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\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)
\(=3\left(2x-y\right)+\left(2x-y\right)^2\)
\(=\left(3+2x-y\right)\left(2x-y\right)\)
\(B=9x^2-\left(y^2-4y+4\right)\)
\(=9x^2-\left(y-2\right)^2\)
\(=\left(3x+y-2\right)\left(3x-y+2\right)\)
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 )
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
x2 - 4xy + 4y2 - x + 2y
= ( x - 2y )2 - ( x - 2y )
= ( x - 2y )( x - 2y - 1 )
Ta có:
A = (x + 2)2 + (x - 3)2 = x2 + 4x + 4 + x2 - 6x + 9 = 2x2 - 2x + 13 = 2(x2 - x + 1/4) + 25/2 = 2(x - 1/2)2 + 25/2
Ta luôn có: (x - 1/2)2 \(\ge\) 0 \(\forall\)x ----> 2(x - 1/2)2 \(\ge\) 0 \(\forall\)x
=> 2(x - 1/2)2 + 25/2 \(\ge\) 25/2 \(\forall\)x
Dấu "=" xảy ra khi: (x - 1/2)2 = 0 <=> x - 1/2 = 0 <=> x = 1/2
Vậy Amin = 25/2 tại x = 1/2
B = x2 - 4x + y2 - 8y + 6 = (x2 - 4x + 4) + (y2 - 8y + 16) - 14 = (x - 2)2 + (y - 4)2 - 14
Ta luôn có: (x - 2)2 \(\ge\)0 \(\forall\)x
(y - 4)2 \(\ge\)0 \(\forall\)y
=> (x - 2)2 + (y - 4)2 - 14 \(\ge\) -14 \(\forall\)x,y
hay B \(\ge\)-14 \(\forall\)x, y
Dấu "=" xảy ra khi : \(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}x-2=0\\x-4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Vậy Bmin = -14 tại x = 2 và y = 4
a) \(x^5+x+1\)
\(=\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)
\(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
b) \(6x^2-13x+6\)
\(=\left(6x^2-9x\right)-\left(4x-6\right)\)
\(=3x\left(2x-3\right)-2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(3x-2\right)\)
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
WTF đăng một loạt vầy ai dám làm @@
Mấy bài này trong sách bài tập cx có bài mẫu
tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)