\(M=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+b^2+2ab-2ab\right)+6a^2b^2\left(a+b\right)\)

\(M=a^2+2ab+b^2-3ab+3ab-6a^2b^2+6a^2b^2\)

\(M=\left(a+b\right)^2=1\)

ngu lắm sơn à

bạn Nguyễn Xuân Bảo có làm đc ko mà nói bạn đăng bài ngu :)) đây là trang học toán thì bạn ấy đăng bài ko bt làm lên thì đã sao :>

\(1,M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay \(a+b=1\) vào ta được:

\(1\left(1-3ab\right)+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)

\(=1\)

Vậy ......................

a/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)

\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)

Suy ra Min B = 20 <=> x = 1/3

a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)

\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)

\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)

Vì \(\left(x^2-5x\right)^2\ge0\)

=> \(\left(x^2-5x\right)^2-36\ge-36\)

Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)

b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)

Vì \(-\left(3x+1\right)^2\le0\)

=> \(-\left(3x+1\right)+20\le20\)

Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)

B = 19 - 6x - 9x²

= - (9x² + 6x + 1 - 20)

= - [(3x + 1)² - 20]

(3x + 1)² lớn hơn hoặc bằng 0

(3x + 1)² + 20 lớn hơn hoặc bằng 20

- [(3x + 1)² + 20] nhỏ hơn hoặc bằng - 20

Vậy Max B = - 20 khi x = -1/3

\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

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

\(=1-3ab+3ab\left(a+b\right)^2\)

= 1

\(M=a^3+b^2+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(M=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left(\left(a+b\right)^2-2ab\right)+6a^2b^2\left(a+b\right)\)

\(M=1^3-3ab.1+3ab\left(1^2-2ab\right)+6a^2b^2.1\)

\(M=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)

vậy \(M=1\) khi \(a+b=1\)

M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)

M = (a + b).(a² - ab + b²) + 3ab[a² + b² + 2ab(a + b)]

M = a² - ab + b² + 3ab(a² + b² + 2ab)

M = a² - ab + b² + 3ab(a + b)²

M = a² - ab + b² + 3ab

M = a² + 2ab + b²

M = (a + b)² = 1

ad bdt cosi ta co can((2-a)(2-b)) =b<= (2-a+2-b)/2 = (4-a-b)/2  TT...                                                                                                            cong ve vs ve ta dc Q<= 12-2(a+b+c) ....Q <=4. dau = xra khi a=b=c=2/3

       \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Rightarrow\left(2x^2+4xy+2y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)

Khi đó: \(A=\left(-1+1\right)^{2014}+\left(-1+2\right)^{2015}+\left(1-1\right)^{2016}\)

\(=0+1+0=1\)

M = a³ + b³ + 3ab(a² + b²) + 6a²b²(a + b)

= (a + b)(a² - ab + b²) + 3ab((a + b)² - 2ab) + 6a²b²(a + b)

= (a + b)((a + b)² - 3ab) + 3ab((a + b)² - 2ab) + 6a²b²(a + b)

= 1 - 3ab + 3ab(1 - 2ab) + 6a²b²

= 1 - 3ab + 3ab - 6a²b² + 6a²b² = 1

câu 2: Ta có:

\(pt\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\Leftrightarrow2a^2+2b^2+2c^2-2ac-2ab-2bc=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow a=b=c}\)