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

\(x^2-3x+5\left(x-3\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

Vậy \(x=3\)hoặc \(x=-5\)

\(x^2-3x+5\left(x-3\right)=0\) 

\(x^2-3x+5x-15=0\) 

\(x^2+2x-15=0\)  

\(x^2-3x+5x-15=0\) 

\(x\left(x-3\right)+5\left(x-3\right)=0\) 

\(\left(x+5\right)\left(x-3\right)=0\) 

\(\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\) 

\(\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

a) x² - xy - 20y²

= x² + 4xy - 5xy - 20y²

= x( x + 4y ) - 5y( x + 4y )

= ( x + 4y )( x - 5y )

b) x³ - x²y - 3xy² + 2y³

= x³ + x²y - 2x²y - xy² - 2xy² + 2y³

= ( x³ + x²y - xy² ) - ( 2x²y + 2xy² - 2y³ )

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

= ( x² + xy - y² )( x - 2y )

\(3\frac{2}{5}=3+\frac{2}{5}=3+0,4=3,4\)

\(\frac{2}{5}=0,4\)

\(\Rightarrow3\frac{2}{5}=3,4\)

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)

\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)

\(=\frac{3}{2^2}=\frac{3}{4}\)

Bài làm :

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{\left(2.3\right)^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{2^{10}.3^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{2^8.3^9.\left(2^2.3-5\right)}{3^8.2^{10}.\left(3+2^2\right)}\)

\(=\frac{3.7}{2^2.7}\)

\(=\frac{3}{4}\)

Học tốt

11 . 12 + 33 . 36 + 55 . 60

= 11 . 12 + 11 . 3 . 12 . 3 + 11 . 5 . 12 . 5

= (11 + 12) . 1 . (11 + 12) . 3 . (11 + 12) . 5

= 23 . 1 . 23 . 3 . 23 . 5

= 23 . (1 + 3 + 5)

= 23 . 9

= 207

\(11\cdot12+33\cdot36+55\cdot60\) 

\(=11\cdot12+3\cdot11\cdot3\cdot12+5\cdot11\cdot5\cdot12\) 

\(=11\cdot12\left(1+9+25\right)\) 

\(=132\cdot35\) 

\(=4620\)

33 x 55 + 33 x 66 - 33 x 21

= 33 x ( 55 + 66 - 21 )

= 33 x 100

= 3300

\(33.55+33.66-33.21=33.\left(55+66-21\right)=33.100=3300\)