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Ta có : \(4x^2+2y^2+2z^2-4xy-4xz+2yz-6y-10z+34=0\)
\(\Leftrightarrow\left(4x^2+y^2+z^2-4xy-4xz+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)
\(\Leftrightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)
Do \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\\\left(y-3\right)^2\ge0\\\left(z-5\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=y+z\\y=3\\z=5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}}\)
Khi đó \(P=\left(4-4\right)^{2018}+\left(3-4\right)^{2018}+\left(5-4\right)^{2018}\)
\(=0+\left(-1\right)^{2018}+1^{2018}\)
\(=2\)
Bài 3 : Ta có : \(A=\frac{2}{5}xy\left(x^2y-5x+10y\right)\)
\(A=\frac{2}{5}xy\cdot x^2y+\frac{2}{5}xy\left(-5x\right)+\frac{2}{5}xy\cdot10y\)
\(A=\frac{2}{5}x^3y^2-2x^2y+4xy^2\)
Chọn C
Bài 4 : \(\left(x-2\right)\left(x+5\right)=x\left(x+5\right)-2\left(x+5\right)\)
\(=x^2+5x-2x-10\)
\(=x^2+3x-10\)
Chọn B
Bài 3 :
Ta có: A = 2/5xy( x2y -5x + 10y )
= 2/5xy.x2y - 2/5xy.5x + 2/5xy.10y
= 2/5x3y2 - 2x2y + 4xy2.
Chọn đáp án C
Bài 4 :
Ta có ( x - 2 )( x + 5 )
= x( x + 5 ) - 2( x + 5 )
= x2 + 5x - 2x - 10 = x2 + 3x - 10.
Chọn đáp án B.
Hok tốt
1) x2 + 7y2 - 4xy - 2x - 2y + 4 = 0
\(\Leftrightarrow\)[ x2 - 2x.( 2y + 1 ) + 4y2 + 4y +1 ] - 4y2 - 4y - 1 + 7y2 - 2y +4 = 0
\(\Leftrightarrow\) [ x2 - 2x.( 2y +1 ) + ( 2y +1 )2 ] + 3y2 - 6y +3 = 0
\(\Leftrightarrow\) ( x - 2y - 1 )2 + 3.( y2 - 2y + 1 ) = 0
\(\Leftrightarrow\)( x - 2y - 1 )2 + 3.( y - 1 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-2y-1\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x-2y-1=0\\y-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=2y+1\\y=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=3\\y=1\end{cases}}\)
Vậy x = 3 , y = 1 thì x2 + 7y2 - 4xy - 2x - 2y + 4 = 0
2) 11x2 + y2 - 6xy - 14x + 2y +9 = 0
\(\Leftrightarrow\)[ y2 - 2y.( 3x - 1 ) + 9x2 - 6x +1 ] + 2x2 - 8x + 8 = 0
\(\Leftrightarrow\)[ y2 - 2y.( 3x - 1 ) + ( 3x - 1 )2 ] + 2.( x2 - 4x + 4 ) = 0
\(\Leftrightarrow\)( y - 3x + 1 )2 + 2.( x - 2 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(y-3x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y-3x+1=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=3x-1\\x=2\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}y=5\\x=2\end{cases}}\)
Vậy x = 2 , y = 5 thì 11x2 + y2 - 6xy - 14x + 2y + 9 = 0
\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(3+2x-y\right)\left(2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x+y-2\right)\left(3x-y+2\right)\)
A = ( 6x - 3y ) + (4x2 - 4xy + y2 )
A = 3.( 2x - y) + [ ( 2x )2 - 2.2.x.y + y2 ]
A = 3.( 2x - y ) + ( 2x - y )2
A = ( 2x - y ).(3 + 2x - y )
B = 9x2 - ( y2 - 4y + 4 )
B = ( 3x )2 - ( y - 2 )2
B = ( 3x - y + 2 ).( 3x + y - 2 )
C = - 25x2 + y2 - 6y + 9
C = ( y2 - 2.3.y + 32 ) - ( 5x )2
C = ( y - 3 )2 - ( 5x )2
C = (y - 3 - 5x ).( y - 3 +5x )
D = x2 - 4x - y2 -- 8y - 12
D = ( x2 - 4x + 4 ) - 4 - y2 - 8y -12
D = ( x - 2.2x + 22 ) - ( y2 + 2.4.y + 42 )
D = ( x - 2 )2 - ( y + 4 )2
D = ( x - 2 + y + 4 ).( x - 2 - y - 4 )
D = ( x + y + 2 ).( x - y - 6 )
1. ( 2x + y )( 4x2 - 2xy + y2 ) - 8x3 - y3 - 16
= [ ( 2x )3 + y3 ] - 8x3 - y3 - 16
= 8x3 + y3 - 8x3 - y3 - 16
= -16 ( đpcm )
2. ( 3x + 2y )2 + ( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 3x + 2y )2 - 18x2 - 8y2 + 3
= 2( 9x2 + 12xy + 4y2 ) - 18x2 - 8y2 + 3
= 18x2 + 24xy + 8y2 - 18x2 - 8y2 + 3
= 24xy + 3 ( có phụ thuộc vào biến )
3. ( -x - 3 )3 + ( x + 9 )( x2 + 27 ) + 19
= -x3 - 9x2 - 27x - 27 + x3 + 9x2 + 27x + 243 + 19
= -27 + 243 + 19 = 235 ( đpcm )
4. ( x - 2 )3 - x( x + 1 )( x - 1 ) + 13( x - 4 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 13x - 52
= x3 - 6x2 + 12x - 8 - x3 + x + 13x - 52
= -6x2 + 26x - 60 ( có phụ thuộc vào biến )
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
\(49\left(y-4\right)^2-9y^2-36y-36\)
\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)
\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)
\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)
\(=\left(10y-22\right)\left(4y-34\right)\)
a) \(VP=x^2+2xy+y^2-2xy=x^2+y^2=VT\left(DPCM\right)\)
b) \(VT=\left(x+y\right)\left[\left(x+y\right)-\left(x-y\right)\right]=\left(x+y\right)\left(x+y-x+y\right)\\ =2y\left(x+y\right)=VP\left(DPCM\right)\)
c) \(VT=\left[\left(2x+y\right)-z\right]^2=\left(2x+y\right)^2-2z\left(2x+y\right)+z^2\\ =4x^2+4xy+y^2-4xz-2yz+z^2=VP\left(DPCM\right)\)
d) \(VP=x^2+2xy+y^2-4xy=x^2-2xy+y^2\\ =\left(x-y\right)^2=VT\left(DPCM\right)\)
a: \(x^2+y^2\)
\(=x^2+y^2+2xy-2xy\)
\(=\left(x+y\right)^2-2xy\)
b: \(\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-x+y\right)\)
\(=2y\left(x+y\right)\)
c: \(\left(2x+y-z\right)^2\)
\(=\left(2x+y\right)^2-2z\left(2x+y\right)+z^2\)
\(=4x^2+4xy+y^2-4xz-2yz+z^2\)
d: \(\left(x+y\right)^2-4xy\)
\(=x^2+2xy+y^2-4xy\)
\(=x^2-2xy+y^2=\left(x-y\right)^2\)