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a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)
b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)
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Ta có:\(x=18\Rightarrow\hept{\begin{cases}x+1=19\\-x-1=-19\end{cases}}\)
Thay vào BT D ta được:
\(D=x^{12}+\left(-x-1\right)x^{11}+\left(x+1\right)x^{10}+\left(-x-1\right)x^9+...+\left(x+1\right)x^2+\left(-x-1\right)x+1\)
\(=x^{12}-x^{12}-x^{11}+x^{11}+x^{10}-x^{10}-x^9+...+x^3+x^2-x^2-x+1\)
\(=1-x\)
`#040911`
`-12(11 - 18) - (x + 1)^2 = 59`
`\Rightarrow -12. (-7) - (x + 1)^2 = 59`
`\Rightarrow 84 - (x + 1)^2 = 59`
`\Rightarrow (x + 1)^2 = 84 - 59`
`\Rightarrow (x + 1)^2 =25`
`\Rightarrow (x + 1)^2 =`\(\left(\pm5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5-1\\x=-5-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-6;4\right\}.\)