Chắc là có điều kiện \(a,b\)là số nguyên. 

\(a^2+1=b^2\Leftrightarrow\left(b-a\right)\left(b+a\right)=1\)

\(\Rightarrow\hept{\begin{cases}b-a=1\\b+a=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=1\end{cases}}\)hoặc \(\hept{\begin{cases}b-a=-1\\b+a=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}b=-1\\a=0\end{cases}}\)

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

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

=> \(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{\left(a+b+c\right)c}\)

Khi a + b = 0 

=> (a + b)(b + c)(c + a) = 0 (1)

Khi a + b \(\ne\)0

=> ab = -(a + b + c).c

=> ab + ac + bc + c² = 0

=> a(b + c) + c(b + c) = 0

=> (a + c)(b + c) = 0

=> (a + b)(a + c)(b + c) = 0 (2)

Từ (1)(2) => (a + b)(a + c)(b + c) = 0

Khi đó Q = (a³ + b³)(b⁵ + c⁵)(a⁷ + c⁷)

= (a + b)(a² - ab + b²)(b + c)(b⁴ - b³c - b²c² - bc³ - c⁴)(a + c)(a⁶ - a⁵b - a⁴b² - a³b³ - a²b⁴ - ab⁵ - b⁶)

= (a + b)(b + c)(c + a)(a² - ab + b²)(b⁴ - b³c - b²c² - bc³ - c⁴)(a⁶ - a⁵b - a⁴b² - a³b³ - a²b⁴ - ab⁵ - b⁶) 

= 0

Sửa lại chỗ a⁷ + c⁷ =  (a + c)(a⁶ - a⁵c - a⁴c² - a³c³ - a²c⁴ - ac⁵ - c⁶)

Mình làm ý tổng quát nhé. 

\(\frac{MA}{MB}=\frac{m}{n}\Leftrightarrow MA=\frac{m}{n}MB\)

\(\frac{AM}{AB}=\frac{AM}{AM+MB}=\frac{\frac{m}{n}MB}{\frac{m}{n}MB+MB}=\frac{\frac{m}{n}}{\frac{m}{n}+1}=\frac{m}{m+n}\)

\(\frac{MB}{AB}=\frac{AB-MA}{AB}=1-\frac{MA}{AB}=1-\frac{m}{m+n}=\frac{n}{m+n}\)

x² - y² + 3x + 10y - 23 = 0

=> (x² + 3x + 9/4) - (y² - 10y + 25) = 0,25

=> (x + 3/2)² - (y - 5)² = 0,25

=> 4[(x + 3/2)² - (y - 5)²] = 0,25.4

=> [2(x + 3/2)]² - [2(y - 5)²] = 1

=> (2x + 3)² - (2y - 10)² = 1

=> (2x + 2y - 7)(2x - 2y + 13) = 1

Lập bảng xét các trường hợp

Với x = -1 ; y = 5 => tm đề bài

Với x = -2 ; y = 5 => tm đề bài

Vậy các cặp (x;y) thỏa mãn là (-1;5) ; (-2;5)

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

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

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

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

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

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

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

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

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

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

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