Áp dụng BĐT BSC và BĐT Cosi:

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\ge17\left(x+y+z\right)+\frac{2.\left(1+1+1\right)^2}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{18}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{17}{x+y+z}+\frac{1}{x+y+z}\)

\(\ge2\sqrt{17\left(x+y+z\right).\frac{17}{x+y+z}}+\frac{1}{1}\)

\(=35\)

\(\Rightarrow17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Áp dụng bất đẳng thức AM-GM kết hợp giả thiết x + y + z ≤ 1 ta có :

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=17x+17y+17z+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)

\(=\left(18x+\frac{2}{x}\right)+\left(18y+\frac{2}{y}\right)+\left(18z+\frac{2}{z}\right)-\left(x+y+z\right)\)

\(\ge2\sqrt{18x\cdot\frac{2}{x}}+2\sqrt{18y\cdot\frac{2}{y}}+2\sqrt{18z\cdot\frac{2}{z}}-1=12\cdot3-1=35\)( đpcm )

Dấu "=" xảy ra <=> x=y=z=1/3

Sai thi thong cam ...

\(\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow\left(x+y+z\right)^3=x^3+y^3+z^3+3yz+3xz^2+3yx^2+3zx^2+3xy^2+3xyz+3zy^2+3yz^2\)

\(\Leftrightarrow\left(x+y+z\right)^3=x^3+3x^2y+3y^2x+y^3+3x^2z+6xyz+3y^2z+3xz^2+z^3\)

\(\Leftrightarrow\left(x+y+z\right)^3=\left(x+y\right)^3+3z\left(x+y\right)^2+3z^2\left(x+y\right)+z^3\)

\(\Leftrightarrow\left(x+y+z\right)^3=\left(x+y+z\right)^3\)

\(M=x^2-2xy+4y^2+12xy+22\)

\(M=\left(x^2-2xy+y^2\right)+\left(3y^2+12y+12\right)+10\)

\(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=-2\) 

( Chỗ \(M=\left(x-y\right)^2+3\left(x+2\right)^2+10\ge10\forall x;y\) bạn phân tích từng cái đã nhá, mình làm tắt ) 

\(A=x^2+2x+9y^2-6y+2018\)

\(=x^2+2x+1+9y^2-6y+1+2016\)

\(=\left(x+1\right)^2+\left(3y-1\right)^2+2016\ge2016\forall x;y\)

Dấu ''='' xảy ra khi x = -1 ; y = 1/3 

Vậy GTNN của A bằng 2016 tại x = -1 ; y = 1/3 

HOC dot

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

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

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

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

\(\left(3x+y-2\right)^2=9x^2+y^2+4+6xy-12x-4y\)

a) -4x² + 8x - 4

= - (4x² - 8x + 4)

= - (2x - 2)²

b) -x5² + 10 x - 5

= - 5(x² - 2x + 1)

= - 5(x - 1)²

-4x^2+8x-4

=-4.(x^2-2x+1)

=-4.(x-1)^2

\(A=n\left(2n-3\right)-2n\left(n+2\right)\)

\(A=n\left(2n-3-2n-4\right)\)

\(A=-7n\)

\(\Rightarrow A\text{ }⋮\text{ }7\)

n(2n-3)-2n(n+2)

=2n²-3n-2n²-4n

= - 7n

Mà -7n ⋮ 7 với mọi n

vậy n(2n-3)-2n(n+2) luôn chia hết cho 7 với mọi n

k mình nha

a) \(\left(a^2-4\right)\left(a^2+4\right)\)

\(=a^4-8\)

c) \(\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\)

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

d) \(\left(a-b+c\right)\left(a+b+c\right)\)

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

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

=\(x-\left(2-y\right)\)

mik lm tắt có gì sai cho mik xin lỗi

( a² - 4 )( a² + 4 ) = a⁴ - 16

( x³ - 3y )( x³ + 3y ) = x⁶ - 9y²

( a - b )( a + b )( a² + b² )( a⁴ + b⁴ ) = ( a² - b² )( a² + b² )( a⁴ + b⁴ ) = ( a⁴ - b⁴ )( a⁴ + b⁴ ) = a⁸ - b⁸

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

( x + 2 - y )( x - 2 - y ) = ( x - y )² - 2² = x² - 2xy + y² - 4

\(\frac{3^2-1}{5^2-1}\times\frac{7^2-1}{9^2-1}\times\frac{11^2-1}{13^2-1}\times...\times\frac{43^2-1}{45^2-1}\)

\(=\frac{2\times4}{4\times6}\times\frac{6\times8}{8\times10}\times\frac{10\times12}{12\times14}\times...\times\frac{42\times44}{44\times46}\)

\(=\frac{2\times4\times6\times8\times...\times42\times44}{4\times6\times8\times10\times...\times44\times46}\)

\(=\frac{2}{46}=\frac{1}{23}\)

Ta có: 5^2n-25=25^n-25

=(...25)-25=(...00) chia hết cho 100 nha

5^2n - 25 : 100

= 25^n - 25 

= (.....25)-25 =(.....00) chia hết cho 100 nha bạn 

chúc bạn học tốt ạ