tìm x sao cho x+2/10^10+x+2/11^11+x+2/12^12=x+2/13^13+x+2/14^14
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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
tìm x
\(A=\frac{x+2}{10}+\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)
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:
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)
\(9,\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow x^2+5x-x-5=0\\ \Leftrightarrow\left(x+5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ 12,\Leftrightarrow\left(x+1\right)^2-36=0\\ \Leftrightarrow\left(x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\\ 13,\Leftrightarrow x^3-25x-x^3-8=17\\ \Leftrightarrow-25x=25\Leftrightarrow x=-1\\ 14,\Leftrightarrow x\left(2x^2+8x-3x-12\right)=0\\ \Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=\dfrac{3}{2}\end{matrix}\right.\)
X=-2
Tick rồi mk ns cách làm cho,hứa