Bài 3: Những hằng đẳng thức đáng nhớ

Hãy lấy máy tình ra và tính cái hiệu đó đi.Bạn là thấy 1 điều hiển nhiên rằng:Đề sai

Đề sai rồi bn phải là "+" mới chia hết cho 8000 chứ

Sai đề chăng?

Gíup được câu b thoy.

\(B=\left(5x-2\right)\left(5x+2\right)-\left(5x-1\right)^2\\ =25x^2-4-\left(25x^2-10x+1\right)\\ =25x^2-4-25x^2+10x-1\\ =10x-5=5\left(x-1\right)\)

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

A=2xy+6x+6y

B=\(25x^2-4-25x^2+10x-1\)

B=10x-5

C=\(100^2+98^2+...+2^2-\left(99^2+97^2+...+1\right)\)

C=\(100^2-99^2+98^2-97^2+...+4^2-3^2+2^2-1\)

C=\(\left(100-99\right)\left(100+99\right)+...+\left(2-1\right)\left(2+1\right)\)

C=\(199+195+191+...+7+3\)

C=\(\left(199+3\right)\left(\left(199-3\right):4+1\right)=10100\)

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

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

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

\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ac-3bc-3ab\right]\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ac-3bc-3ab\right)\)

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

\(2x^2-5x+3=2x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(2x-3\right)\)

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

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

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

1)(5x-3y+4z)(5x-3y-4z)=(5x-3y)²-(4z)²

=25x²-30xy+9y²-16z²

Do x²=y²+z²

=>z²=x²-y²

=>(5x-3y+4z)(5x-3y-4z)=25x²-30xy+9y²-16x²+16y²=9x²-30xy+25y²=(3x+5y)²(đpcm)

2)(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)

<=>(a+d)²-(b+c)²=(a-d)²-(b-c)²

<=>(a+d)²-(a-d)²=(b+c)²-(b-c)²

<=>(a+d-a+d)(a+d+a-d)=(b+c-b+c)(b+c+b-c)

<=>4ab=4bc

<=>ad=bc(đpcm)

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

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

\(=x^2+4xy+4y^2-z^2\)

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

a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)

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

b) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

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

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

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

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

Ta có: A = \(\left(x+3y-5\right)^2-6xy+26\)

=> A = \(x^2+9y^2+25+6xy-10x-30y+6xy+26\)

=> A = \(\left(x^2-10x+25\right)+\left(9y^2-30y+25\right)+1\)

=> A = \(\left(x-5\right)^2+\left(3y-5\right)^2+1\)

Vì \(\left\{{}\begin{matrix}\left(x-5\right)^2\ge0\\\left(3y-5\right)^2\ge0\end{matrix}\right.\) => A \(\ge\) 1

=> Dấu bằng xảy ra <=> \(\left\{{}\begin{matrix}x-5=0\\3y-5=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\y=\dfrac{5}{3}\end{matrix}\right.\)

=> GTNN của A =1 khi x = 5; y = \(\dfrac{5}{3}\)