1.

$x(x+2)(x+4)(x+6)+8$

$=x(x+6)(x+2)(x+4)+8=(x^2+6x)(x^2+6x+8)+8$

$=a(a+8)+8$ (đặt $x^2+6x=a$)

$=a^2+8a+8=(a+4)^2-8=(x^2+6x+4)^2-8\geq -8$

Vậy $A_{\min}=-8$ khi $x^2+6x+4=0\Leftrightarrow x=-3\pm \sqrt{5}$

2.

$B=5+(1-x)(x+2)(x+3)(x+6)=5-(x-1)(x+6)(x+2)(x+3)$

$=5-(x^2+5x-6)(x^2+5x+6)$

$=5-[(x^2+5x)^2-6^2]$

$=41-(x^2+5x)^2\leq 41$

Vậy $B_{\max}=41$. Giá trị này đạt tại $x^2+5x=0\Leftrightarrow x=0$ hoặc $x=-5$

3.

Đặt $x+3=a; 7-x=b$ thì $a+b=10$ 

$C=a^4+b^4$

Áp dụng BĐT Bunhiacopxky:

$(a^4+b^4)(1+1)\geq (a^2+b^2)^2$

$\Rightarrow C\geq \frac{(a^2+b^2)^2}{2}$
$(a^2+b^2)(1+1)\geq (a+b)^2=100$

$\Rightarrow a^2+b^2\geq 50$

$\Rightarrow C\geq \frac{50^2}{2}=1250$

Vậy $C_{\min}=1250$

Giá trị này đạt tại $a=b=5\Leftrightarrow x=2$

b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

\(A=\left|x+1\right|-3\\ min_A=-3.khi.x+1=0\Leftrightarrow x=-1\\ B=-\left|x-\dfrac{3}{7}\right|-\dfrac{1}{4}\\ max_B=-\dfrac{1}{4}.khi.\left(x-\dfrac{3}{7}\right)=0\Leftrightarrow x=\dfrac{3}{7}\)

a)

A = |x + 1| - 3 ≥ 0 - 3 = -3

Dấu "=" xảy ra khi x + 1 = 0 hay x = -1

Do đó A đạt GTNN là -3 khi x = -1

b)

\(B=-\left|x-\dfrac{3}{7}\right|-\dfrac{1}{4}\le-0-\dfrac{1}{4}=-\dfrac{1}{4}\)

Dấu "=" xảy ra khi khi \(x-\dfrac{3}{7}=0\) hay \(x=\dfrac{3}{7}\)

Do đó B đạt GTLN là \(-\dfrac{1}{4}\) khi \(x=\dfrac{3}{7}\)

a: \(M-\dfrac{3}{2}=\dfrac{x+7}{\sqrt{x}+3}-\dfrac{3}{2}\)

\(=\dfrac{2x+14-3\sqrt{x}-9}{2\left(\sqrt{x}+3\right)}\)

\(=\dfrac{2x-3\sqrt{x}+5}{2\left(\sqrt{x}+3\right)}>0\)

=>M>3/2

b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)

\(=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\cdot\sqrt{\dfrac{16}{\sqrt{x}+3}\cdot\left(\sqrt{x}+3\right)}-6=2\cdot4-6=2\)

Dấu = xảy ra khi (căn x+3)^2=16

=>căn x+3=4

=>x=1

a) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{8\sqrt{x}+24}{x-9}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+3\sqrt{x}+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+8}{\sqrt{x}-3}\) (đpcm)

b) Mình không biết làm bạn thông cảm.

a: \(B=\dfrac{\sqrt{x}}{x+\sqrt{x}}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{x+1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: B=2/7

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

=>\(2\left(x+\sqrt{x}+1\right)=7\sqrt{x}\)

=>\(2x+2\sqrt{x}-7\sqrt{x}+2=0\)

=>\(2x-5\sqrt{x}+2=0\)

=>\(\left(2\sqrt{x}-1\right)\cdot\left(\sqrt{x}-2\right)=0\)

=>\(\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

a.

\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)

Dấu "=" xảy ra khi \(x=2013\)

b.

\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)

\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)

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

\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)

a) 2ˣ + 2ˣ⁺³ = 72

2ˣ.(1 + 2³) = 72

2ˣ.9 = 72

2ˣ = 72 : 9

2ˣ = 8

2ˣ = 2³

x = 3

b) Để số đã cho là số nguyên thì (x - 2) ⋮ (x + 1)

Ta có:

x - 2 = x + 1 - 3

Để (x - 2) ⋮ (x + 1) thì 3 ⋮ (x + 1)

⇒ x + 1 ∈ Ư(3) = {-3; -1; 1; 3}

⇒ x ∈ {-4; -2; 0; 2}

Vậy x ∈ {-4; -2; 0; 2} thì số đã cho là số nguyên

c) P = |2x + 7| + 2/5

Ta có:

|2x + 7| ≥ 0 với mọi x ∈ R

|2x + 7| + 2/5 ≥ 2/5 với mọi x ∈ R

Vậy GTNN của P là 2/5 khi x = -7/2

\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

\(B=\dfrac{7}{-\left(x-5\right)^2-5}\ge-\dfrac{7}{5}\)

\(B_{min}=-\dfrac{7}{5}\) khi \(x=5\)