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Các cậu cứu tớ đi ... híc 
Đề cậu viết khó nhìn qá :)
Bài 1 :
Ta có :
\(a+b+c=2014\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{9}\)
\(\Leftrightarrow2014\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=2014.\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{2014}{a+b}+\dfrac{2014}{b+c}+\dfrac{2014}{c+a}=\dfrac{2014}{9}\)
Mà \(a+b+c=2014\) nên :
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2014}{9}\)
\(\Leftrightarrow\left(\dfrac{a+b}{a+b}+\dfrac{c}{a+b}\right)+\left(\dfrac{b+c}{b+c}+\dfrac{a}{b+c}\right)+\left(\dfrac{c+a}{c+a}+\dfrac{b}{c+a}\right)=\dfrac{2014}{9}\)
\(\Leftrightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{2014}{9}\)
\(\Leftrightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}=\dfrac{1987}{9}\)
\(\Leftrightarrow S=\dfrac{1987}{9}\)
Ta có: \(x^2=yz\)
\(\Rightarrow\frac{x^2+y^2}{x^2+z^2}=\frac{yz+y^2}{yz+z^2}=\frac{y\left(y+z\right)}{z\left(y+z\right)}=\frac{y}{z}\)
đpcm
Tham khảo nhé~
Bài 1:
Ta có: \(\left|x-2017\right|+\left|x+2018\right|=\left|2017-x\right|+\left| x+2018\right|\)
Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có:
\(\left|2017-x\right|+\left|x+2018\right|\ge\left|2017-x+x+2018\right|=4035\)
Dấu "=" sảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}2017-x\ge0\\x+2018\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le2017\\x\ge-2018\end{matrix}\right.\Rightarrow-2018\le x\le2017\)
Vậy.....................
Bài 2:
Ta có:
\(\left\{{}\begin{matrix}\dfrac{1}{2!}=\dfrac{1}{1.2}\\\dfrac{1}{3!}=\dfrac{1}{2.3}\\.....\\\dfrac{1}{2017!}< \dfrac{1}{2016.2017}\end{matrix}\right.\)
\(\Rightarrow1+\dfrac{1}{1!}+\dfrac{1}{2!}+....+\dfrac{1}{2017!}< 1+1+\dfrac{1}{1.2}+...+\dfrac{1}{2016.2017}\)
Ta lại có:
\(1+1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2016.2017}\)
\(=2+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(=2+1-\dfrac{1}{2017}=3-\dfrac{1}{2017}\)
\(\Rightarrow1+1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{2016.2017}< 3\)
Do đó: \(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+....+\dfrac{1}{2017!}< 3\)(đpcm)
Chúc bạn học tốt!!!
\(\left(x-3\right).\left(x-2015\right)< 0\)
\(\Rightarrow\left(x-3\right)và\left(x-2015\right)\) phải khác dấu
\(\Rightarrow\left(x-3\right)< \left(x-2015\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x-3>0\\x-2015< 0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>3\\x< 2015\end{matrix}\right.\)
\(\Rightarrow3< x< 2015\)
\(\Rightarrow x\in\left\{4;5;6;7;8;...;2013;2014\right\}\)
( ko bt đúng hay sai nx )
thám tử
\(\left(x-3\right)\left(x-2015\right)< 0\)
Với mọi \(x\in R\) thì:
\(x-2015< x-3\)
Khi đó: \(\left\{{}\begin{matrix}x-2015< 0\\x-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2015\\x>3\end{matrix}\right.\)
Nên \(3< x< 2015\)
a/ xem lại đề
b/đặt: \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)
\(\Rightarrow x=12k;y=9k;z=5k\)
\(\Rightarrow xyz=12k\cdot9k\cdot5k=540k^3=20\)
\(\Rightarrow k^3=\dfrac{1}{27}\Rightarrow k=\dfrac{1}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}x=12k=12\cdot\dfrac{1}{3}=4\\y=9k=9\cdot\dfrac{1}{3}=3\\z=5k=5\cdot\dfrac{1}{3}=\dfrac{5}{3}\end{matrix}\right.\)
Vậy........
c/ Áp dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{12x-15y}{7}=\dfrac{20z-12x}{9}=\dfrac{15y-20z}{11}=\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}=\dfrac{0}{27}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{12x-15y}{7}=0\\\dfrac{20z-12x}{9}=0\\\dfrac{15y-20z}{11}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x-15y=0\\20z-12x=0\\15y-20z=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)=> \(12x=15y=20z\)
\(\Rightarrow\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{20}}\)
A/dụng t/c của dãy tỉ số = nhau có:
\(\dfrac{x}{\dfrac{1}{12}}=\dfrac{y}{\dfrac{1}{15}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x+y+z}{\dfrac{1}{12}+\dfrac{1}{15}+\dfrac{1}{20}}=\dfrac{48}{\dfrac{1}{5}}=240\)
\(\Rightarrow\left\{{}\begin{matrix}x=240\cdot\dfrac{1}{12}=20\\y=240\cdot\dfrac{1}{15}=16\\z=240\cdot\dfrac{1}{20}=12\end{matrix}\right.\)
Vậy......
a) sai đề bn nhé:
\(\frac{x}{2}\) = \(\frac{y}{3}\); \(\frac{y}{4}\) = \(\frac{z}{5}\) và x2 - y2 = -16
\(\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|=0\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|y+\dfrac{2}{3}\right|\ge0\forall y\\\left|x^2+xz\right|\ge0\forall x;z\end{matrix}\right.\) \(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|y+\dfrac{2}{3}\right|+\left|x^2+xz\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{2}{3}\\z=-\dfrac{1}{2}\end{matrix}\right.\)