\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n^2+n+2n+2}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(n+1\right).\left(n+2\right)}\)

\(\Leftrightarrow\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{\left(n+2\right)-\left(n+1\right)}{\left(n+2\right).\left(n+1\right)}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x+1}-\frac{1}{x+2}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+2}< \frac{1}{2}\left(đpcm\right)\)

n là số lẻ nên \(n=2k+1\)

Ta có : \(n^2=\left(2k+1\right)^2=4k^2+4k+1=4k\left(k+1\right)+1\)

Mà \(4k\left(k+1\right)⋮4\) nên\(n^2:4\)dư 1

Ta sử dụng ẩn phụ:

\(\hept{\begin{cases}a=x+y-z\\b=y+z-x\\c=x+z-y\end{cases}}\)=> x+y+z=a+b+c

Khi đó :

A= (x+y+z)^3-(x+y-z)^3-(-x+y+z)^3-(x-y+z)^3=(a+b+c)^3+a^3+b^3+c^3=3(a+b)(b+c)(c+a)=3*2y*2z*2x=24xyz

Ta có :\(\left(a-b\right)^2\ge0\forall a,b\)

         \(\Leftrightarrow a^2-2ab+b^2\ge0\)

Mà \(a^2+b^2=c^2\left(Py-ta-go\right)\)

\(\Rightarrow c^2-2ab\ge0\)

\(\Leftrightarrow c^2\ge2ab\)

\(\Leftrightarrow2c^2\ge a^2+b^2+2ab\)( Do c²=a²+b²)

\(\Leftrightarrow2c^2\ge\left(a+b\right)^2\)

\(\Leftrightarrow c\sqrt{2}\ge a+b\)( ĐPCM ) 

Ta có a+b \(\le\)c√2

<=> (a+b) ²\(\le\)(c√2)²

<=> a²+2ab+b²\(\le\)2c²

<=> a²+2ab+b² \(\le\)2(a²+b²) = 2a²+2b²

<=> 0 \(\le\)a²-2ab+b² = (a-b)² ( luôn đúng)

=> a+b \(\le\)c√2

Áp dụng BĐT cô-si, ta được:

\(\hept{\begin{cases}\frac{a}{\sqrt{b}}+\sqrt{b}\ge2\sqrt{a}\\\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{b}\end{cases}}\)

=>  \(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}+\sqrt{a}+\sqrt{b}\ge2\left(\sqrt{a}+\sqrt{b}\right)\)

=> \(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\ge\sqrt{a}+\sqrt{b}\) (đpcm)

Vậy....

Biến đổi tương đương ta được :

\(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\ge\sqrt{a}+\sqrt{b}\)

\(\Leftrightarrow\sqrt{a}+\sqrt{b}\le\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{ab}}\)

\(\Leftrightarrow\sqrt{a}+\sqrt{b}\le\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{ab}}\)

\(\Leftrightarrow\sqrt{ab}\le a-\sqrt{ab}+b\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)( đúng với đk )

\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

    \(=\sqrt{a-2+2\sqrt{a-2}.2+4}+\sqrt{a-2-2\sqrt{a-2}.2+4}\)

      \(=\sqrt{\left(\sqrt{a-2}\right)^2+2\sqrt{a-2}.2+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2\sqrt{a-2}.2+2^2}\)

        \(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)

         \(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)

Ta có:

\(A=\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

\(A=\sqrt{\left(\sqrt{a-2}\right)^2+2.2\sqrt{a-2}+2^2}+\sqrt{\left(\sqrt{a-2}\right)^2-2.2\sqrt{a-2}+2^2}\)

\(A=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)

\(A=|\sqrt{a-2}+2|+|\sqrt{a-2}-2|\)   (1)

=> Điều kiên: a - 2 >= 0  <=> a >= 2

(1)  => \(A=\sqrt{a-2}+2+|\sqrt{a-2}-2|\)(Do số hạng đầu luôn lớn hơn 0 nên bỏ trị tuyệt đối)

TH1:  \(\sqrt{a-2}-2\ge0\Rightarrow A=\sqrt{a-2}+2+\sqrt{a-2}-2\)

           \(\sqrt{a-2}\ge2\Rightarrow A=2\sqrt{a-2}\)

          \(a\ge6\Rightarrow A=2\sqrt{a-2}\)

           (nhận)

TH2: \(\sqrt{a-2}-2\le0\Rightarrow A=\sqrt{a-2}+2-\sqrt{a-2}+2\)

            \(\sqrt{a-2}\le2\Rightarrow A=4\)

             \(2\le a\le6\Rightarrow A=4\) (Do đkxđ)

Vậy....

a) Ta có: \(A=x^2+4x+7=x^2+2.x.2+2^2+3=\left(x+2\right)^2+3\ge3\)

Dấu "=" xảy ra <=> x + 2 =0 => x = -2

Vậy A_Min = 3 khi và chỉ khi x = -2

b) \(B=x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra <=> x - 1/2 = 0 <=> x = 1/2

Vậy B_Min = 3/4 khi và chỉ khi x = 1/2

c) \(C=x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra <=> x+1/2 = 0 <=> x = -1/2

Vậy C_Min = 3/4 khi và chỉ khi x = -1/2

e) \(E=x+\sqrt{x}+1=\left(\sqrt{x}\right)^2+2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" không xảy ra

g) \(G=x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra <=> \(\sqrt{x}-\frac{1}{2}=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

Vậy G_Min = 3/4 khi x = 1/4

min hết à bạn