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Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
\(a,\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x\left(x-3\right)\)
\(=x^3-6x^2+12x-27-x^3+x+6x^2-18x\)
\(=-5x-27\)
\(b,\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-\left(8x^3-y^3\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y+z-x-y\right)^2\)
\(=z^2\)
a)
=\(x^3-6x^2+12x+8-27-x^3+x+6x^2-18x\)
=-5x-19
b)
=\(8x^3+y^3-8x^3+y^3\)
=\(2y^3\)
c)
=(x+y+z-x-y)\(^2\) +x+y
=\(z^2+x+y\)
hc tốt
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
a. 3(x - y)2 - 2(x + y)2 - (x - y)(x + y)
= 3(x2 - 2xy + y2) - 2(x2 + 2xy + y2) - (x2 - y2)
= 3x2 - 6xy + 3y2 - 2x2 - 4xy - 2y2 - x2 + y2
= - 10xy + 2y2
b. 2(2x + 5)2 - 3(4x + 1)(1 - 4x)
= 2(4x2 + 20x + 25) - 3(1 + 4x)(1 - 4x)
= 8x2 + 40x + 50 - 3(1 - 16x2)
= 8x2 + 40x + 50 - 3 + 48x2
= 56x2 + 40x + 47
pn coi kt lại nhé
Ở câu a, hàng thứ ba gần cuối phần...-x2+y2 hình như sai kìa bạn ơi phải là x2-y2 chớ
b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)+\left(2x+y\right)\left(4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2+4x^2+2xy+y^2\right)\)
\(=\left(2x+y\right)\left(8x^2+2y^2\right)\)
\(=\left(2x+y\right)\left(4x+y\right).2xy\)
Bài 1 :
a) \(A=\left(x-3\right)^2-\left(x-5\right)\left(x+5\right)\)
\(A=x^2-6x+9-x^2+25\)
\(A=34-6x\)
b) \(B=2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
Dễ thấy đây là HĐT thứ 1
\(B=\left(x+y+x-y\right)^2\)
\(B=\left(2x\right)^2\)
\(B=4x^2\)
Bài 2 :
a) \(2x\left(x+5\right)-x^2-5x=0\)
\(2x\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(x+5\right)\left(2x-2\right)=0\)
\(2\left(x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)
b) \(4x\left(x-2013\right)-x+2013=0\)
\(4x\left(x-2013\right)-\left(x-2013\right)=0\)
\(\left(x-2013\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2013=0\\4x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2013\\x=\frac{1}{4}\end{cases}}}\)