Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
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, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
a) x2 - 4x + 2 = (x2 - 4x + 4) - 2 = (x - 2)2 - 2 = \(\left(x-2+\sqrt{2}\right)\left(x-2-\sqrt{2}\right)\)
b) x2 - 12x + 11 = x2 - x - 11x + 11 = x(x - 1) - 11(x - 1) = (x - 1)(x - 11)
c) 3x2 + 6x - 9 = 3x2 - 3x + 9x - 9 = 3x(x - 1) + 9(x - 1) = (3x + 9)(x - 1) = 3(x + 3)(x - 1)
d) 2x2 - 6x + 2 = 2(x2 - 3x + 1) = 2(x2 - 3x + 9/4 - 5/4) = 2[(x - 3/2)2 - 5/4] = \(2\left(x-\frac{3}{2}+\sqrt{\frac{5}{4}}\right)\left(x-\frac{3}{2}-\sqrt{\frac{5}{4}}\right)\)
1.
a) \(x^2-4x+2=\left(x^2-4x+4\right)-2=\left(x-2\right)^2-2=\left(x-2-\sqrt{2}\right)\left(x-2+\sqrt{2}\right)\)
b) \(x^2-12x+11=\left(x^2-12x+36\right)-25=\left(x-6\right)^2-5^2=\left(x-6-5\right)\left(x-6+5\right)=\left(x-11\right)\left(x-1\right)\)
c) \(3x^2+6x-9=3\left(x^2+2x-3\right)=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2-6x+2=2\left(x^2-3x+1\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\right)=2\left[\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\right]\)
\(=2\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)\)
a) 7x+7y=7(x+y)
b) 2x2y-6xy2=2xy(x-3y)
c)3x(x-1)+7x2(x-1)=x(x-1)(3+7x)
d)3x(x-4)+5x2(4-x)=(x-4)(3x-5x2)
=x(x-4)(3-5x)
e)6x4-9x3=3x3(2x-3)
f)5y8-15y6=5y6(y2-3)
Bài 1
a, x2 + 4x + 3
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)