c/m với mọi x thuộc R: \(\left(\frac{12}{5}\right)^2+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)
Chứng minh rằng với mọi \(x\in R\)ta có :
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)
Áp dụng BĐT AM-GM ta có:
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{9^x}=2\cdot3^x\)
\(\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge2\sqrt{25^x}=2\cdot5^x\)
\(\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2\sqrt{16^x}=2\cdot4^x\)
Cộng theo vế ta có: \(2VT\ge2VP\Leftrightarrow VT\ge VP\)
CMR với mọi x ∈ R , ta có: \(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\text{≥}3^x+4^x+5^x\)
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x\ge2\sqrt{\left(\frac{12}{5}\right)^x.\left(\frac{15}{4}\right)^x}=2.3^x;\left(\frac{20}{3}\right)^x+\left(\frac{12}{5}\right)^x\ge2.4^x\)
Cộng các vế tương ứng => đpcm
CMR: moi \(x\in R\) ta có:
\(\left(\frac{12}{5}\right)^x+\left(\frac{15}{4}\right)^x+\left(\frac{20}{3}\right)^x\ge3^x+4^x+5^x\)
Tìm x,biết
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
Với x ∉ -2,-5,-10,-17
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Với x∉1,3,8,20
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
Tìm x:
a) \(\frac{3}{\left(x+2\right)\cdot\left(x+5\right)}\)+\(\frac{5}{\left(x+5\right)\cdot\left(x+10\right)}\)+\(\frac{7}{\left(x+10\right)\cdot\left(x+17\right)}\)= \(\frac{x}{\left(x+2\right)\cdot\left(x+17\right)}\)
Với x không thuộc (-2;-5;-10;-17)
b) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}\)+\(\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}\)+\(\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}\)-\(\frac{1}{20}\)= \(\frac{-3}{4}\)
Với x không thuộc (1;3;8;20)
c)\(\frac{x+1}{2019}\)+\(\frac{x+2}{2018}\)= \(\frac{x-3}{2017}\)\(\frac{x-4}{2016}\)
Bài 2 :
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
b, \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{\left(x-20\right)}=\frac{-3}{4}\)
Tìm x biết: \(\frac{2}{\left(x-1\right).\left(x-3\right)}+\frac{5}{\left(x-3\right).\left(x-8\right)}+\frac{12}{\left(x-8 \right).\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)\(\frac{-3}{4}\)với x\(\notin\){1;3;8;20}
Vì các phân số đều có dạng
=>phần này dễ rùi bn tự lm nhé tích trung tỉ ngoại tỉ
Tìm x \(\frac{2}{\left(x-1\right)\times\left(x-3\right)}+\frac{5}{\left(x-3\right)\times\left(x-8\right)}+\frac{12}{\left(x-8\right)\times\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)-3/4
TÌM x BIẾT:
a,\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
với x\(\notin\){-2;-5;-10;-17}
b,\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
với x\(\notin\){1;3;8;20}
c, TÌM X BIẾT:
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
GIÚP MÌNH CHÚT NHA MÌNH CẦN NGAY. THANKS!