\(\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}\ge\frac{1}{1+xy}\)

\(\Leftrightarrow x^3y+xy^3-x^2y^2-2xy+1\ge0\)

\(\Leftrightarrow\left(x^3y-2x^2y^2+xy^3\right)+\left(x^2y^2-2xy+1\right)\ge0\)

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

Đặt t = x - 7

pt <=> ( t + 1 )⁴ + ( t - 1 )⁴ = 16 ( bạn tự khai triển = hệ thức Newton )

<=> 2t⁴ + 12t² - 14 = 0

<=> t⁴ + 6t² - 7 = 0

<=> t⁴ - t² + 7t² - 7 = 0

<=> t²( t² - 1 ) + 7( t² - 1 ) = 0

<=> ( t² + 7 )( t² - 1 ) = 0

<=> ( t² + 7 )( t - 1 )( t + 1 ) = 0

=> [ ( x - 7 )² + 7 ]( x - 8 )( x - 6 ) = 0 (1)

Vì ( x - 7 )² + 7 ≥ 7 > 0 ∀ x

nên (1) <=> ( x - 8 )( x - 6 ) = 0

<=> x = 8 hoặc x = 6

Vậy ...

cái đó là tam giác pascal chứ nhỉ

\(Q\left(x\right)⋮P\left(x\right)\Leftrightarrow Q\left(x\right)=P\left(x\right)R\left(x\right)\)

\(P\left(x\right)=x^2-1=0\Leftrightarrow x=\pm1\)

Suy ra \(Q\left(\pm1\right)=0\)

\(Q\left(1\right)=1+a+b+2=0\)

\(Q\left(-1\right)=1-a+b+2=0\)

Ta có hệ: 

\(\hept{\begin{cases}a+b=-3\\-a+b=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=-3\end{cases}}\)

5 + 2x = x - 5

<=> 2x - x = -5 - 5

<=> x = -10

Vậy pt có nghiệm x = -10

Trả lời:

5 + 2x = x - 5

<=> 2x - x = -5 - 5

<=> x = -10

Vậy S = { -10 }

\(\frac{3}{a^2+b^2}+\frac{2}{ab}=3\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{1}{2ab}\ge\frac{3.4}{a^2+b^2+2ab}+\frac{1}{\frac{\left(a+b\right)^2}{2}}\)

\(=\frac{12}{1}+\frac{1}{\frac{1}{2}}=12+2=14\)

Dấu = xảy ra khi: \(a=b=\frac{1}{2}\)

Ta có: 

\(P=\frac{5}{x^2+y^2}+\frac{3}{xy}\)

\(P=\frac{1}{\frac{x^2+y^2}{5}}+\frac{1}{\frac{2}{5}xy}+\frac{1}{2xy}\)

\(\ge\frac{\left(1+1\right)^2}{\frac{x^2+y^2}{5}+\frac{2xy}{5}}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\)

\(=\frac{4}{\frac{\left(x+y\right)^2}{5}}+\frac{1}{\frac{3^2}{2}}=\frac{4}{\frac{3^2}{5}}+\frac{2}{9}=\frac{22}{9}\)

Dấu "=" xảy ra khi: x = y = 3/2

\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{xy}=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{x^2+2xy+y^2}{xy}\)

\(=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{x}{y}+2+\frac{y}{x}\ge\frac{\left(x+y\right)^2}{x^2+y^2}+2+2\sqrt{\frac{x}{y}\cdot\frac{y}{x}}=\frac{\left(x+y\right)^2}{x^2+y^2}+4\)(AM-GM)

dự đoán MinS = 6 <=> x=y nhưng mà chưa xử được \(\frac{\left(x+y\right)^2}{x^2+y^2}\ge2\):v căng đét :D

Ta có: 

\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{xy}\)

\(S=\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{2xy}+\frac{\left(x+y\right)^2}{2xy}\)

\(\ge\frac{\left(x+y+x+y\right)^2}{x^2+y^2+2xy}+\frac{4xy}{2xy}\)

\(=\frac{4\left(x+y\right)^2}{\left(x+y\right)^2}+2=6\)

Dấu "=" xảy ra khi: x = y

Ta có:

\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)

\(P=\left(\frac{\sqrt{ab}}{a+b}+\frac{a+b}{4\sqrt{ab}}\right)+\frac{3\left(a+b\right)}{4\sqrt{ab}}\)

\(\ge2\sqrt{\frac{\sqrt{ab}}{a+b}\cdot\frac{a+b}{4\sqrt{ab}}}+\frac{3\cdot2\sqrt{ab}}{4\sqrt{ab}}\) (BĐT Cauchy)

\(=2\cdot\frac{1}{2}+\frac{3}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi: a = b

Vậy \(Min_P=\frac{5}{2}\Leftrightarrow a=b\)

\(P=\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}=\frac{3}{4}\frac{a+b}{\sqrt{ab}}+\frac{1}{4}\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\)

Ta có: 

\(a+b\ge2\sqrt{ab}\Leftrightarrow\frac{a+b}{\sqrt{ab}}\ge2\).

\(\frac{1}{4}\frac{a+b}{\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}\ge2\sqrt{\frac{1}{4}\frac{\left(a+b\right)\sqrt{ab}}{\sqrt{ab}\left(a+b\right)}}=1\).

Suy ra \(P\ge\frac{3}{4}.2+1=\frac{5}{2}\).

Dấu \(=\)xảy ra khi \(a=b\).