a.

Vơi mọi x, y ta luôn có:

\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\) (1)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge x^2+y^2+2xy\)

\(\Leftrightarrow x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2>\dfrac{1}{2}.1=\dfrac{1}{2}\) (đpcm)

b. 

Sử dụng kết quả (1), ta có: 

\(\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a^2+b^2}{ab}\ge\dfrac{2ab}{ab}=2\) (đpcm)

2đpcm bạn nhé 

Chúc Bạn Học Tốt.

Ta có:\(\frac{x+10}{x+5}=\frac{x+5+5}{x+5}=1+\frac{5}{x+5}\)

      \(\Rightarrow5⋮x+5\)

             Hoặc \(x+5\inƯ\left(5\right)\)

 Vậy Ư(5) là:[1,-1,5,-5]

     Do đó ta có bảng sau:

               Vậy x=-10;-6;-4;0

Giải:

Ta có:

\(x+10⋮x+5\)

\(\Rightarrow\left(x+5\right)+5⋮x+5\)

\(\Rightarrow5⋮x+5\)

\(\Rightarrow x+5\in\left\{\pm1;\pm5\right\}\)

+) \(x+5=1\Rightarrow x=-4\)

+) \(x+5=-1\Rightarrow x=-6\)

+) \(x+5=5\Rightarrow x=0\)

+) \(x+5=-5\Rightarrow x=-10\)

Vậy \(x\in\left\{-4;-6;0;-10\right\}\)

Ta có :

x+10 chia hết cho x+5

=> x+10 - (x+5) chia hết cho x+5

=> x+10 - x - 5 chia hết cho x+5

=> 5 chia hết cho x+5

=> x+5 ϵ Ư₍₅₎

mà Ư₍₅₎ = { 1,5,(-1),(-5) }

=> ta có bảng giá trị :

x=1/5 hoặc x= -1/5

\(\left|x\right|=\dfrac{1}{5}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(\left|x\right|=\dfrac{1}{5}\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{5}\left(x\ge\dfrac{1}{5}\right)\left(TM\right)\\x=-\dfrac{1}{5}\left(x< \dfrac{1}{5}\right)\left(TM\right)\end{matrix}\right.\)

Câu 1:

\(2x^3-3x^2+x+a\)

\(=2\left(x^3-6x^2+12x-8\right)+9\left(x^2-4x+4\right)+13\left(x-2\right)+\left(6+a\right)\)

\(=2\left(x-2\right)^3+9\left(x-2\right)^2+13\left(x-2\right)+\left(6+a\right)\)chia hết cho \(x-2\)khi và chỉ khi :

\(6+a=0\Leftrightarrow a=-6\). Vậy \(a=-6\).

Câu 2:

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

\(\Leftrightarrow x^2+x-\left(3x^2+11x+10\right)=-4x^2+1\)

\(\Leftrightarrow x^2+x-3x^2-11x-10+4x^2-1=0\)

\(\Leftrightarrow2x^2-10x-11=0\)

\(\Delta'=\left(-5\right)^2-2\left(-11\right)=47>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt:

\(x=\frac{5+\sqrt{47}}{2}\)hoặc \(x=\frac{5-\sqrt{47}}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{5+\sqrt{47}}{2};\frac{5-\sqrt{47}}{2}\right\}\)

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2013\right)=4+1007\cdot2013\)

\(\Leftrightarrow2014x+2027091=2027095\)

\(\Leftrightarrow2014x=4\)

hay \(x=\dfrac{2}{1007}\)

-7 - 5.| 2x + 1 | = x 

=> -5.| 2x + 1 | = x+7 (*)

Xét 2x + 1 ≥ 0 => x ≥ \({{-1} \over 2}\)

=> Phương trình (*) có dạng: 

-5(2x+1) = x+7

<=> -10x - 5 = x  + 7

<=> -11x = 12<=> \(x = {{-12} \over 11}\)( Loại)

Xét 2x + 1 < 0 => x < \({{-1} \over 2}\)

=> Phương trình (*) có dạng: 

-5( -2x - 1 ) = x + 7

<=>10x + 5 = x + 7

<=> 9x = 2

<=>  \(x = {{2} \over 9}\)(Loại)

Vậy pt vô nghiệm

\(\dfrac{x}{x+1}+1\)

\(=\dfrac{x+x+1}{x+1}\)

\(=\dfrac{2x+1}{x+1}\)

ĐKXĐ : x \(\ne-1\)

\(=\dfrac{x}{x+1}+\dfrac{x+1}{x+1}\)

\(=\dfrac{2x+1}{x+1}\)

=x+x+1

= 2x+1

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)

\(=\left(\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\right).\left(1-4x\right)\)

\(=\left(\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\right)\left(1-4x\right)\)

\(=\dfrac{-4\sqrt{x}.\left(4x-1\right)}{4x-1}=-4\sqrt{x}\)

\(Q=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\left(dkxd:x\ge0;x\ne\dfrac{1}{4}\right)\)

\(=\left[\dfrac{2\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\right]\cdot\left(1-4x\right)\)

\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{4x-1}\cdot\left[-\left(4x-1\right)\right]\)

\(=4\sqrt{x}\cdot\left(-1\right)\)

\(=-4\sqrt{x}\)