Bài 2: 

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

Xem lại đề đi rồi chúng mình nói chuyện :))

\(A=\dfrac{5x^2}{x^2}-\dfrac{x}{x^2}+\dfrac{1}{x^2}=\dfrac{1}{x^2}-\dfrac{1}{x}+5=\left(\dfrac{1}{x^2}-\dfrac{1}{x}+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(\dfrac{1}{x}-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

\(A_{min}=\dfrac{19}{4}\) khi \(\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\)

Lời giải:
a. 
$C=16-3(x^2+4x+4)=16-3(x+2)^2$
Vì $(x+3)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow C\leq 16-3.0=16$

Vậy $C_{\max}=16$ khi $x=-2$

b.

$D=-x^2+5x=2,5^2-(x^2-5x+2,5^2)$

$=6,25-(x+2,5)^2\leq 6,25-0=6,25$

Vậy $D_{\max}=6,25$ khi $x=-2,5$

c.

$M=2x-x^2=1-(x^2-2x+1)=1-(x-1)^2\leq 1-0=1$
Vậy $M_{\max}=1$ khi $x=1$

a: Ta có: \(C=-3x^2-12x+4\)

\(=-3\left(x^2+4x-\dfrac{4}{3}\right)\)

\(=-3\left(x^2+4x+4-\dfrac{16}{3}\right)\)

\(=-3\left(x+2\right)^2+16\le16\forall x\)

Dấu '=' xảy ra khi x=-2

b: Ta có: \(D=-x^2+5x\)

\(=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

c: Ta có: \(M=-x^2+2x\)

\(=-\left(x^2-2x+1-1\right)\)

\(=-\left(x-1\right)^2+1\le1\forall x\)

Dấu '=' xảy ra khi x=1

ĐK: \(x\ge0\)

+) Với x = 0 => A = 0

+) Với x khác 0

Ta có: \(\frac{1}{A}=\frac{3}{4}\sqrt{x}-\frac{3}{4}+\frac{3}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)-\frac{3}{4}\ge\frac{3}{4}.2-\frac{3}{4}=\frac{3}{4}\)

=> \(A\le\frac{4}{3}\)

Dấu "=" xảy ra <=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\)<=> x = 1

Vậy max A = 4/3 tại x = 1

Còn có 1 cách em quy đồng hai vế giải đenta theo A thì sẽ tìm đc cả GTNN và GTLN 

\(A=-2x^2+5x-8=-2\left(x^2-\frac{5}{2}x+4\right)\)

\(=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}+\frac{39}{16}\right)=-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\)

Vì: \(-2\left(x-\frac{5}{2}\right)^2-\frac{39}{8}\le\frac{39}{8}\forall x\)

GTLN  của bt là 39/8 tại \(-2\left(x-\frac{5}{2}\right)^2=0\Rightarrow x=\frac{5}{2}\)

cn lại lm tg tự  nha bn

Answer:

a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)

\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)

\(\Rightarrow5x+2x+2-12=0\)

\(\Rightarrow7x-10=0\)

\(\Rightarrow x=\frac{10}{7}\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)

\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)

\(\Rightarrow\frac{3}{2}x=-6\)

\(\Rightarrow x=-4\)

c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)

\(\Rightarrow9x-6-6x-6\ge0\)

\(\Rightarrow3x-12\ge0\)

\(\Rightarrow x\ge4\)

d) \(\left(x+1\right)^2< \left(x-1\right)^2\)

\(\Rightarrow x^2+2x+1< x^2-2x+1\)

\(\Rightarrow4x< 0\)

\(\Rightarrow x< 0\)

e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)

\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)

\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)

\(\Rightarrow6x\le24\)

\(\Rightarrow x\le4\)

f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)

\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)

\(\Rightarrow9x-6-6x-6\le0\)

\(\Rightarrow3x\le12\)

\(\Rightarrow x\le4\)

A= |5x - 1| - 3

Ta thấy:\(\left|5x-1\right|\ge0\)

\(\Rightarrow\left|5x-1\right|-3\ge0-3=-3\)

\(\Rightarrow A\ge-3\)

Dấu = khi x=1/5

Vậy...

B= |2x - 7| + 12

Ta thấy: \(\left|2x-7\right|\ge0\)

\(\Rightarrow\left|2x-7\right|+12\ge0+12=12\)

\(\Rightarrow B\ge12\)

Dấu = khi x=7/2

C và D fai là tìm Max 

C=-|5 - 3x| + 2005

Ta thấy :\(-\left|5-3x\right|\le0\)

\(\Rightarrow-\left|5-3x\right|+2005\le0+2005=2005\)

\(\Rightarrow C\le2005\)

Dấu = khi x=5/3

Vậy...

D= 29 - |7 + 3x|

Ta thấy:\(-\left|7+3x\right|\le0\)

\(\Rightarrow29-\left|7+3x\right|\le29-0=29\)

\(\Rightarrow D\le29\)

Dấu = khi x=-7/3

Vậy....