\(-x^2+x-\dfrac{1}{2}\)

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

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

\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}< 0\)

\(\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)+x^3+\dfrac{1}{x^3}}\)

\(=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+2+\dfrac{1}{x^6}\right)}{\left(x+\dfrac{1}{x}\right)+\left(x^3+\dfrac{1}{x^3}\right)}\)

\(=\dfrac{\left[\left(x+\dfrac{1}{x}\right)^3\right]^2-\left(x^3+\dfrac{1}{x^3}\right)^2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\)

\(=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)\)

\(=3x+\dfrac{3}{x}\)

\(=3\left(x+\dfrac{1}{x}\right)\)

Áp dụng bất đẳng thức \(x+\dfrac{1}{x}\ge2\forall x>0\)

\(\Rightarrow3\left(x+\dfrac{1}{x}\right)\ge6\)

\(\Rightarrowđpcm\)

Akai Haruma Ace Legona Unruly Kid

ai đi ngang qua cứu e vs :((

\(S=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)}\)

☘ Đặt \(M=\left(x+\dfrac{1}{x}\right)^3+\left(x^3+\dfrac{1}{x^3}\right)\)

\(=\left(x+\dfrac{1}{x}\right)^3+\left[\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)\right]\)

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

☘ Đặt \(N=\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2\)

\(=\left(x+\dfrac{1}{x}\right)^6-\left[\left(x^2+\dfrac{1}{x^2}\right)^3-3\left(x^2+\dfrac{1}{x^2}\right)\right]-2\)

\(=\left(x+\dfrac{1}{x}\right)^6-\left\{\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^3-3\left[\left(x+\dfrac{1}{x}\right)^2-2\right]\right\}-2\)

\(=\left(x+\dfrac{1}{x}\right)^6-\left[\left(x+\dfrac{1}{x}\right)^6-6\left(x+\dfrac{1}{x}\right)^4+12\left(x+\dfrac{1}{x}\right)^2-8-3\left(x+\dfrac{1}{x}\right)^2+6\right]-2\)

\(=6\left(x+\dfrac{1}{x}\right)^4-9\left(x+\dfrac{1}{x}\right)^2\)

☘ Đặt \(x+\dfrac{1}{x}=a\)

\(\Rightarrow S=\dfrac{6a^4-9a^2}{2a^3-3a}=3a\)

Áp dụng bất đẳng thức AM - GM

\(\Rightarrow a=x+\dfrac{1}{x}\ge2\)

\(\Rightarrow S\ge6\) (đpcm)

Vậy \(S\ge6\Leftrightarrow x=1\)

♬♫♪ Cách này hơi dài. Nhưng thật sự, chưa nghĩ được cách khác ngắn hơn. Thông cảm nhé.

câu B nhé , vẽ hàm số là sẽ thấy

Lời giải:

a)

Áp dụng bất đẳng thức AM-GM:

\(x^3+x^2+x+1\geq 4\sqrt[4]{x^3.x^2.x.1}=4\sqrt[4]{x^6}\)

\(\Rightarrow (x^3+x^2+x+1)^2\geq 16\sqrt{x^6}\)

\(\Leftrightarrow (x^3+x^2+x+1)^2\geq 16x^3\) (đpcm)

Dấu bằng xảy ra khi \(x=1\)

b)

Áp dụng BĐT AM-GM:

\(\frac{b+c}{a}.1\leq \left(\frac{\frac{b+c}{a}+1}{2}\right)^2=\frac{1}{4}\left(\frac{b+c+a}{a}\right)^2\)

\(\Rightarrow \frac{a}{b+c}\geq 4\left(\frac{a}{a+b+c}\right)^2\Leftrightarrow \sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}\)

Thực hiện tương tự với cac phân thức còn lại và cộng theo vế thu được:

\(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq \frac{2a+2b+2c}{a+b+c}=2\)

Dấu bằng xảy ra khi

\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=1\Rightarrow a+b+c=2a=2b=2c\)

\(\Rightarrow a=b=c\Rightarrow \frac{b+c}{a}=2\neq 1\) (vô lý)

Do đó dấu bằng không xảy ra

Vì vậy: \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2\)

2.

\(4n^3+n+3=4n^3+2n^2+2n-2n^2-n-1+4=2n\left(2n^2+n+1\right)-\left(2n^2+n+1\right)+4\)-Để \(\left(4n^3+n+3\right)⋮\left(2n^2+n+1\right)\) thì \(4⋮\left(2n^2+n+1\right)\)

\(\Leftrightarrow2n^2+n+1\in\left\{1;-1;2;-2;4;-4\right\}\) (do n là số nguyên)

*\(2n^2+n+1=1\Leftrightarrow n\left(2n+1\right)=0\Leftrightarrow n=0\) (loại) hay \(n=\dfrac{-1}{2}\) (loại)

*\(2n^2+n+1=-1\Leftrightarrow2n^2+n+2=0\) (phương trình vô nghiệm)

\(2n^2+n+1=2\Leftrightarrow2n^2+n-1=0\Leftrightarrow n^2+n+n^2-1=0\Leftrightarrow n\left(n+1\right)+\left(n+1\right)\left(n-1\right)=0\Leftrightarrow\left(n+1\right)\left(2n-1\right)=0\)

\(\Leftrightarrow n=-1\) (loại) hay \(n=\dfrac{1}{2}\) (loại)

\(2n^2+n+1=-2\Leftrightarrow2n^2+n+3=0\) (phương trình vô nghiệm)

\(2n^2+n+1=4\Leftrightarrow2n^2+n-3=0\Leftrightarrow2n^2-2n+3n-3=0\Leftrightarrow2n\left(n-1\right)+3\left(n-1\right)=0\Leftrightarrow\left(n-1\right)\left(2n+3\right)=0\)\(\Leftrightarrow n=1\left(nhận\right)\) hay \(n=\dfrac{-3}{2}\left(loại\right)\)

-Vậy \(n=1\)

1. \(x^2+y^2=z^2\)

\(\Rightarrow x^2+y^2-z^2=0\)

\(\Rightarrow\left(x-z\right)\left(x+z\right)+y^2=0\)

-TH1: y lẻ \(\Rightarrow x-z;x+z\) đều lẻ.

\(x+3z-y=x+z-y+2x\) chia hết cho 2. \(\Rightarrow\)Hợp số.

-TH2: y chẵn \(\Rightarrow\)1 trong hai biểu thức \(x-z;x+z\) chia hết cho 2.

*Xét \(\left(x-z\right)⋮2\):

\(x+3z-y=x-z+4z-y\) chia hết cho 2. \(\Rightarrow\)Hợp số.

*Xét \(\left(x+z\right)⋮2\):

\(x+3z-y=x+z+2z-y\) chia hết cho 2 \(\Rightarrow\)Hợp số.

a: \(=\dfrac{x^3-x^2+x+3\left(x^2-1\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

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

b: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

=>A>0 với mọi x<>-1

\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)

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

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

\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)