Ta có: x=9

nên x+1=10

Ta có: \(x^{14}-10x^{13}+10x^{12}-...+10x^2-10x+10\)

\(=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...+x^3+x^2-x^2-x+x+1\)

=1

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)

Nên x + 1 = 0 => x = -1

b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)

\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)

Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)

Nên x  +15 = 0 => x = -15

a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1

\(a)\) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

Nên \(x+1=0\)

\(\Rightarrow\)\(x=-1\)

Vậy \(x=-1\)

Chúc bạn học tốt ~ 

A=\(\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}\)\(+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}\)\(+\frac{x+2}{13}\)-\(\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\left(x+2\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

x+2=0 vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\)khác\(0\)

x=0-2=-2

Vậy x=-2

Bạn xem lại đề nha \(\dfrac{x+2}{15}\)

\(\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}\)

=> \(\left(x+2\right)\left(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

=> x=-2 do  \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\)\(\ne0\)

Câu 9:

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(9,x^3-2x^2-x+2=0\\ \Rightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)

\(10,\) giống 9

\(11,x^2+4x-5=0\\ \Rightarrow\left(x^2-x\right)+\left(5x-5\right)=0\\ \Rightarrow x\left(x-1\right)+5\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(12,2x^2+4x+2=72\\ \Rightarrow2x^2+4x-70=0\\ \Rightarrow x^2+2x-35=0\\ \Rightarrow\left(x^2-5x\right)+\left(7x-35\right)=0\\ \Rightarrow x\left(x-5\right)+7\left(x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

\(13,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\\ \Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=17\\ \Rightarrow x^3-25x-x^3-8=17\\ \Rightarrow-25x=25\\ \Rightarrow x=-1\)

\(14,2x^3+5x^2-12x=0\\ \Rightarrow x\left(2x^2+5x-12\right)=0\\ \Rightarrow x\left[\left(2x^2+8x\right)-\left(3x+12\right)\right]=0\\ \Rightarrow x\left[2x\left(x+4\right)-3\left(x+4\right)\right]=0\\ \Rightarrow x\left(2x-3\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\\x=-4\end{matrix}\right.\)