Ta sẽ chứng minh BĐT phụ sau : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Áp dụng BĐT Cauchy dạng Engel , ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{3^2}{a+b+c}=\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Trong đó : \(\hept{\begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}}\) , Ta có :

\(\left(x+y+y+z+x+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge9\)

\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge4,5\)

\(\Leftrightarrow\frac{x+y+z}{x+y}+\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}\ge4,5\)

\(\Leftrightarrow1+\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{x+z}\ge4,5\)

\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{z+y}\ge1,5\)

\(\Rightarrow P_{min}=1,5\) " = " \(\Leftrightarrow x=y=z\)

Chúc bạn học tốt !!!

khong them viet ct :v

(x + 3)(x + 4)(x + 5)(x + 6) = 24

x^4 + 6x^3 + 5x^3 + 30x^2 + 4x^3 + 24x^2 + 20x^2 + 120x + 3x^2 + 18x^2 + 15x^2 + 90x + 12x^2 + 72x + 60x + 360 = 24

x^4 + 18x^3 + 119x^2 + 342x + 360 = 24

x^4 + 18x^3 + 119x^2 + 342x + 360 - 24 = 0

x^4 + 18x^3 + 119x^2 + 342x + 336 = 0

(x^3 + 16x^2 + 87x + 168)(x + 2) = 0

(x^2 + 9x + 24)(x + 7)(x + 2) = 0

vi x^2 + 9x + 24 # 0 nen:

=> x + 7 = 0 => x = -7

     x + 2 = 0 => x = -2

=> x = -7; -2

Ta có: \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+2xy-xz-yz-3xy\right)\)

\(=0\)( do x+y+z =0 )

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\left(đpcm\right)\)

Ta có :

 \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

Chúc bạn học tốt !!!

\(x^2+4x=5\)

\(\Leftrightarrow x^2+4x+4=9\)

\(\Leftrightarrow\left(x+2\right)^2=9\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)

Vậy \(x=1\) hoặc \(x=-5\)

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\(x^2+4x=5\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Delta=4^2+4.5=36,\sqrt{\Delta}=6\)

pt có 2 nghiệm

\(x_1=\frac{-4+6}{2}=1\);\(x_2=\frac{-4-6}{2}=-5\)

\(A=2x^2+4y^2+4xy+10x+12y+18\)

\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)

\(A=\left(x+2y^2\right)+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)

\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)

Do \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)

\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)

" = " \(\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)

\(\Rightarrow A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)

Chúc bạn học tốt !!!

Áp dụng BĐT Cauchy dang Engel , ta có :

\(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}=\frac{16}{a+b}\)

Dấu " = " xay ra \(\Leftrightarrow\hept{\begin{cases}x=y=2\\a=b\end{cases}}\)

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Ta có : \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow2\left(ab+bc+ca\right)=-2017\)

\(\Rightarrow ab+bc+ca=-\frac{2017}{2}\)

\(\Rightarrow\left(ab+bc+ca\right)^2=1017072,25\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=1017072,25\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^21017072,25\)

Ta có : 

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2017^2\)

\(\Rightarrow a^4+b^4+c^4+2.1017072,25=2017^2\)

\(\Rightarrow a^4+b^4+c^4=3051216,75\)

Chúc bạn học tốt !!!

\(A=\frac{4x}{x^2-2x}+\frac{3}{2-x}+\frac{12x}{x^3-4x}\)

\(A=\frac{4x}{x\left(x-2\right)}-\frac{3}{x-2}+\frac{12x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x\left(x+2\right)-3x\left(x+2\right)+12x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x\left(x+2\right)+12x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x^2+2x+12x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x^2+14x}{x\left(x-2\right)\left(x+2\right)}\)