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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\hept{\begin{cases}\frac{a+b}{3}=\frac{b+c}{4}\Rightarrow4a+4b=3b+3c\Rightarrow4a+b-3c=0\left(1\right)\\\frac{b+c}{4}=\frac{c+a}{5}\Rightarrow5b+5c=4c+4a\Rightarrow4a-5b-c=0\Rightarrow4a=5b+c\left(2\right)\\\frac{c+a}{5}=\frac{a+b}{3}\Rightarrow3c+3a=5a+5b\Rightarrow2a+5b-3c=0\Rightarrow3c=2a+5b\left(3\right)\end{cases}}\)
Thay (2) vào (1) ta có: 3b=c
Thay (3) và (1) ta có: 2b=a
Vậy M=10a+b-7c+2017=10.2b+b-7.3b+2017=21b-21b+2017=0+2017=2017
![](https://rs.olm.vn/images/avt/0.png?1311)
`Answer:`
\(\frac{a+b}{3}=\frac{b+c}{3}=\frac{c+a}{10}\)
\(\Rightarrow\frac{a+b}{3}=\frac{b+c}{3}\)
\(\Rightarrow a+b=b+c\)
\(\Rightarrow a=c\)
Mặt khác ta có: \(\frac{b+c}{3}=\frac{c+a}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{c+c}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{2c}{10}\)
\(\Rightarrow\frac{b+c}{3}=\frac{c}{5}\)
\(\Rightarrow5\left(b+c\right)=3c\)
\(\Rightarrow5b+5c=3c\)
\(\Rightarrow5b=-2c\)
\(\Rightarrow b=-\frac{2}{5}c\)
Có `M=11a+20b-4c+2020`
`=>M=11c+20(-2/5c)-4c+2020`
`=>M=11c-8c-4c+2020`
`=>M=-c+2020`
![](https://rs.olm.vn/images/avt/0.png?1311)
1. a, \(2^{x+2}.3^{x+1}.5^x=10800\)
\(2^x.2^2.3^x.3.5^x=10800\)
\(\Rightarrow\left(2.3.5\right)^x.12=10800\)
\(\Rightarrow30^x=\frac{10800}{12}=900\)
\(\Rightarrow30^x=30^2\)
\(\Rightarrow x=2\)
b,\(3^{x+2}-3^x=24\)
\(\Rightarrow3^x\left(3^2-1\right)=24\)
\(\Rightarrow3^x.8=24\)\(\Rightarrow3^x=3^1\Rightarrow x=1\)
2, c, Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Dấu bằng xảy ra khi \(ab\ge0\)
Ta có: \(\left|x-2017\right|=\left|2017-x\right|\)
\(\Rightarrow\left|x-1\right|+\left|2017-x\right|\ge\left|x-1+2017-x\right|\)\(=\left|2016\right|=2016\)
Dấu bằng xảy ra khi \(\left(x-1\right)\left(2017-x\right)\ge0\)\(\Rightarrow2017\ge x\ge1\)
Vậy \(Min_{BT}=2016\)khi \(2017\ge x\ge1\)
d, Áp dụng BĐT \(\left|a\right|-\left|b\right|\le\left|a-b\right|\forall a,b\inℝ\)
Dấu bằng xảy ra khi \(b\left(a-b\right)\ge0\)
Ta có \(B=\left|x-2018\right|-\left|x-2017\right|\le\left|x-2018-x+2017\right|\)
\(\Rightarrow B\le1\)
Dấu bằng xảy ra khi \(\left(x-2017\right)\left[\left(x-2018\right)-\left(x-2017\right)\right]\ge0\)
\(\Rightarrow x\le2017\)
Vậy \(Max_B=1\) khi \(x\le2017\)
để BT \(\frac{5}{\sqrt{2x+1}+2}\) nguyên thì \(\sqrt{2x+1}+2\inƯ\left(5\right)\)
suy ra \(\sqrt{2x+1}+2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{2x+1}\in\left\{-7;-3;-1;3\right\}\)
Mà \(\sqrt{2x+1}\ge0\) nên \(\sqrt{2x+1}\)chỉ có thể bằng 3
\(\Rightarrow2x+1=9\Rightarrow x=4\)( thỏa mãn điều kiện \(x\ge-\frac{1}{2}\))
Đây là cách lớp 9. Mk đang phân vân ko biết giải theo cách lớp 7 thế nào!!!!
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{3}=\frac{b+c}{5}=\frac{c+a}{10}=\frac{a+b-b-c-c-a}{-12}=\frac{c}{6}\)
\(\Rightarrow\frac{a+b}{3}=\frac{c}{6}\Rightarrow\left(a+b\right).6=3c\Rightarrow6a+6b=3c\Rightarrow3a+3b=c\Rightarrow a+b=\frac{c}{3}\)
\(\frac{b+c}{5}=\frac{c}{6}\Rightarrow6b+6c=5c\Rightarrow6b=-c\Rightarrow b=\frac{-c}{6}\)
\(\frac{c+a}{10}=\frac{c}{6}\Rightarrow6c+6a=10c\Rightarrow6a=4c\Rightarrow3a=2c\Rightarrow a=\frac{2c}{3}\)
thay vào M ta có:
\(\frac{22c}{3}+\frac{-20c}{6}-c+2017=4c-c+2017=3c+2017\)
p/s: ko chắc :))