\(\dfrac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}=1\)

Áp dụng BĐT Bunhiacopxki:

\(\left(\dfrac{2}{3}+1+\dfrac{1}{3}\right)\left(\dfrac{3}{2}x^2+\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}\right)\ge\left(\sqrt{\dfrac{2}{3}.\dfrac{3}{2}x^2}+\sqrt{1.\left(y+\dfrac{z}{2}\right)^2}+\sqrt{\dfrac{1}{3}.\dfrac{3z^2}{4}}\right)^2\)

\(\Leftrightarrow2.1\ge\left(x+y+\dfrac{z}{2}+\dfrac{z}{2}\right)^2=\left(x+y+z\right)^2\)

\(\Rightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(\frac{3x^2}{2}+y^2+z^2+yz=1\)

\(\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)

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

\(\Rightarrow\left(x+y+z\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)

\(x+y\le2\Rightarrow-\left(x+y\right)\ge-2\)

Do đó:

\(A=2\left(x+\dfrac{1}{x}\right)+2\left(y+\dfrac{1}{y}\right)-\left(x+y\right)\ge2.2\sqrt{x.\dfrac{1}{x}}+2.2\sqrt{y.\dfrac{1}{y}}-2=6\)

\(A_{min}=6\) khi \(x=y=1\)

Gọi vận tốc xe đi từ A đến B là x ( x> 0 ) 

xe đi từ B đến A là x - 3 

Theo bài ra ta có pt \(2x+2x-6=46\Leftrightarrow4x=52\Leftrightarrow x=13\left(tm\right)\)

Vậy vận tốc xe đi từ A đến B là 13 km/h 

vận tốc xe đi từ B đến A là 10 km/h 

Đặt \(2y=a\)thì ta được

\(P=\frac{1}{x^2+a^2}+\frac{1}{xa}=\left(\frac{1}{x^2+a^2}+\frac{1}{2xa}\right)+\frac{1}{2xa}\)

\(\ge\frac{4}{x^2+a^2+2ax}+\frac{2}{\left(x+a\right)^2}=\frac{6}{\left(x+a\right)^2}\ge\frac{6}{4}=\frac{3}{2}\)

\(Q=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\left(\dfrac{1}{ab}+ab\right)+\dfrac{1}{2ab}\)

\(Q\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{ab}{ab}}+\dfrac{2}{\left(a+b\right)^2}\)

\(Q\ge\dfrac{6}{\left(a+b\right)^2}+2\ge\dfrac{6}{2^2}+2=\dfrac{7}{2}\)

\(Q_{min}=\dfrac{7}{2}\) khi \(a=b=1\)

\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)

\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)

Cũa mị:>>>

Tham khảo ạ !!!

A = 100² - 99² + 98² - 97² + ...... + 2² - 1²

= ( 100 - 99 ) ( 100 + 99 ) + ( 98 - 97 ) ( 98 + 97 ) + ......... + ( 2 - 1 ) ( 2 + 1 )

= 1 + 2 + 3 + ......... + 99 + 100

= ( 100 + 1 ) . 100 : 2 = 5050 

 B = 3 ( 2² + 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1²

= ( 2² - 1 ) ( 2² + 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2⁴ - 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2⁸ - 1 ) ( 2⁸ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2¹⁶ - 1 ) ( 2¹⁶ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2³² - 1 ) ( 2³² + 1 ) ( 2⁶⁴ + 1 ) + 1

= ( 2⁶⁴ - 1 ) ( 2⁶⁴ + 1 ) + 1

= 2¹²⁸ - 1 + 1 

= 2¹²⁸

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

= a² + b² + c² + 2ab + 2bc + 2ca + a² + ab - ac + ab + b² - bc - ac - bc + c² - 2 ( a² + 2ab + b² )

= a² + b² + c² + 2ab + 2bc + 2ca + a² + ab - ac + ab + b² - bc - ac - bc + c² - 2a² - 4ab - 2b²

= 2c²

Theo hệ quả Ta lét \(\dfrac{AK}{AC}=\dfrac{NK}{BC}\Rightarrow NK=\dfrac{AK.BC}{AC}=4\)

có HK//BC

=> ak/ac = x/bc
=>3/(3+1,5) = x/6
=>3/4,5 = x/6

=> x = 4 or NK = 4

\(\left(\dfrac{7}{2}x-7\right)\left(21x-63\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x-7=0\\21x-63=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x=7\\21x=63\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

TH1: 2

TH2: 3

ta giả sử rằng : 

\(\hept{\begin{cases}m=a_1^2+a_2^2\\n=b_1^2+b_2^2\end{cases}}\text{ với }a_1,a_2,b_1,b_2\text{ là các số tự nhiên}\)

khi đó : \(mn=\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)=a_1^2b_1^2+a_2^2b_2^2+2a_1a_2b_1b_2+a_1^2b_2^2+a_2^2b_1^2-2a_1a_2b_1b_2\)

\(=\left(a_1b_1+a_2b_2\right)^2+\left(a_1b_2-a_2b_1\right)^2\)

Vậy mn cũng là tổng của hai số chính phương