\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\left(\frac{1}{1-x}+\frac{1}{1+x}\right)+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\left(\frac{2}{1-x^2}+\frac{2}{1+x^2}\right)+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\left(\frac{4}{1-x^4}+\frac{4}{1+x^4}\right)+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\left(\frac{8}{1-x^8}+\frac{8}{1+x^8}\right)+\frac{16}{1+x^{16}}\)

\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)

\(=\frac{32}{1-x^{32}}\)

Ta có : \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{1+x+1-x}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2\left(1-x^2+1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4\left(1-x^4+1+x^4\right)}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8\left(1-x^8+1+x^8\right)}{1-x^{16}}+\frac{16}{1+x^{16}}\)

\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)

\(=\frac{16\left(1-x^{16}+1+x^{16}\right)}{1+x^{32}}\)

\(=\frac{32}{1+x^{32}}\)

Vì \(a\ne1,b\ne1,c\ne1\)\(\Rightarrow a-1\ne0,b-1\ne0,c-1\ne0\)

Ta có : \(B=\frac{\left(a-1\right)^2}{\left(b-1\right)\left(c-1\right)}+\frac{\left(b-1\right)^2}{\left(c-1\right)\left(a-1\right)}+\frac{\left(c-1\right)^2}{\left(a-1\right)\left(b-1\right)}\)

\(=\frac{\left(a-1\right)^3}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}+\frac{\left(b-1\right)^3}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}+\frac{\left(c-1\right)^3}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)

\(=\frac{\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\left(1\right)\)

Lại có : \(\left(a-1\right)+\left(b-1\right)+\left(c-1\right)=\left(a+b+c\right)-3=3-3=0\)

Ta chứng minh tính chất sau : Nếu \(x+y+z=0\)thì \(x^3+y^3+z^3=3xyz\)

Thật vậy :

Ta có : \(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y+z\right)^3-3\left(x+y\right)z-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)[\left(x+y+z\right)^2-3\left(x+y\right)z-3xy]=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2yz+2zx-3zx-3yz-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)luôn đúng , do \(x+y+z=0\)

Áp dụng vào , khi đó : \(\left(1\right)\Leftrightarrow\)\(\frac{3\left(a-1\right)\left(b-1\right)\left(c-1\right)}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)

Vì \(a-1\ne0,b-1\ne0,c-1\ne0\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)\ne0\)

\(\Rightarrow B=3\)

Vậy \(B=3\)

\(B=\frac{\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)

Đặt \(a-1=x,b-1=y,z-1=z\)thì \(x+y+z=0\).

\(B=\frac{x^3+y^3+z^3}{xyz}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz}{xyz}=\frac{3xyz}{xyz}=3\)

tao chơi hayyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy tao đó

Áp dụng bđt: a² + b² > = (a + b)²/2

Cm đúng <=> 2a² + 2b² - a² - 2ab - b² > = 0

<=> (a - b)² > = 0 (luôn đúng với mọi a,b

Khi đó, ta có: A = \(\left(1+\frac{1}{x}\right)^2+\left(1+\frac{1}{y}\right)^2\ge\frac{\left(2+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

Áp dụng bđt: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

CM đúng <=> (a + b)² > = 4ab

<=> (a - b)² > = 0 (luôn đúng với mọi a,b)

Ta lại có: A \(\ge\frac{\left(2+\frac{4}{x+y}\right)^2}{2}=\frac{\left(2+\frac{4}{1}\right)^2}{2}=18\)

Dấu"=" xảy ra <=> x = y = 1/2

Vậy minA = 18/ <=> x = y = 1/2

Đặt \(d=\left(2n+1,2n^2-1\right)\).

\(\hept{\begin{cases}2n+1⋮d\\2n^2-1⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2n^2+n⋮d\\2n^2-1⋮d\end{cases}\Rightarrow}\left[\left(2n^2+n\right)-\left(2n^2-1\right)\right]⋮d\)

\(\Rightarrow\left(n+1\right)⋮d\Rightarrow\left[2\left(n+1\right)-\left(2n+1\right)\right]⋮d\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

\(\Rightarrow\left(2n+1,2n^2-1\right)=1\)

Suy ra đpcm. 

Ta có: \(A=\frac{x^2+25x+144}{x}=x+\frac{144}{x}+25\)

Các số dương : x và \(\frac{144}{x}\) có tích k đổi nên tổng nhỏ nhất và chỉ khi  \(x=\frac{144}{x}\)=> x=12

Vậy Min A = 49 khi và chỉ khi x=12

\(A=\frac{\left(x+16\right)\left(x+9\right)}{x}=\frac{x^2+25x+144}{x}=x+25+\frac{144}{x}\)

Vì \(x>0\)\(\Rightarrow\) Áp dụng bđt Cô si ta có:

\(x+\frac{144}{x}\ge2\sqrt{x.\frac{144}{x}}=2.\sqrt{144}=2.12=24\)

Dấu " = " xảy ra \(\Leftrightarrow x=\frac{144}{x}\)\(\Leftrightarrow x^2=144\)\(\Leftrightarrow x=12\)( do \(x>0\))

\(\Rightarrow A\ge25+24=49\)

Vậy \(minA=49\)\(\Leftrightarrow x=12\)

( 2x - 1 )( x - 3 ) - 2x( x + 5 ) = 0

⇔ 2x² - 7x + 3 - 2x² - 10x = 0

⇔ -17x + 3 = 0

⇔ x = 3/17

\(\left(2x-1\right)\left(x-3\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow2x^2-6x-x+3-2x^2-10x=0\)'

\(\Leftrightarrow-17x+3=0\Leftrightarrow x=\frac{3}{17}\)

Ta có: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{xz}+\frac{1}{yz}\right)\)

\(\left(\sqrt{3}\right)^2=P+\frac{2\left(z+y+x\right)}{xyz}\) 

Mà x+y+z=xyz

=> P+2=3=>P=1

Vậy P=1

 2x² -6x -x +3-2x²-10x=0

-17x+3=0

-17x=-3

x=\(\frac{17}{3}\)