Cho 1 mũ 2 + 2 mũ 2 + 3 mũ 2 +......+10 mũ 2 = 385 Tính nhanh: P= 3 mũ 2 + 6 mũ 2 +9 mũ 2 +.....30 mũ 2

Ta có : P = 3² + 6² + 9² + .... + 30²

= 3²(1² + 2² + 3² + .... + 10²)

= 9.385 

= 3465

Vậy P = 3465

P = 3² + 6² + 9² + ... + 30²

= ( 1.3 )² + ( 2.3 )² + ( 3.3 )² + ... + ( 10.3 )²

= 1².3² + 2².3² + 3².3² + ... + 10².3²

= 3²( 1² + 2² + 3² + ... + 10² )

= 9.385 = 3465

a) Ta có \(\hept{\begin{cases}\left(x-2y\right)^2\ge0\forall x;y\\\left(y+1\right)^6\ge0\forall y\end{cases}}\Rightarrow\left(x-2y\right)^2+\left(y+1\right)^6\ge0\forall x;y\)

=> (x - 2y)² + (y + 1)⁶ = 0

<=> \(\hept{\begin{cases}x-2y=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\y=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}\)

b) \(\left(\frac{2x}{3}\right)^2+10x=0\)

=> \(\frac{4x^2}{9}+10x=0\)

=> \(x\left(\frac{4x}{9}+10\right)=0\)

=> \(\orbr{\begin{cases}x=0\\\frac{4x}{9}+10=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\frac{4x}{9}=-10\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-22,5\end{cases}}\)

Vậy \(x\in\left\{0;-22,5\right\}\)

11÷2=5,5

11 chia đôi bằng 1

A = 31 - 43 - (-53) + 721 - 92 - 361 + 151

A = (-12 + - 51) + 721 + 151 - 92 - 361

A = (-63) + 780 - 361

A = 780 - 424

A = 356

\(3^{2n+1}+5.2^{3n+1}\)

Với \(n=1\)thì \(3^5+5.2^4=243+80=323⋮19\)

Gải sử \(3^{2k+1}+5.2^{3k+1}⋮19\)

Xét \(3^{3k+5}+5.2^{3k+4}=3^{3k+2}.3^3+5.2^{3k+1}.2^3\)

\(=27\left(3^{3k+2}+5.2^{3k+1}\right)-19.3^{2k+1}⋮19\)

Đặt \(S_n=3^{2n+1}+40n-67\)

Xét \(n=1\Rightarrow S_n=0⋮64\)

Giả sử n đúng với \(n=k\left(k\inℤ^+\right)\)tức là ta có :

\(S_k=3^{2k+1}+40k-67⋮64\). Ta cần chứng minh n đúng với \(n=k+1\).

Tức cần chứng minh \(S_{k+1}=2^{2\left(k+1\right)+1}+40\left(k+1\right)-67⋮64\)

Thật vậy ta có : \(S_{k+1}=2^{2\left(k+1\right)+1}+40\left(k+1\right)-67\)

\(=9\cdot2^{2k+1}+40k-27\)

\(=9\cdot\left(2^{2k+1}+40k-67\right)-320k+576\)

\(=9\cdot S_k-320k+576⋮64\)

Vậy n đúng với \(n=k+1\)

Do đó \(S_n=3^{2n+1}+40n-67⋮64\forall n\inℤ^+\)

Với \(n=1\)thì \(3^3+40-67=0⋮64\)

Giả sử \(3^{2k+1}+40k-67⋮64\)

Xét \(3^{2k+3}+40\left(k+1\right)-67\)

\(=9\left(3^{2k+1}+40k-67\right)+64\left(9-5k\right)⋮64\)

\(\)

\(S_n=\frac{1.2.3.4...n\left(n+1\right)\left(n+2\right)...2n}{1.2.3.4...n}\)

\(=\frac{1.3...\left(2n-1\right).2.4...\left(2n-2\right)2n}{1.2.3.4...n}\)

\(=\frac{1.3...\left(2n-1\right).2^n.1.2...n}{1.2...n}\)

\(=2^n.1.3...\left(2n-1\right)⋮2n\)

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

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

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

\(=\left(x+y+1\right)^2+\left(x-2\right)^2-3\ge-3\forall x,y\)

Dấu"="xảy ra khi \(\orbr{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}y=-3\\x=2\end{cases}}}\)

Vậy.....

A = 2x² + 2xy + y² - 2x + 2y + 2

= ( x² + 2xy + y² + 2x + 2y + 1 ) + ( x² - 4x + 4 ) - 3

= [ ( x + y )² + 2( x + y ) + 1² ] + ( x - 2 )² - 3

= ( x + y + 1 )² + ( x - 2 )² - 3 ≥ -3 ∀ x, y

Dấu "=" xảy ra <=> x = 2 ; y = -3

=> MinA = -3 <=> x = 2 ; y = -3

B thì nhờ các cao nhân khác ._. Em tịt rồi