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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) \(\dfrac{3x-2}{2xy}+\dfrac{7x+2}{2xy}\)
\(=\dfrac{\left(3x-2\right)+\left(7x+2\right)}{2xy}\)
\(=\dfrac{3x-2+7x+2}{2xy}\)
\(=\dfrac{10x}{2xy}\)
\(=\dfrac{5}{y}\)
b) \(\dfrac{5x+y^2}{x^2y}+\dfrac{x^2-5y}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}+\dfrac{x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)+x\left(x^2-5y\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3+x^3-5xy}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
c) \(\dfrac{3x-2}{2xy}-\dfrac{7x-y}{2xy}\)
\(=\dfrac{\left(3x-2\right)-\left(7x-y\right)}{2xy}\)
\(=\dfrac{3x-2-7x+y}{2xy}\)
\(=\dfrac{-2-4x+y}{2xy}\)
d) \(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\) MTC: \(x^2y^2\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
e) \(\dfrac{16xy}{3x-1}.\dfrac{3-9x}{12xy^3}\)
\(=\dfrac{16xy\left(3-9x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{4\left(3-9x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4\left(9x-3\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-4.3\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-12}{3y^2}\)
\(=\dfrac{-4}{y^2}\)
f) \(\dfrac{8xy}{3x-1}:\dfrac{12xy^3}{5-15x}\)
\(=\dfrac{8xy}{3x-1}.\dfrac{5-15x}{12xy^3}\)
\(=\dfrac{8xy\left(5-15x\right)}{12xy^3\left(3x-1\right)}\)
\(=\dfrac{2\left(5-15x\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2\left(15x-5\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-2.5\left(3x-1\right)}{3y^2\left(3x-1\right)}\)
\(=\dfrac{-10}{3y^2}\)
dài quá mình làm 3 câu đầu thôi nhé!
a)7x^2-14xy
=7x(x-2y)
b) 3x^2-6xy+3y^2
=3(x^2-2xy+y^2)
c) x^2-4z^2-2xy+y^2
=(x^2-2xy+y^2)-4z^2
=(x-y-2z)(x-y+2z)
=3(x-y)^2
c)
6.
\(A=x^2-2x+2\\ =x^2-2x+1+1\\ =\left(x-1\right)^2+1\\ \left(x-1\right)^2\ge0\forall x\\ \left(x-1\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
Vậy \(Min_A=1\) khi \(x=1\)
5,
a,
\(x\left(x-2\right)+x-2=0\\ \left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b,
\(5x\left(x-3\right)-x+3=0\\ 5x\left(x-3\right)-\left(x-3\right)=0\\ \left(5x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
c,
\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\\ 3x^2-15x-\left(2x-2+3x^2-3x\right)\\ 3x^2-15x-\left(3x^2-x-2\right)=30\\ 3x^2-15x-3x^2+x+2=30\\ -14x+2=30\\ -14x=28\\ x=-2\)
Vậy \(x=-2\)
d,
\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\\ x^2+5x+6-\left(x^2+3x-10\right)=0\\ x^2+5x+6-x^2-3x+10\\ 8x+16=0\\ 8x=-16\\ x=-2\)
Vậy \(x=-2\)
\(B=7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
\(E=x^2+7x+12\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)
\(F=x^2-9x+18\)
\(=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)\)
\(=\left(x-3\right)\left(x-6\right)\)
\(H=8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=4x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
1 ) Thực hiện phép tính :
a ) \(-\frac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
b ) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2-x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2x-2-x^3\)
c ) \(\left(x^3+5x^2-2x+1\right)\left(x-7\right)\)
\(=x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\)
\(=x^4-2x^3-37x^2+15x-7\)
d ) \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2+y^3\)
e ) \(\left[\left(x^2-2xy+2y^2\right)\left(x+2y\right)-\left(x^2-4y^2\right)\left(x-y\right)\right]2xy\)
( để xem lại )
2 Tìm x
a ) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=3\)
b ) Sai đề
c ) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
( Để xem lại )
mình chép đúng theo đề cô cho mà sao lại sai được ,hay cô cho sai đề
a) x2 +2xy + x + 2y
= (x2 +2xy) + (x + 2y)
= x(x + 2y) + (x + 2y)
= (x + 2y)(x + 1)
b) 7x2 - 7xy - 5x + 5y
= (7x2 - 7xy) - (5x - 5y)
= 7x(x - y) - 5(x - y)
= (x - y)(7x - 5)
c) 3(x + 2) - y(x + 2)
= (x + 2)(3 - y)
d) 5(x - 3) + y(3 - x)
= 5(x - 3) - y(x - 3)
= (x - 3)(5 - y)
e) (x2 - 8)2 + 36
= x4 - 16x2 + 64 + 36
= x4 - 16x2 + 100
= x4 + 20x2 - 36x2 + 100
= (x4 + 20x2 + 100) - 36x2
= (x2 + 10)2 - (6x)2
= (x2 + 10 - 6x)(x2 + 10 + 6x)