\(T=\frac{1}{a^2+b^2+3}+\frac{1}{2ab}\)

\(T=\frac{1}{a^2+b^2+3}+\frac{1}{5ab}+\frac{3}{10ab}\)

Ta có: \(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{x+y}{2}}=\frac{4}{x+y}\left(x,y>0\right)\)

          \(2ab\le a^2+b^2\Leftrightarrow4ab\le\left(a^2+b^2+2ab\right)\Leftrightarrow2ab\le\frac{\left(a+b\right)^2}{2}\)

Áp dụng:

\(T\ge\frac{4}{a^2+b^2+3+5ab}+\frac{3}{5.\frac{\left(a+b\right)^2}{2}}\ge\frac{4}{\left(a+b\right)^2+3+1,5.\frac{\left(a+b\right)^2}{2}}+\frac{3}{5.\frac{2^2}{2}}=\frac{4}{2^2+3+1,5.\frac{2^2}{2}}+\frac{3}{5.2}=\frac{4}{10}+\frac{3}{10}=\frac{7}{10}\)Dấu " = " xảy ra \(\Leftrightarrow a=b=1\)( lát giải thích sau )

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2ab\\\frac{1}{a^2+b^2+3}=\frac{1}{5ab}\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\a^2+b^2+3=5ab\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\\left(a+b\right)^2-2ab+3=5ab\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\4+3=5ab+2ab\end{cases}\Leftrightarrow}\hept{\begin{cases}a=b\\7=7ab\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\ab=1\end{cases}}\Leftrightarrow a=b=1\)

Bổ sung thêm:

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( x,y>0)

Dấu " = '" xảy ra <=> x=y

\(2ab\le a^2+b^2\)

Dấu " = '" xảy ra <=> a=b

Giả sử x>0

\(x^2+x+3-\left(x-1\right)^2=x^2+x+3-x^2+2x-1=3x+2>0\)

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

\(\Rightarrow\left(x-1\right)^2< x^2+x+3< \left(x+2\right)^2\)

\(\Rightarrow y^2=\orbr{\begin{cases}x^2\\\left(x+1\right)^2\end{cases}}\)

Với \(y^2=x^2\)

\(\Rightarrow x^2+x+3=x^2\Leftrightarrow x+3=0\Leftrightarrow x=-3\)(loại)

\(y^2=\left(x+1\right)^2\)

\(\Rightarrow x^2+x+3=x^2+2x+1\)

\(\Rightarrow2=x\)(t/m)

Thay x = 2 \(\Rightarrow y^2=4+2+3=9\Leftrightarrow y=\pm3\)

Vậy \(x=2;y=\pm3\left(tm\right)\)

vì sao lại giả sử x>0?

Vô lý
Ta có AB=CD=4cm  mà sao AB=AD+BC=10cm???

a) x(x+1)(x^2+x+1)=42

=> (x^2+x)(x^2+x+1)=42 (1)

Đặt x^2+x=t

=> x^2+x+1=t+1

=> pt (1) có dạng: t(t+1)=42

=> t^2+t=42

=> 4t^2+4t=168

=> 4t^2+4t+1=169

=> (2t+1)^2=(+-13)^2

Xong tìm t và tự tìm nốt x

b) x(x+1)(x+2)(x+3)=24

=> x(x+3)(x+1)(x+2)=24

=> (x^2+3x)(x^2+3x+2)=24

Đặt x^2+3x+1=t

=> x^2+3x=t-1 và x^2+3x+2=t+1

Xong thay vào tìm t và tự tìm x.

a, \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt x^2+x=a

=>\(a^2+a=42\)

\(a^2+a-42=0\)

\(a^2+7a-6a-42=0\)

\(\left(a+7\right)\left(a-6\right)=0\)

\(\left(x^2+x+7\right)\left(x^2+x-6\right)=0\)

\(\left(x^2+x+7\right)\left(x-2\right)\left(x+3\right)=0\)

x^2+x+7>0

=>(x-2)(x-3)=0

=>x=2,3

b,x(x+1)(x+2)(x+3)=24

[x(x+3)][(x+1)(x+2)]=24

(x^2+3x)(x^2+3x+2)=24

Đặt x^2+3x=a

=>a(a+2)-24=0

=>a^2+2a-24=0

=>a^2+6a-4a-24=0

=>(a-4)(a+6)=0

=>(x^2+3x-4)(x^2+3x+6)=0

=>(x-1)(x+4)(x^2+3x+6)=0

vì (x^2+3x+6)>0

=>(x-1)(x+4)=0

\(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+2017}{3}+\frac{x+2016}{4}\)

\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+2017}{3}+1+\frac{x+2016}{4}+1\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{3}-\frac{x+2020}{4}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)=0\)

Mà \(\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{3}-\frac{1}{4}\right)\ne0\)

\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)

Vậy...

\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)

\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)

\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)

                                    |________________A________________|

Do A > 0

nên x - 2014 = 0

<=> x = 2014