-x^3-x^2-x+5=5\)

Trả lời:

a, a² - 9 ( b - c )² 

= a² - [ 3 ( b - c ) ]² 

= a² - ( 3b - 3c )² 

= ( a - 3b + 3c )( a + 3b - 3c )

b, 4 ( x + y )² - z² 

= [ 2 ( x + y ) ]² - z² 

= ( 2x + 2y )² - z² 

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

\(x^{n-1}\left(x+y\right)-y\left(x^{n-1}+y^{n-1}\right)\)

\(=x^{n-1}x+x^{n-1}y-yx^{n-1}-yy^{n-1}\)

\(=x^{n-1+1}+\left(x^{n-1}y-x^{n-1}y\right)-y^{n-1+1}\)

\(=x^n-y^n\)

4(x+y)² - z²

= 2² (x+y)² - z²

= (2x+2y)- z²

= (2x + 2y-z)(2x + 2y +z) 

ABCD là hình thang có 2 đáy là AB và CD nên AB//CD

=> ∠A + ∠D= 180⁰ (hai góc trong cùng phía)

Mà ∠A - ∠D= 60⁰

=> ∠A= (180⁰ + 60⁰):2= 120⁰

∠D = 120⁰ - 60⁰ = 60⁰

∠A - ∠B= 30⁰

=> ∠B = 120⁰ - 30⁰ = 90⁰

Áp dụng định lí tứ giác ta có: ∠A + ∠B + ∠C+ ∠D =360⁰

=> ∠C=  360⁰ - 90⁰ - 60⁰ -120⁰= 90⁰

Vậy ∠A= 120⁰ ; ∠B= 90⁰ ; ∠C = 90⁰ ; ∠D = 60⁰