Theo BĐT AM-GM, ta có

\(\frac{x^2}{y-2}\)+4(y-2) \(\ge\)4x

\(\frac{y^2}{x-2}\)+4(x-2)\(\ge\)4y

Cộng vế theo vế của 2 bất đẳng thức trên ta được:

\(\frac{x^2}{y-2}\)+\(\frac{y^2}{x-2}\)+4x+4y-16\(\ge\)4x+4y

<=>\(\frac{x^2}{y-2}\)+\(\frac{y^2}{x-2}\)​​\(\ge\)16

Dấu đẳng thức xảy ra khi x=y=4

Không biết cách làm đúng k nữa :D

Đặt: \(\hept{\begin{cases}a+bc=7^x\\b+ac=7^y\end{cases}}\)

TH1: Nếu \(7^x=7^y\)khi đó: n chẵn

\(\Leftrightarrow a+bc=b+ac\)

\(\Leftrightarrow\left(a-b\right)\left(1-c\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\c=1\end{cases}}\)

TH2:Nếu: \(7^x>7^y\)(*)

\(\Leftrightarrow a+bc>b+ac\)

\(\Leftrightarrow\left(a-b\right)\left(1-c\right)>0\)

\(\hept{\begin{cases}a>b\\c< 1\end{cases}\left(ktm\right)}\)hoặc: \(\hept{\begin{cases}a< b\\c>1\end{cases}\left(tm\right)}\)(1)

Đồng thời phải thỏa mãn điều kiện: \(a+bc⋮b+ac\)

\(\Leftrightarrow\left(a-b\right)\left(1-c\right)⋮b+ac\)

\(\Leftrightarrow\orbr{\begin{cases}a-b⋮b+ac\\1-c⋮b+ac\end{cases}\Leftrightarrow}\orbr{\begin{cases}a+ac⋮b+ac\\a\left(1-c\right)⋮b+ac\end{cases}\Leftrightarrow\orbr{\begin{cases}a+ac⋮b+ac\\a+b⋮b+ac\end{cases}}}\)(2)

Vì a,b,c thuộc N* nên:

\(\left(1\right)\Leftrightarrow\hept{\begin{cases}a+ac< b+ac\\ac+b>a+b\end{cases}}\)

Mặt khác: \(a+ac;a+b\ne0\)

Nên (2) sai

Dẫn đến (*) sai

Tương tự xét: \(7^x< 7^y\)(loại)

Vậy n chẵn

k cho tui

kiểm tra lại đề bài bn ơi

\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{2}{x-1}\)

Xét phương trình trên có:

\(\Delta'=\left(m-2\right)^2-\left(m^2-2m+4\right)=m^2-4m+4-m^2+2m-4=-2m\)

Để phương trình trên có hai nghiệm phân biệt \(x_1;x_2\)điều kiện là:

\(\Delta'>0\Leftrightarrow-2m>0\Leftrightarrow m< 0\)

Với m<0. Áp dụng định lí Vi ét ta có:

\(\hept{\begin{cases}x_1+x_2=-2\left(m-2\right)\\x_1.x_2=m^2-2m+4\end{cases}}\)

=> \(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1.x_2=4\left(m-2\right)^2-2\left(m^2-2m+4\right)=2m^2-12m+8\)

Ta có:

\(\frac{2}{x_1^2+x_2^2}-\frac{1}{x_1x_2}=\frac{1}{15m}\)

<=> \(\frac{2}{2m^2-12m+8}-\frac{1}{m^2-2m+4}=\frac{1}{15m}\)(điều kiện: \(2m^2-12m+8\ne0\))

<=> \(\frac{1}{m^2+4-6m}-\frac{1}{m^2+4-2m}=\frac{1}{15m}\)

<=> \(\frac{4m}{\left(m^2+4-6m\right)\left(m^2+4-2m\right)}=\frac{1}{15m}\)

<=> \(60m^2=\left(m^2+4\right)^2-8m\left(m^2+4\right)+12m^2\)

<=> \(\left(m^2+4\right)^2-8m\left(m^2+4\right)-48m^2=0\)

<=> \(\left(\frac{m^2+4}{m}\right)^2-8\frac{m^2+4}{m}-48=0\)

Đặt t=\(\frac{m^2+4}{m}< 0\)

Ta có phương trình ẩn t:

\(t^2-8t-48=0\Leftrightarrow\orbr{\begin{cases}t=-4\\t=12\left(loai\right)\end{cases}}\)

Với t=-4 ta có:

\(\frac{m^2+4}{m}=-4\Leftrightarrow m^2+4m+4=0\Leftrightarrow\left(m+2\right)^2=0\Leftrightarrow m=-2\)( tmđk)

vậy m=-2

\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+bc+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=a^3+b^3+c^3+3\left(abc+c^2a+b^2c+bc^2+a^2b+ca^2+ab^2+abc\right)\)

\(=a^3+b^3+c^3+3\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\right]\)

\(=a^3+b^3+c^3+3\left(a+b+c\right)\left(ab+bc+ca\right)\)

\(\Rightarrow\)\(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)\)

Lại có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a-b+b-c+c-a\right)^2\)

\(-2\left[\left(a-b\right)\left(b-c\right)+\left(b-c\right)\left(c-a\right)+\left(c-a\right)\left(a-b\right)\right]\)

\(=-2\left(ab-ca-b^2+bc+bc-ab-c^2+ca+ca-bc-a^2+ab\right)\)

\(=2\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)\)

\(\Rightarrow\)\(P=\frac{\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)}{2\left(a+b+c\right)^2-6\left(ab+bc+ca\right)}\)

\(=\frac{\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}{2\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]}=\frac{a+b+c}{2}\)

\(P=\)\(\left(\frac{\sqrt{a}}{\sqrt{a}+1}-\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{1}{a-1}\right):\frac{a}{2+2\sqrt{a}}.\)

\(=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}+\frac{1}{a-1}\right):\frac{a}{2\left(\sqrt{a}+1\right)}\)

\(=\frac{a-\sqrt{a}-a-\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\div\frac{a}{2\left(\sqrt{a}+1\right)}\)

\(=\frac{1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{2\left(\sqrt{a}+1\right)}{a}=\frac{2}{a\left(\sqrt{a}-1\right)}\)

\(b,P=\frac{2}{a\left(\sqrt{a}-1\right)}=\frac{2}{25\left(5-1\right)}=\frac{2}{25.4}=\frac{1}{50}\)

Vậy \(P=\frac{1}{50}\)tại \(a=25\)

Với mọi a nguyên dương , 

Ta có: 

\(\frac{1}{\sqrt{a}}=\frac{2}{2\sqrt{a}}>\frac{2}{\sqrt{a}+\sqrt{a+1}}=2\left(\sqrt{a-1}-\sqrt{a}\right)=2\sqrt{a-1}-2\sqrt{a}\)

Biểu thức:

\(B=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}\)

\(>2\sqrt{2}-2\sqrt{1}+2\sqrt{3}-2\sqrt{2}+2\sqrt{4}-2\sqrt{3}+...+2\sqrt{25}-2\sqrt{24}\)

\(=-2\sqrt{1}+2\sqrt{25}=-2+10=8\)

Vậy B>8