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a
\(xy+3x-7y-21\\ =\left(xy+3x\right)-\left(7y+21\right)\\ =x\left(y+3\right)-7\left(y+3\right)\\ =\left(y+3\right)\left(x-7\right)\)
b
\(2xy-15-6x+5y\\ =\left(2xy-6x\right)-\left(15-5y\right)\\ =2x\left(y-3\right)-5\left(3-y\right)\\ =2x\left(y-3\right)+5\left(y-3\right)\\ =\left(y-3\right)\left(2x+5\right)\)
c Đề phải là \(\left(2x^2y+2xy^2-x-y\right)\) mới phân tích được: )
\(=2xy\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(2xy-1\right)\)
d
\(7x^3y-3xyz-21x^2+9z\\ =\left(7x^3y-21x^2\right)-\left(3xyz-9z\right)\\ =7x^2\left(xy-3\right)-3z\left(xy-3\right)\\ =\left(xy-3\right)\left(7x^2-3z\right)\)
e
\(4x^2-2x-y^2-y\\ =\left(2x\right)^2-y^2-\left(2x+y\right)\\ =\left(2x-y\right)\left(2x+y\right)-\left(2x+y\right)\\ =\left(2x+y\right)\left(2x-y-1\right)\)
f
\(9x^2-25y^2-6x+10y\\ =\left(3x\right)^2-\left(5y\right)^2-\left(6x-10y\right)\\ =\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)\\ =\left(3x-5y\right)\left(3x+5y-2\right)\)
a: =x(y+3)-7(y+3)
=(y+3)(x-7)
b: \(=2xy-6x+5y-15\)
=2x(y-3)+5(y-3)
=(y-3)(2x+5)
c: \(=2xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(2xy-1\right)\)
d: \(=xy\left(7x^2-3z\right)-3\left(7x^2-3z\right)\)
=(7x^2-3z)(xy-3)
e: =4x^2-y^2-2x-y
=(2x-y)(2x+y)-(2x+y)
=(2x+y)(2x-y-1)
f: =(3x-5y)(3x+5y)-2(3x-5y)
=(3x-5y)(3x+5y-2)
x2 + 2y2 + 2xy - 6x - 2y + 13 = 0
<=> ( x2 + 2xy + y2 - 6x - 6y + 9 ) + ( y2 + 4y + 4 ) = 0
<=> [ ( x2 + 2xy + y2 ) - ( 6x + 6y ) + 9 ] + ( y + 2 )2 = 0
<=> [ ( x + y )2 - 2( x + y ).3 + 32 ] + ( y + 2 )2 = 0
<=> ( x + y - 3 )2 + ( y + 2 )2 = 0
Ta có : \(\hept{\begin{cases}\left(x+y-3\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(x+y-3\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Dấu "=" xảy ra <=> x = 5 ; y = -2
Thế x = 5 ; y = -2 vào A ta được :
\(A=\frac{5^2-7\cdot5\cdot\left(-2\right)+52}{5-\left(-2\right)}=\frac{25+70+52}{7}=\frac{147}{7}=21\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a) 6x2 - 12x
= 6x(x - 2)
b) x2 + 2x + 1 - y2
= (x2 + 2x + 1) - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
c) x + y + z + x2 + xy + xz
= (x + x2) + (y + xy) + (z + xz)
= x(1 + x) + y(1 + x) + z(1 + x)
= (x + y + z)(x + 1)
d) xy + xz + y2 + yz
= (xy + xz) + (y2 + yz)
= x(y + z) + y(y + z)
= (x + y)(x + z)
e) x3 + x2 + x + 1
= (x3 + x2) + (x + 1)
= x2(x + 1) + (x + 1)
= (x2 + 1)(x + 1)
f) xy + y - 2x - 2
= (xy + y) - (2x + 2)
= y(x + 1) - 2(x + 1)
= (y - 2)(x + 1)
g) x3 + 3x - 3x2 - 9
= (x3 - 3x2) + (3x - 9)
= x2(x - 3) + 3(x - 3)
= (x2 + 3)(x - 3)
h) x2 - y2 - 2x - 2y
= (x2 - y2) - (2x + 2y)
= (x + y)(x - y) - 2(x + y)
= (x + y)(x - y - 2)
i) 7x2 - 7xy - 5x = 5y
mk thấy con này sai sai ý
a) \(xy+3x-7y-21\)
\(\Leftrightarrow\left(xy+3x\right)-\left(7y+21\right)\)
\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)\)
\(\Leftrightarrow\left(x-7\right)\left(y+3\right)\)
b) \(2xy-15-6x+5y\)
\(\Leftrightarrow\left(2xy-6x\right)-\left(15-5y\right)\)
\(\Leftrightarrow x\left(2y-6\right)-5\left(3-y\right)\)
\(\Leftrightarrow2x\left(y-3\right)+5\left(y-3\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(y-3\right)\)
\(B=7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
\(E=x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(F=x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
\(H=8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=4x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
Sửa đề: B=-2x^2+xy+2y^2-3-5x+2y
a: A+B+C
=x^2-3xy-y^2+2x-3y+1-2x^2+xy+2y^2-3-5x+2y+C
=-x^2-2xy+y^2-3x-y-2+3x^2+7y^2-4xy-6x+4y+5
=2x^2+8y^2-6xy-9x+3y+3
b: 7A-B-C-9
=7A-9-(x^2+9y^2-3xy-11x+6y+2)
=7x^2-7y^2-21xy+14x-21y+7-x^2-9y^2+3xy-11x-6y-2-9
=6x^2-16y^2-18xy+3x-27y-4
\(3x^2+2y^2=7xy\)
\(\Leftrightarrow3x^2-7xy+2y^2=0\)
\(\Leftrightarrow3x^2-6xy-xy+2y^2=0\)
\(\Leftrightarrow3x\left(x-2y\right)-y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(3x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-y=0\\x-2y=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x=y\\x=2y\end{matrix}\right.\)
+) TH1 : \(y=3x\)
\(\Leftrightarrow A=\dfrac{3x+y}{7y-x}+\dfrac{6x-9y}{2x+y}\)
\(=\dfrac{3x+3x}{7.3x-x}+\dfrac{6x-9.3x}{2x+3x}\)
\(=\dfrac{9x}{20x}+\dfrac{-21x}{5x}\)
\(=-\dfrac{15}{4}\)
+) TH2 : \(x=2y\)
\(\Leftrightarrow A=\dfrac{3x+y}{7y-x}+\dfrac{6x-9y}{2x+y}\)
\(=\dfrac{3.2y+y}{7y-2y}+\dfrac{6.2y-9y}{2.2y+y}\)
\(=\dfrac{7y}{5y}+\dfrac{3y}{5y}\)
\(=2\)
Vậy...
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