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Không vào đc bạn ơi

( 2x + 4 )( 8x - 3 ) - ( 4x + 1 )²

= 16x² - 6x + 32x - 12 - ( 16x² + 8x + 1 )

= 16x² + 26x - 12 - 16x² - 8x - 1

= 18x - 13

\(\left(2x+4\right)\left(8x-3\right)-\left(4x+1\right)^2\)   

\(=16x^2-6x+32x-12-\left(16x^2+8x+1\right)\)   

\(=16x^2+26x-12-16x^2-8x-1\)   

\(=18x-13\)

Ta có: \(\frac{4n^3+11n^2+5n+5}{n+2}=\frac{\left(n+2\right)\left(4n^2+3n-1\right)+7}{n+2}=4n^2+3n-1+\frac{7}{n+2}\)

Để 4n³ + 11n² + 5n + 5 chia hết cho n + 2 thì \(\frac{7}{n+2}\inℤ\Rightarrow7⋮n+2\Rightarrow n+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta lập bảng giá trị:

Vậy \(n\in\left\{-1;-3;5;-9\right\}\)thì 4n³ + 11n² + 5n + 5 chia hết cho n + 2

Ta có: \(ab\left(x-5\right)-a^2\left(5-x\right)\)

\(=ab\left(x-5\right)+a^2\left(x-5\right)\)

\(=a\left(x-5\right)\left(a+b\right)\)

ab( x - 5 ) - a²( 5 - x )

= ab( x - 5 ) + a²( x - 5 )

= ( x - 5 )( ab + a² )

= a( x - 5 )( b + a )

x³ - x + y³ - y

= (x³ + y³) - (x + y)

= (x + y)(x² + xy + y²) - (x + y)

= (x + y)(x² + xy + y² - 1)

\(x^3-x+y^3-y\)   

\(=x^3+y^3-x-y\)   

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

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

a³ - a² + 9a - 9

= a²(a - 1) + 9(a - 1)

= (a² + 9)(a - 1)

a3-a2+9a-9=a2(a-1)+9(a-1)=(a2+9)(a-1)

Bài làm :

\(x^2.\left(x-3\right)-4x+12\)

\(=x^2.\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x^2-4\right)\left(x-3\right)\)

\(x^2\left(x-3\right)-4x+12\)   

\(=x^2\left(x-3\right)-4\left(x-3\right)\)   

\(=\left(x-3\right)\left(x^2-4\right)\)   

\(\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

Gọi M là trung điểm BD khi đó ta có:

MB=MD=MA;

CM vuông góc BD ta có:

CD^2=CM^2+MD^2  r dùng pitago là ra