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Rõ ràng \(x=y=z=0\)   là nghiệm của hệ

Với \(xyz\ne0\), Ta có

\(y=\frac{2x^2}{x^2+1}\le\frac{2x^2}{2x}=x\)

\(z=\frac{3y^3}{y^4+y^2+1}\le\frac{3y^3}{3y^2}=y\)

\(x=\frac{4z^4}{z^6+z^4+z^2+1}\le\frac{4z^4}{4z^3}=z\)

Suy ra \(y\le x\le z\le y\Rightarrow x=y=z\)

Từ pt thứ nhất của hệ suy ra 

\(\frac{2x^2}{x^2+1}=x\Leftrightarrow2x=1=x^2\)( vì \(x\ne0\))\(\Leftrightarrow x=1\)

Vậy hệ pt có hai nghiệm \(\left(0,0,0\right)\)và \(\left(1,1,1\right)\)

\(4x^3-3x=\sqrt{1-x^2}\Leftrightarrow\left(4x^3-3x\right)^2=1-x^2\)

\(\Leftrightarrow16x^6-24x^4+9x^2=1-x^2\Leftrightarrow16x^6-24x^4+10x^2-1=0\) vonghiem 

Đặt \(\left(a,b,c\right)\rightarrow\left(\frac{x}{y},\frac{y}{z},\frac{z}{x}\right)\)

\(VT=\Sigma_{cyc}\frac{1}{\sqrt{\frac{x}{z}+\frac{x}{y}+2}}=\Sigma_{cyc}\frac{\sqrt{yz}}{\sqrt{xy+xz+2yz}}\)

\(\Rightarrow VT^2\le\left(1+1+1\right)\left(\Sigma_{cyc}\frac{yz}{xy+xz+2yz}\right)\)\(\le\frac{3}{4}\left[\Sigma_{cyc}yz\left(\frac{1}{xy+yz}+\frac{1}{xz+yz}\right)\right]=\frac{9}{4}\)

Đẳng thức xảy ra khi a = b = c = 1

Bài 1: Bổ đề: \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

\(P=\frac{1}{\sqrt{2}}\left(\sqrt{4a^2+2ab+4b^2}+\sqrt{4b^2+2bc+4c^2}+\sqrt{4c^2+2ca+4a^2}\right)\)

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

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

\(=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{5}{2}\left(a+b\right)^2}+\sqrt{\frac{5}{2}\left(b+c\right)^2}+\sqrt{\frac{5}{2}\left(c+a\right)^2}\right)\)

\(=\frac{1}{\sqrt{2}}.\frac{\sqrt{5}}{\sqrt{2}}+\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)\(=\frac{\sqrt{5}}{2}.2\left(a+b+c\right)=\sqrt{5}.2020\)

Dấu "=" xảy ra khi \(a=b=c=\frac{2020}{3}\)

( x + y )³ - ( x - y )³

= [ ( x + y ) - ( x - y ) ][ ( x + y )² + ( x + y )( x - y ) + ( x - y )² ]

= ( x + y - x + y )( x² + 2xy + y² + x² - y² + x² - 2xy + y² )

= 2y( 3x² + y² )

(x+y)^3-(x-y)^3=(x+y-x+y)[(x+y)^2-2(x+y)(x-y)+(x-y)^2   áp dụng HĐT nhá

                       =2y(x^2+2xy+y^2 -2x^2 + 2y^2 + x^2 -2xy +y^2)

                       = 2y(4y^2)

                      =8y^3

Cái này nak mk đang rút gọn luôn r nhá

( a + 2 )³ - a( a - 3 )²

= a³ + 6a² + 12a + 8 - a( a² - 6a + 9 )

= a³ + 6a² + 12a + 8 - a³ + 6a² - 9a

= 12a² + 3a + 8

cách của symbolab:

\(\left(a+2\right)^3-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)

\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)

\(=12a^2+3a+8\)