\(A=x^2+2x+9y^2-6y+2018\)

\(=x^2+2x+1+9y^2-6y+1+2016\)

\(=\left(x+1\right)^2+\left(3y-1\right)^2+2016\ge2016\forall x;y\)

Dấu ''='' xảy ra khi x = -1 ; y = 1/3 

Vậy GTNN của A bằng 2016 tại x = -1 ; y = 1/3 

HOC dot

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\) 

\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ac\)

\(\left(x-2y+1\right)^2=x^2+4y^2+1-4xy-4y+2x\)

\(\left(3x+y-2\right)^2=9x^2+y^2+4+6xy-12x-4y\)

a) -4x² + 8x - 4

= - (4x² - 8x + 4)

= - (2x - 2)²

b) -x5² + 10 x - 5

= - 5(x² - 2x + 1)

= - 5(x - 1)²

-4x^2+8x-4

=-4.(x^2-2x+1)

=-4.(x-1)^2

\(A=n\left(2n-3\right)-2n\left(n+2\right)\)

\(A=n\left(2n-3-2n-4\right)\)

\(A=-7n\)

\(\Rightarrow A\text{ }⋮\text{ }7\)

n(2n-3)-2n(n+2)

=2n²-3n-2n²-4n

= - 7n

Mà -7n ⋮ 7 với mọi n

vậy n(2n-3)-2n(n+2) luôn chia hết cho 7 với mọi n

k mình nha

a) \(\left(a^2-4\right)\left(a^2+4\right)\)

\(=a^4-8\)

c) \(\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\)

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

d) \(\left(a-b+c\right)\left(a+b+c\right)\)

=\(a^2-\left(b+c\right)^2\)

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

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

mik lm tắt có gì sai cho mik xin lỗi

( a² - 4 )( a² + 4 ) = a⁴ - 16

( x³ - 3y )( x³ + 3y ) = x⁶ - 9y²

( a - b )( a + b )( a² + b² )( a⁴ + b⁴ ) = ( a² - b² )( a² + b² )( a⁴ + b⁴ ) = ( a⁴ - b⁴ )( a⁴ + b⁴ ) = a⁸ - b⁸

( a - b + c )( a + b + c ) = ( a + c )² - b² = a² - b² + c² + 2ac

( x + 2 - y )( x - 2 - y ) = ( x - y )² - 2² = x² - 2xy + y² - 4

\(\frac{3^2-1}{5^2-1}\times\frac{7^2-1}{9^2-1}\times\frac{11^2-1}{13^2-1}\times...\times\frac{43^2-1}{45^2-1}\)

\(=\frac{2\times4}{4\times6}\times\frac{6\times8}{8\times10}\times\frac{10\times12}{12\times14}\times...\times\frac{42\times44}{44\times46}\)

\(=\frac{2\times4\times6\times8\times...\times42\times44}{4\times6\times8\times10\times...\times44\times46}\)

\(=\frac{2}{46}=\frac{1}{23}\)

Ta có: 5^2n-25=25^n-25

=(...25)-25=(...00) chia hết cho 100 nha

5^2n - 25 : 100

= 25^n - 25 

= (.....25)-25 =(.....00) chia hết cho 100 nha bạn 

chúc bạn học tốt ạ 

Hạ \(DH\perp AB,CK\perp AB\).

Tam giác \(AHD\)vuông tại \(H\)có\(\widehat{ADH}=30^o\)nên \(AH=\frac{1}{2}AD=1\left(cm\right)\)

\(HD=\sqrt{AD^2-AH^2}=\sqrt{2^2-1^2}=\sqrt{3}\left(cm\right)\)

Tương tự \(BK=1\left(cm\right)\).

\(DC=HK=AB-AH-BK=2,5\left(cm\right)\)

\(S_{ABCD}=\frac{AB+CD}{2}.DH=\frac{4,5+2,5}{2}.\sqrt{3}=\frac{7\sqrt{3}}{2}\left(cm^2\right)\)

Sửa đề: (x-3)(x² + 3x+9) - x(x-4)(x+4)=2

<=> (x³ - 3³) - x(x² - 4²) =2

<=> x³ -27 - x³ + 16x =2

<=> 16x = 2 + 27

<=> 16x = 29

<=> x = 29/16

Vậy x = 29/16

( x - 3 )( x² + 3x + 9 ) + x( x - 4 )( x + 4 ) = 2

<=> x³ - 27 + x( x² - 4 ) - 2 = 0

<=> x³ - 27 + x³ - 4x - 2 = 0

<=> 2x³ - 4x - 29 = 0

đến đây chịu:)

Trả lời:

1, \(\left(x-2\right)^3-\left(x+1\right)\left(x^2-x+1\right)+6\left(x-1\right)^2\)

\(=x^3-6x^2+12x-8-\left(x^3+1\right)+6\left(x^2-2x+1\right)\)

\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)

\(=-2\)

2, \(-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)

\(=\)\(-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)

\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1\)

\(=27\)

xin lỗi nha, ý thứ nhất kết quả là - 3, mình đánh máy nhầm