a) Ta có: \(\left(2x+\frac{1}{4}\right)^4\ge0\Rightarrow\left(2x+\frac{1}{4}\right)^4+6\ge6\)

Dấu "=" xảy ra khi \(2x+\frac{1}{4}=0\Rightarrow2x=\frac{-1}{4}\Rightarrow x=\frac{-1}{8}\)

Vậy E_min = 6 \(\Leftrightarrow x=\frac{-1}{8}\)

b) Ta có: \(\left(5-3x\right)^2\ge0\Rightarrow\left(5-3x\right)^2-2013\ge-2013\)

Dấu "=" xảy ra khi \(5-3x=0\Rightarrow3x=5\Rightarrow x=\frac{5}{3}\)

Vậy E_min = -2013 \(\Leftrightarrow x=\frac{5}{3}\)

Mấy bài còn lại làm tương tự.

6

-2013

2013

-1

2014

2016

a, \(\dfrac{6}{2x+1}\Rightarrow2x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

c, \(\dfrac{x-3}{x-1}=\dfrac{x-1-2}{x-1}=1-\dfrac{2}{x-1}\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

tương tự .... 

a) C = 20013 - $\left|5-2x\right|$

$do\; \backslash (-\backslash left|5-2x\backslash right|\backslash le0\backslash forall\; x\backslash )$

=> 20013-\(\left|5-2x\right|\le20013\)

=>A≤20013

=> GTLN C =20013 khi 5-2x=0

=> 2x=5

=> x=\(\dfrac{5}{2}\)

vậy GTLN C = 20013 khi x=\(\dfrac{5}{2}\)

b) D = 7 - \(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\)

do \(-\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le0\forall x\)

=> 7-\(\left|\dfrac{2}{3}+\dfrac{1}{4}x\right|\le7\)

=> D≤7

=> GTLN D =7 khi \(\dfrac{2}{3}+\dfrac{1}{4}x=0\)

=> x=-\(\dfrac{8}{3}\)

a.

\(A=x^2+\dfrac{2021}{x}=x^2+\dfrac{2021}{2x}+\dfrac{2021}{2x}\ge3\sqrt[3]{\dfrac{2021^2}{4x^2}}=3\sqrt[3]{\dfrac{2021^2}{4}}\)

Dấu "=" xảy ra khi \(x=\sqrt[3]{\dfrac{2021}{3}}\)

b.

\(B=4\left(x-1\right)+\dfrac{25}{x-1}+4\ge2\sqrt{\dfrac{100\left(x-1\right)}{x-1}}+4=24\)

Dấu "=" xảy ra khi \(x=\dfrac{7}{2}\)

c.

\(C=3x+\dfrac{16}{x^3}=x+x+x+\dfrac{16}{x^3}\ge4\sqrt[4]{\dfrac{16x^3}{x^3}}=8\)

\(A_{min}=8\) khi \(x=2\)

d.

\(D=x+\dfrac{1}{x}=\left(\dfrac{x}{4}+\dfrac{1}{x}\right)+\dfrac{3}{4}.x\ge2\sqrt{\dfrac{x}{4x}}+\dfrac{3}{4}.2=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=2\)

e.

\(E=\dfrac{9\left(x-2\right)+18}{2-x}+\dfrac{2}{x}=2\left(\dfrac{1}{x}+\dfrac{9}{2-x}\right)-9\ge\dfrac{2.\left(1+3\right)^2}{x+2-x}-9=7\)

\(E_{min}=7\) khi \(x=\dfrac{1}{5}\)

f.

\(F=\dfrac{3}{1-x}+\dfrac{4}{x}\ge\dfrac{\left(\sqrt{3}+2\right)^2}{1-x+x}=7+4\sqrt{3}\)

Dấu "=" xảy ra khi \(x=4-2\sqrt{3}\)

a: Để \(\dfrac{3x-2}{4}\) không nhỏ hơn \(\dfrac{3x+3}{6}\) thì \(\dfrac{3x-2}{4}>=\dfrac{3x+3}{6}\)

=>\(\dfrac{6\left(3x-2\right)}{24}>=\dfrac{4\left(3x+3\right)}{24}\)

=>18x-12>=12x+12

=>6x>=24

=>x>=4

b: Để \(\left(x+1\right)^2\) nhỏ hơn \(\left(x-1\right)^2\) thì \(\left(x+1\right)^2< \left(x-1\right)^2\)

=>\(x^2+2x+1< x^2-2x+1\)

=>4x<0

=>x<0

c: Để \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\) không lớn hơn \(\dfrac{x^2}{7}-\dfrac{2x-3}{5}\) thì

\(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}< =\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)

=>\(\dfrac{2x-3+5x\left(x-2\right)}{35}< =\dfrac{5x^2-7\cdot\left(2x-3\right)}{35}\)

=>\(2x-3+5x^2-10x< =5x^2-14x+21\)

=>-8x-3<=-14x+21

=>6x<=24

=>x<=4

a) \(=x^2-2x+x-2=x^2-x-2\)

b) \(=2x^2-8x-3x+12=2x^2-11x+12\)

c) \(=\left(x^2+2x-3\right)\left(x-2\right)=x^3-2x^2+2x^2-4x-3x+6=x^3-7x+6\)

d) \(=x^2+\dfrac{1}{2}x-5x-\dfrac{5}{2}=x^2-\dfrac{9}{2}x-\dfrac{5}{2}\)

e) \(=3x^2+\dfrac{3}{2}x+x+\dfrac{1}{2}=3x^2+\dfrac{5}{2}x+\dfrac{1}{2}\)

f) \(=5x^3-\dfrac{2}{3}x^4y-10x+\dfrac{4}{3}x^2y+30-4xy\)

a) x^2 - x - 2

b) 2x^2 - 11x + 12

c) x^3 - 7x + 6

d) x^2 - 9x/2 - 5/2

e) 3x^2 + 5x/2 + 3/2

f) 5x^3 - 2/3.x^4.y - 10x + 4/3.x^2.y - 4xy + 30

Mình đã trả lời trong bài viết 54 phút trước của bạn rồi mà nhỉ?

Huhu, mik không biết giải mong bạn thông cảm!

câu B bài cuối là D= 1 phần 2|x-1|+3 nha mọi ng

MỌI NGƯỜI GIÚP MIK CÂU NÀY VỚI Q= |x-3|-2.|-5x|

GẤP LẸ LÊN