Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}=k\)
\(\Rightarrow x=3k;y=4k;z=2k\)
Mà \(x^3-y^3+z^3=-29\)
\(\Rightarrow\left(3k\right)^3-\left(4k\right)^3+\left(2k\right)^3=-29\)
\(\Rightarrow27k^3-64k^3+8k^3=-29\)
\(\Rightarrow-29k^3=-29\)
\(\Rightarrow k^3=1\)
\(\Rightarrow k=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\\z=2\end{matrix}\right.\)
#DatNe

Theo đầu bài ra ta có :

x/3=y/4=z/2=x^3/27= x^3/64= z^3/8 và x^3-y^3+z^3 =-29

áp dụng tc dãy tỉ số = nhau nên ta có :

x^3/27=z^3/64= z^3/8=x^3-y^3+z^3/ 27-64+8=-29/-29=1

x/3=1 => x=3

y/4=1=>x=4

x/2=1=>x=2

vậy x=3 ; y=4 ;z=2

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Ta có:\(\frac{x}{y}=\frac{9}{7}\Rightarrow\frac{x}{9}=\frac{y}{7}\left(1\right)\)

          \(\frac{y}{z}=\frac{7}{3}\Rightarrow\frac{y}{7}=\frac{z}{3}\left(2\right)\)

                  Từ (1) và (2) suy ra:\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)

Áp dụng t/c dãy tỉ số bằng nhau ta đc:

       \(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=-\frac{15}{5}=-3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{9}=-5\\\frac{y}{7}=-5\\\frac{z}{3}=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-45\\y=-35\\z=-15\end{cases}}}\)

Ta có:

\(\frac{x}{y}=\frac{9}{7}\)=> \(\frac{x}{9}=\frac{y}{7}\)(1)

\(\frac{y}{z}=\frac{7}{3}\)=>\(\frac{y}{7}=\frac{z}{3}\)(2)

Từ (1) (2)

=>\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=-\frac{15}{5}=-3\)

=>\(\frac{x}{9}=-3\)=>x=-27

    \(\frac{y}{7}=-3\)=>y=-21

     \(\frac{z}{3}=-3\)=>z=-9

Vậy x=-27 ; y=-21 ; z=-9

a) 

Ta có: \(\hept{\begin{cases}\frac{x}{y}=\frac{9}{7}\Leftrightarrow\frac{x}{9}=\frac{y}{7}\\\frac{y}{z}=\frac{7}{3}\Leftrightarrow\frac{y}{7}=\frac{z}{3}\end{cases}\Leftrightarrow\frac{x}{9}=\frac{y}{7}=\frac{z}{3};x-y+z=-15}\)

Tính chất dãy tỉ số bằng nhau:

\(\frac{x}{9}=\frac{y}{7}=\frac{z}{3}=\frac{x-y+z}{9-7+3}=\frac{-15}{5}=-3\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{9}=-3\Leftrightarrow x=-3.9=-27\\\frac{y}{7}=-3\Leftrightarrow y=-3.7=-21\\\frac{z}{3}=-3\Leftrightarrow z=-3.3=-9\end{cases}}\)

Vậy \(\left(x,y,z\right)=\left(-27;-21;-9\right)\)

b) \(\frac{x}{5}=\frac{y}{7}=\frac{z}{3};x^2+y^2-z^2=582\)

Tính chất dãy tỉ số bằng nhau:

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}=\frac{x^2+y^2-z^2}{5^2+7^2-3^2}=\frac{585}{65}=9\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=9\Leftrightarrow x=9.5=45\\\frac{y}{7}=9\Leftrightarrow y=9.7=63\\\frac{z}{3}=9\Leftrightarrow z=9.3=27\end{cases}}\)

Vậy \(\left(x,y,z\right)=\left(45;63;27\right)\)

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x^3}{125}=\dfrac{y^3}{64}=\dfrac{z^3}{8}=\dfrac{x^3-y^3+z^3}{125-64+8}=\dfrac{69}{69}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt[3]{125}=5\\y=\sqrt[3]{64}=4\\z=\sqrt[3]{8}=2\end{matrix}\right.\)

1 + 2xy - x² - y²

= 1 - ( x² - 2xy + y² )

= 1² - ( x - y )²

= [ 1 - ( x - y ) ][ 1 + ( x - y ) ]

= ( y - x + 1 )( x - y + 1 )

a² + b² - c² - d² - 2ab + 2cd

= ( a² - 2ab + b² ) - ( c² - 2cd + d² )

= ( a - b )² - ( c - d )²

= [ ( a - b ) - ( c - d ) ][ ( a - b ) + ( c - d ) ]

= ( a - b - c + d )( a - b + c - d )

a³b³ - 1

= ( ab )³ - 1³

= ( ab - 1 )[ ( ab )² + ab.1 + 1² ]

= ( ab - 1 )( a²b² + ab + 1 )

x²( y - z ) + y²( z - x ) + z²( x - y )

= z²( x - y ) + x²y - x²z + y²z + y²x

= z²( x - y ) + ( x²y - y²x ) - ( x²z - y²z )

= z²( x - y ) + xy( x - y ) - z( x² - y² )

= z²( x - y ) + xy( x - y ) - z( x + y )( x - y )

= ( x - y )[ z² + xy - z( x + y ) ]

= ( x - y )( z² + xy - zx - zy )

= ( x - y )[ ( z² - zx ) - ( zy - xy ) ]

= ( x - y )[ z( z - x ) - y( z - x ) ]

= ( x - y )( z - x )( z - y )

D

`D ( 10x^6 y^4 z^2)`.

D.

Giải:

1) \(\left(x^2-y\right)^3\)

\(=x^6-3x^4y+4x^2y^2-y^3\)

Vậy ...

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

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

\(=x^3-3x^2+16x-2^3+3\left(x^2-4x-4\right)y+3\left(x-2\right)y^2+y^3\)

\(=x^3-3x^2+16x-2^3+3x^2-12x-12y+3\left(xy^2-2y^2\right)+y^3\)

\(=x^3-3x^2+16x-2^3+3x^2-12x-12y+3xy^2-6y^2+y^3\)

\(=x^3+4x-8-12y+3xy^2-6y^2+y^3\)

Vậy ...

3) \(\left(z+y^2\right)^3\)

\(=z^3+3z^2y^2+3zy^4+y^6\)

Vậy ...

4) \(\left(x-y+z\right)^3\)

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

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

\(=x^3-3x^2y+3xy^2-y^3+3x^2-6xy+3y^2z+3xz^2-3yz^2+z^3\)

\(=-3x^2y+3xy^2-y^3+4x^2-6xy+3y^2z+3xz^2-3yz^2+z^3\)

Vậy ...