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Bài này sử dụng tính chất cơ bản: \(\left|A\right|\pm A\ge0\) với mọi A

a.

\(A=\left|-x-3\right|+\left|4x+1\right|+\left|3x+5\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|3x+5\right|+5x+2=\left|3x-2\right|+\dfrac{3}{2}.\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|3x-2\right|+\left|2x+\dfrac{10}{3}\right|+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+5x+2\)

\(A\ge\left|5x+\dfrac{4}{3}\right|+5x+\dfrac{4}{3}+\dfrac{1}{2}\left|2x+\dfrac{10}{3}\right|+\dfrac{2}{3}\ge\dfrac{2}{3}\)

\(A_{min}=\dfrac{2}{3}\) khi \(2x+\dfrac{10}{3}=0\Rightarrow x=-\dfrac{5}{3}\)

b. Tương tự

\(B\ge\left|5x+7\right|+\left|x+\dfrac{5}{4}\right|+3\left|x+\dfrac{5}{4}\right|-6x+5\)

\(B\ge\left|6x+\dfrac{33}{4}\right|-\left(6x+\dfrac{33}{4}\right)+3\left|x+\dfrac{5}{4}\right|+\dfrac{53}{4}\ge\dfrac{53}{4}\)

\(B_{min}=\dfrac{53}{4}\) khi \(x=-\dfrac{5}{4}\)

Lời giải:

a. Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:

\(A=|-x-3|+|4x+1|+|3x+5|+5x+2\)

\(\geq |-x-3+4x+1|+|3x+5|+5x+2=|3x-2|+|3x+5|+5x+2\)

Nếu $x\geq \frac{2}{3}$ thì:

$A\geq 3x-2+3x+5+5x+2=11x+5\geq 11.\frac{2}{3}+5=\frac{37}{3}$

Nếu $\frac{-5}{3}\leq x< \frac{2}{3}$ thì:

$A\geq 2-3x+3x+5+5x+2=9+5x\geq 9+5.\frac{-5}{3}=\frac{2}{3}$

Nếu $x< \frac{-5}{3}$ thì:

$A\geq 2-3x-3x-5+5x+2=-1-x>\frac{2}{3}$

Từ 3 TH trên suy ra $A_{\min}=\frac{2}{3}$ khi $x=\frac{-5}{3}$

b. Áp dụng BĐT $|a|+|b|\geq |a+b|$ thì:

\(B\geq |2x+3+3x+4|+|4x+5|-6x+5=|5x+7|+|4x+5|-6x+5\)

Nếu $x\geq \frac{-5}{4}$ thì:

$B\geq 5x+7+4x+5-6x+5=3x+17\geq 3.\frac{-5}{4}+17=\frac{53}{4}$ 

Nếu $\frac{-7}{5}\leq x< \frac{-5}{4}$ thì:

$B\geq 5x+7-4x-5-6x+5=-5x+7> -5.\frac{-5}{4}+7=\frac{53}{4}$

Nếu $x< \frac{-7}{5}$ thì:

$B\geq -5x-7-4x-5-6x+5=-15x-7> -15.\frac{-7}{5}-7=14$

Từ 3 TH trên suy ra $B_{\min}=\frac{53}{4}$. Giá trị này đạt tại $x=\frac{-5}{4}$

1.

\(G=\dfrac{2}{x^2+8}\le\dfrac{2}{8}=\dfrac{1}{4}\)

\(G_{max}=\dfrac{1}{4}\) khi \(x=0\)

\(H=\dfrac{-3}{x^2-5x+1}\) biểu thức này ko có min max

2.

\(D=\dfrac{2x^2-16x+41}{x^2-8x+22}=\dfrac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\dfrac{3}{\left(x-4\right)^2+6}\ge2-\dfrac{3}{6}=\dfrac{3}{2}\)

\(D_{min}=\dfrac{3}{2}\) khi \(x=4\)

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

\(E_{min}=-1\) khi \(x=0\)

\(G=\dfrac{3\left(x^2-4x+5\right)-5}{x^2-4x+5}=3-\dfrac{5}{\left(x-2\right)^2+1}\ge3-\dfrac{5}{1}=-2\)

\(G_{min}=-2\) khi \(x=2\)

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

3,26 + 4/5 =?

làm nhanh lên giúp mình nhé

3,26+4/5=3,26+0,8=3,34

K cho mình cái

\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x+16+11\)

\(\Leftrightarrow-2x-1=12x+27\Leftrightarrow-14x-28=0\Leftrightarrow x=-2\)