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\(a,x^3-1=-28\\ \Leftrightarrow x^3=-27\\ \Leftrightarrow x^3=\left(-3\right)^3\\ \Leftrightarrow x=-3\\ b,\left(y-1\right)^2-32=-23\\ \Leftrightarrow\left(y-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}y-1=3\\y-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=4\\y=-2\end{matrix}\right.\\ c,15-16:\left|x\right|=-1\\ \Leftrightarrow16:\left|x\right|=16\\ \Leftrightarrow\left|x\right|=1\\ \Leftrightarrow x=\pm1\)
A<1/1*2+1/2*3+...+1/2021*2022
=>A<1-1/2+1/2-1/3+...+1/2021-1/2022<1
Đặt \(A=\dfrac{10^{1987}+1}{10^{1988}+1};B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
Ta có: \(A=\dfrac{10^{1987}+1}{10^{1988}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{1988}+10}{10^{1988}+1}=1+\dfrac{9}{10^{1988}+1}\)
Ta có: \(B=\dfrac{10^{1989}+1}{10^{1990}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{1990}+10}{10^{1990}+1}=1+\dfrac{9}{10^{1990}+1}\)
Ta có: \(10^{1988}+1< 10^{1990}+1\)
\(\Leftrightarrow\dfrac{9}{10^{1988}+1}>\dfrac{9}{10^{1990}+1}\)
\(\Leftrightarrow1+\dfrac{9}{10^{1988}+1}>1+\dfrac{9}{10^{1990}+1}\)
hay A>B
\(\Leftrightarrow n+1\in\left\{1;3\right\}\)
hay \(n\in\left\{0;2\right\}\)
\(2x+1⋮x-1\)
=>\(2x-2+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
2x+1⋮x−12x+1⋮x-1
⇔(2x−2)+3⋮x−1⇔(2x-2)+3⋮x-1
⇔2(x−1)+3⋮x−1⇔2(x-1)+3⋮x-1
Mà x−1⋮x−1x-1⋮x-1
⇒2(x−1)⋮x−1⇒2(x-1)⋮x-1
⇒3⋮x−1⇒3⋮x-1
⇔x−1∈Ư(3)={±1;±3}⇔x-1∈Ư(3)={±1;±3}
⇔x∈{0;2;4;−2}⇔x ∈{0;2;4;-2}
Vậy x∈{0;±2;4}x ∈{0;±2;4} thì 2x+1⋮x−1
Ta có:
\(\dfrac{1}{5}>\dfrac{1}{10}\\ \dfrac{1}{6}>\dfrac{1}{10}\\ ...\\ \dfrac{1}{9}>\dfrac{1}{10}\\ \Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}>\dfrac{5}{10}=\dfrac{1}{2}.\)
Tương tự:
\(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}>\dfrac{5}{15}=\dfrac{1}{3}.\\ \dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}>\dfrac{3}{18}=\dfrac{1}{6}.\)
Cộng vế theo vế ta được \(B>\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\left(đpcm\right)\)
khi làm sai
Khi bạn làm sai