D = a + b a − b a 3 + b 3 a 3 − b 3 = a + b a 3 − b 3 − a − b a 3 + b 3

= a + b a − b a 2 + a b + b 2 − a − b a + b a 2 − a b + b 2

= a + b a − b a 2 + a b + b 2 − a 2 + a b − b 2 = 2 a b a + b a − b

D x = 2 a − b 2 ( a 2 + b 2 ) a 3 − b 3 = 2 a 3 − b 3 − 2 a − b a 2 + b 2

= 2 a − b a 2 + a b + b 2 − 2 a − b a 2 + b 2 = 2 a b ( a − b )

D y = a + b 2 a 3 + b 3 2 ( a 2 + b 2 ) = 2 a + b a 2 + b 2 − 2 ( a 3 + b 3 )

= 2 a + b a 2 + b 2 − 2 a + b a 2 − a b + b 2 = 2 a b ( a + b )

Với a ≠ b ;   a , b ≠ 0 ⇒ D ≠ 0 , hệ phương trình có nghiệm duy nhất

x = D x D = 2 a b a − b 2 a b a − b a + b = 1 a + b x = D y D = 2 a b a + b 2 a b a − b a + b = 1 a − b

Đáp án cần chọn là: B

bai 1

=ax5-x5-9xy-4xy-7x

=ax5-(5x+7x)-(9xy+4xy)

=5ax-12x-13xy

2

M=4a+ab-2b+2a-2b+ab

=6a+2ab-4b

n=6a+2b-ab+2a

=8a+2b-ab

m-n=6a+2ab-4b-8a-2b+ab

=3ab-2a-6b

b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1

A nhé

=Sum(A1,C1)*B2

A) nha bạn

vẫn A nha

Xét hiệu \(2a^2+2b^2-\left(a^3+ab^2\right)=\left(2a^2-a^3\right)+\left(2b^2-ab^2\right)\)

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

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

Do \(a^2+b^2\ge0;\forall a;b\) nên:

\(2a^2+2b^2>a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\2-a>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2\ne0\\a< 2\end{matrix}\right.\)

\(2a^2+2b^2=a^3+ab^2\) khi \(\left[{}\begin{matrix}a^2+b^2=0\\2-a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=b=0\\a=2\end{matrix}\right.\)

\(2a^2+2b^2< a^3+ab^2\) khi \(\left\{{}\begin{matrix}a^2+b^2\ne0\\a>2\end{matrix}\right.\) \(\Rightarrow a>2\)

\(2a^2+2b^2\ge a^3+ab^2\) khi \(2-a\ge0\Leftrightarrow a\le2\)