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\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)
\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
\(A=\dfrac{2\left|y+7\right|+13}{2\left|y+7\right|+6}\le\dfrac{13}{6}\)
Dấu "=" xảy ra khi:
\(y=-7\)
\(B=\dfrac{15\left|x+1\right|+32}{6\left|x+1\right|+8}\le\dfrac{32}{8}=4\)
Dấu "=" xảy ra khi: \(x=-1\)
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)
Thay vào ta được:
\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)
Vậy...................
Chúc bạn học tốt!!!
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)
\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)
\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
a,
\(\dfrac{89}{-13}< 0< \dfrac{1}{123}\\ \Rightarrow\dfrac{89}{-13}< \dfrac{1}{123}\)
Vậy \(\dfrac{89}{-13}< \dfrac{1}{123}\)
b,
\(\dfrac{-13}{15}>\dfrac{-15}{15}=-1=\dfrac{-30}{30}>\dfrac{-31}{30}\)
Vậy \(\dfrac{-13}{15}>\dfrac{-31}{30}\)
c,
\(\dfrac{125}{123}=\dfrac{123}{123}+\dfrac{2}{123}=1+\dfrac{2}{123}\\ \dfrac{99}{97}=\dfrac{97}{97}+\dfrac{2}{97}=1+\dfrac{2}{97}\)
Vì \(\dfrac{2}{97}>\dfrac{2}{123}\Rightarrow1+\dfrac{2}{97}>1+\dfrac{2}{123}\Leftrightarrow\dfrac{99}{97}>\dfrac{125}{123}\)
Vậy \(\dfrac{99}{97}>\dfrac{125}{123}\)
d,
\(\dfrac{125}{126}< \dfrac{126}{126}=1=\dfrac{986}{986}< \dfrac{987}{986}\)
Vậy \(\dfrac{125}{126}< \dfrac{987}{986}\)
Đặt: \(\left|x+1\right|=t\ge0\)
Ta có: \(A=\dfrac{15t+32}{6t+8}=\dfrac{6t+8+6t+8+3t+16}{6t+8}\)
\(=\dfrac{6t+8}{6t+8}+\dfrac{6t+8}{6t+8}+\dfrac{3t+16}{6t+8}\)
\(=1+1+\dfrac{3t+16}{6t+8}\)
\(=1+1+\dfrac{3t+4+12}{6t+8}=1+1+\dfrac{3t+4}{6t+8}+\dfrac{12}{6t+8}\)
\(\le1+1+\dfrac{1}{2}+\dfrac{12}{8}=1+1+\dfrac{1}{2}+\dfrac{3}{2}=4\)
Dấu "=" xảy ra khi: \(t=0\) hay \(x=-1\)
t nghĩ đề là max nhé