Vì \(\left(x^2-2^2\right)^2\ge0\Rightarrow\left(x^2-2^2\right)^2+2010\ge2010\)

                                    \(\Leftrightarrow\frac{\left(x^2-2^2\right)+2010}{-2009}\le\frac{2010}{-2009}\)

Vậy Dmax=-2010/2009, dấu = xảy ra khi và chỉ khi \(x^2-2^2=0\Leftrightarrow x=\pm2\)

a) VT = x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³

          = 2x³ + 6xy²

          = 2x( x² + 3y² ) = VP

=> đpcm

b) VT = x³ + 3x²y + 3xy² + y³ - ( x³ - 3x²y + 3xy² - y³ )

          = x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³

          = 3x²y + 2y³

          = 2y( 3x² + y² ) = VP

=> đpcm

a)

 \(VT=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

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

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

b)

\(VT=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)

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

Ta có: \(D=\frac{\left(x^2-2^2\right)+2010}{-2009}=\frac{x^2+2006}{-2009}=\frac{3-x^2}{2009}-1\)

Để D đạt GTLN => \(\frac{3-x^2}{2009}\) đạt GTLN, mà \(3-x^2\le3\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(x=0\)

Vậy Max(D) = \(-\frac{2006}{2009}\) khi x = 0

mình ghi sai đề nha

Ta có :

\(VP=x^3+3x^2y+3xy^2+y^3-3x^2y-3xy^2\)

\(=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=VT\)

\(\RightarrowĐPCM\)

VT = x³ + y³ ( HĐT số 6 )

= x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

= ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

= ( x + y )³ - 3xy( x + y ) = VP

=> đpcm

B=\(\frac{2011-x}{11-x}\) nha ghi sai

Ta có: 

\(B=\frac{2011-x}{11-x}=\frac{2000+\left(11-x\right)}{11-x}=1+\frac{2000}{11-x}\)

Để B đạt GTLN

=> \(\frac{2000}{11-x}\) đạt GTLN

Mà x nguyên nên dấu "=" xảy ra khi: \(11-x=1\Rightarrow x=10\)

Vậy Max(B) = 2001 khi x = 10

Sửa đề : x³ + y³ - xy( x + y ) = ( x + y )( x - y )²

x³ + y³ - xy( x + y )

= x³ + y³ - x²y - xy²

= x³ + 3x²y + 3xy² + y³ - 4x²y - 4xy²

= ( x³ + 3x²y + 3xy² + y³ ) - 4xy( x + y )

= ( x + y )³ - 4xy( x + y )

= ( x + y )[ ( x + y )² - 4xy ]

= ( x + y )( x² + 2xy + y² - 4xy )

= ( x + y )( x² - 2xy + y² )

= ( x + y )( x - y )²

=> đpcm

Bị tự tin quá khả năng nhẩm mồm, sai em xin lỗi ...

a, Ta có \(P\left(x\right)=8x^3+2x^2-3x-3x^3+10-x-2x^2-3\)

\(=5x^3-4x-7\)

\(Q\left(x\right)=9x^3-4x^2+2x-3+2x+3x^2+4x^3-2\)

\(=13x^3-x^2+4x-5\)

b, Ta có : \(P\left(-\frac{1}{2}\right)=5.\left(-\frac{1}{2}\right)^3-4.\left(-\frac{1}{2}\right)-7=-\frac{45}{8}\)

c , \(M\left(x\right)=P\left(x\right)+Q\left(x\right)\) 

  \(5x^3-4x-7+13x^3-x^2+4x-5=18x^3-x^2-12\)

\(N\left(x\right)=P\left(x\right)-Q\left(x\right)\)

\(5x^3-4x-7-13x^3+x^2-4x+5=-8x^3-8x-2+x^2\)

d, Đặt \(5x^3-4x-7=0\)( vô nghiệm )

=56 phân số bất đồng trị của a+b

chắc câu này a đăng lên cho vui :vv

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2< =>\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\left(\frac{2}{xy}-\frac{1}{z^2}\right)+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}+4=4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{xy}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4-4\)

\(< =>\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{zx}=0\)

\(< =>\left(\frac{1}{x^2}+\frac{2}{zx}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(< =>\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0< =>\frac{1}{x}=\frac{1}{y}=-\frac{1}{z}\)

\(< =>x=y=-z\)Thế vào giả thiết ta được : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(< =>\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2< =>\frac{-1}{z}+\frac{-1}{z}+\frac{1}{z}=2\)

\(< =>\frac{-1-1+1}{z}=2< =>2z=-1< =>z=-\frac{1}{2}\)

Suy ra \(x=y=-z=-\left(-\frac{1}{2}\right)=\frac{1}{2}< =>\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{cases}}\)

Nên \(P=\left(x+2y+z\right)^{2019}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2019}=1^{2019}=1\)

dễ 

\(\infty\)

Soái ca 2k6 Làm đi bạn !!

\(\frac{3^{2}+1}{3^{2}-1}+\frac{5^{2}+1}{5^{2}-1}+...+\frac{99^{2}+1}{99^{2}-1}=49+\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}=49+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}=49.49\)