phần A, B bạn làm như bạn nguyễn quang trung còn C,D làm theo mình:

\(C=\frac{2017}{2018}-\left|x-\frac{3}{5}\right|\)

vì \(\left|x-\frac{3}{5}\right|\ge0\forall x\)

nên \(\frac{2017}{2018}-\left|x-\frac{3}{5}\right|\le\frac{2017}{2018}\forall x\)

vậy \(MaxC=\frac{2017}{2018}\Leftrightarrow x=\frac{3}{5}\)

\(D=\left|x-2\right|+\left|y+1\right|+3\)

\(\left|x-2\right|\ge0;\left|y+1\right|\ge0\forall x\)

nên \(\left|x-2\right|+\left|y+1\right|+3\ge3\forall x\)

vậy \(MinA=3\Leftrightarrow x=2;y=-1\)

a ) Ta có : A = \(\left|x+\frac{1}{2}\right|\ge0\forall x\)

Vậy A_min = 0 , khi x = \(-\frac{1}{2}\) 

b) \(B=\left|\frac{3}{7}-x\right|+\frac{1}{9}\)

Mà : \(\left|\frac{3}{7}-x\right|\ge0\forall x\)

Nên : \(B=\left|\frac{3}{7}-x\right|+\frac{1}{9}\ge\frac{1}{9}\forall x\)

Vậy B_min = \(\frac{1}{9}\) kh x = \(\frac{3}{7}\)

\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)

\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)

\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất

Mà \(\left|2018x-2019\right|\ge0\)

\(\Rightarrow\left|2018x-2019\right|+1\ge1\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left|2018x-2019\right|=0\)

\(\Leftrightarrow x=\frac{2019}{2018}\)

Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)

\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)

\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)

\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)

\(\Rightarrow5^x=3^{2x}\)

Mà \(\left(5;3\right)=1\)

\(\Rightarrow x=2x=0\)

trả lời...............................

ok..................................

hk tốt...............................

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!!!!

Ta có: 

\(\frac{1}{1+a}+\frac{2017}{2017+b}+\frac{2018}{2018+c}\le1\)

\(\Leftrightarrow\frac{a}{1+a}\ge\frac{2017}{2017+b}+\frac{2018}{2018+c}\ge2\sqrt{\frac{2017.2018}{\left(2017+b\right)\left(2018+c\right)}}\left(1\right)\)

Tương tự ta cũng có:

\(\hept{\begin{cases}\frac{b}{2017+b}\ge2\sqrt{\frac{2018}{\left(1+a\right)\left(2018+c\right)}}\left(2\right)\\\frac{c}{2018+c}\ge2\sqrt{\frac{2017}{\left(1+a\right)\left(2017+b\right)}}\left(3\right)\end{cases}}\)

Lấy (1), (2), (3) nhân vế theo vế rút gọi ta được

\(abc\ge2\sqrt{2017.2018}.2.\sqrt{2018}.2.\sqrt{2017}=8.2017.2018\)