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a) \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
b) \(x^2+y^2-2xy-25\)
\(=\left(x+y\right)^2-5^2\)
\(=\left(x+y+5\right)\left(x+y-5\right)\)
c) \(4x^2-4xy+y^2=\left(2x-y\right)^2\)
m) \(81-x^2+2xy-y^2\)
\(=9^2-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
k) \(x^2-xy-x+y\)
\(=x\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
a) 2x2 + 4x + xy + 2y
= (2x2 + xy) + (4x + 2y)
= x(2x + y) + 2(2x + y)
= (x + 2)(2x + y)
b) x2 + xy - 7x - 7y
= x(x + y) - 7(x + y)
= (x - y)(x + y)
Trả lời :
Ta có :
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
Hok tốt
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y+2\right)\left(x+y+5\right).\)
b) \(x^2y+xy^2+x+y=2010\)
\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)
\(\Leftrightarrow11\left(x+y\right)+1\left(x+y\right)=2010\)
\(\Leftrightarrow12\left(x+y\right)=2010\)
\(\Leftrightarrow x+y=\frac{335}{2}\)
\(\Leftrightarrow\left(x+y\right)^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+2xy+y^2=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2+22=\frac{112225}{4}\)
\(\Leftrightarrow x^2+y^2=\frac{112137}{4}.\)
Vậy \(x^2+y^2=\frac{112137}{4}.\)
a,\(x^2+2xy+7x+7y+y^2+10=\left(x^2+2xy+y^2\right)+7\left(x+y\right)+10\)
\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)
\(=\left(x+y\right)\left(x+y+2\right)+5\left(x+y+2\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b,\(x^2y+xy^2+x+y=2010\Rightarrow xy\left(x+y\right)+x+y=2010\)
\(\Rightarrow12\left(x+y\right)=2010\Rightarrow x+y=167,5\)
Ta có:\(x^2+y^2=x^2+2xy+y^2-2xy=\left(x+y\right)^2-2xy=\left(167,5\right)^2-2.11=28034,25\)
g. \(x^{^3}+3x^2+3x+1-27z^3\\ =\left(x^{^3}+3x^2+3x+1\right)-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3\right)\left[\left(x+1\right)^2+\left(x+1\right)3z+9z^2\right]\\ =\left(x-2\right)\left(x+2x+1+3zx+3z+9z^2\right)\\ =\left(x-2\right)\left(3x+3zx+3z+9z^2+1\right)\left(x-2\right)3x\left(1+z\right)+3z\left(1+z\right)+1\\ =\left(x-2\right)\left(1+z\right)\left(3x+3z\right)+1\\ =\left(x-2\right)\left(1-z\right)3\left(x+z\right)+1\)
Mk lm hơi tắt, bn chú ý nha:
a,\(x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
=\(\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
=\(\left(x+1\right)^2\left(x^2-x+1\right)\)
b,\(\left(x^4-x^3\right)-\left(x^2-1\right)\)
=\(x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
= \(\left(x-1\right)\left(x^3-x-1\right)\)
c,Đề phải thế này nha:
\(x^2y-xy^2-x+y\)=\(xy\left(x-y\right)-\left(x-y\right)\)
=\(\left(x-y\right)\left(xy-1\right)\)
d,hình như đề sai đó bn, thế này đúng ko?
\(a^2x+a^2y-7x-7y\)=\(a^2\left(x+y\right)-7\left(x+y\right)\)=\(\left(x+y\right)\left(a^2-7\right)\)
e,\(4x^2-x^2-16y^2+4y^2\)
=\((4x^2-16y^2)-\left(x^2-4y^2\right)\)
=\(4\left(x-2y\right)\left(x+2y\right)-\left(x^2-2y\right)\left(x^2+2y\right)\)=\(3\left(x-2y\right)\left(x+2y\right)\)
Cách này nhanh hơn:\(3\left(x^2-4y^2\right)\)
=\(3\left(x-2y\right)\left(x+2y\right)\)
g,\(\left(x+1\right)^3-\left(3z\right)^3\)=
\(\left(x-3z+1\right)[\left(x+1\right)^2+3z\left(x+1\right)+9z^2]\)Nếu thấy đề bn đưa sai thì nhắc mk nhé?
Mong các bn giúp đỡ thêm
Chúc các bn hc tốt
a) Ta có: \(x^2+y=y^2+x\)
\(\Leftrightarrow\left(x^2-y^2\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-y=0\\x+y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=y\left(ktm\right)\\x+y=1\end{cases}}\)
Thay vào ta được:
\(A=\frac{x^2+y^2+xy}{xy-1}=\frac{\left(x+y\right)^2-xy}{xy-1}=\frac{1-xy}{xy-1}=-1\)
b) Kết quả tìm được là các nghiệm vô tỉ nên mong bạn xem lại đề
\(x_1\approx-1,84...\) ; \(x_2\approx-1,15...\) ; \(x_3\approx0,92...\) ; \(x_4\approx3,07...\)
2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
a) \(x^2+7x+7y-y^2\)
\(=\left(x+y\right)\left(x-y\right)+7\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+7\right)\)
b) \(x^2-xy-6y^2\)
\(=-6y^2-3xy+2xy+x^2\)
\(=-3y\left(2y+x\right)+x\left(2y+x\right)\)
\(=\left(x-3y\right)\left(2y+x\right)\)
c) \(x^2-3x^2-6x+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)-3x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-5x+4\right)\)
\(=\left(x+2\right)\left(x^2-4x-x+4\right)\)
\(=\left(x+2\right)\left[x\left(x-4\right)-\left(x-4\right)\right]\)
\(=\left(x+2\right)\left(x-1\right)\left(x-4\right)\)
a)x2+7x+7y-y2=(x-y)(x+y)+7.(x+y)
=(x+y)(x-y+7)
b)x2-xy-6y2=x2-xy-4y2-2y2
=(x-2y)(x+2y)-y(x-2y)
=(x-2y)(x+2y-y)
Ta có: \(xy=2\Rightarrow2xy=4\)
\(x+y=3\Rightarrow\left(x+y\right)^2=9\Rightarrow x^2+2xy+y^2=9\)
\(\Rightarrow x^2+y^2=5\text{ vì }2xy=4\)
\(x^3+y^3=\left(x+y\right)^3-3xy.\left(x+y\right)=\left(x+y\right).\left(x^2-xy+y^2\right)\)
\(\Rightarrow x^3+y^3+7x^2+7y^2=\left(x+y\right).\left(x^2-xy+y^2\right)+7.\left(x^2+y^2\right)\)
\(=3.\left(5-2\right)+7.5=9+35=44\)