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Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
a) = -18x^4*y^4 + 3x^3*y^3 - 6/5x^2*y^4
b) = x^4 + 6x^3 + 9x^2 + 9x^2 - 6x^3 -18x^2 = x^4
bn dùng phương pháp nhân đơn thức vs đa thức đó dễ mà!!!!
chúc bạn học tốt!! ^^
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Đăng từng bài thôi nha bạn
Bài 1 : Năm nay mới lên lớp 8 -_-
Bài 2 :
\(a)\)
* Câu A :
\(A=x^2+4x-7\)
\(A=\left(x^2+4x+4\right)-11\)
\(A=\left(x+2\right)^2-11\ge-11\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=-2\) ( ở đây nhiều bài quá nên mình làm tắt cho nhanh, bạn nhớ trình bày rõ ra nhé )
Vậy GTNN của \(A\) là \(-11\) khi \(x=-2\)
* Câu B :
\(B=2x^2-3x+5\)
\(2B=4x^2-6x+10\)
\(2B=\left(4x^2-6x+1\right)+9\)
\(2B=\left(2x-1\right)^2+9\ge9\)
\(B=\frac{\left(2x-1\right)^2+9}{2}\ge\frac{9}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy GTNN của \(B\) là \(\frac{9}{2}\) khi \(x=\frac{1}{2}\)
* Câu C :
\(C=x^4-3x^2+1\)
\(C=\left(x^4-3x^2+\frac{9}{4}\right)-\frac{5}{4}\)
\(C=\left(x^2-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{\frac{3}{2}}\\x=-\sqrt{\frac{3}{2}}\end{cases}}\)
Vậy GTNN của \(C\) là \(-\frac{5}{4}\) khi \(x=\sqrt{\frac{3}{2}}\) hoặc \(x=-\sqrt{\frac{3}{2}}\)
Chúc bạn học tốt ~
Bài 1:
a) Sửa đề \(x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(x^2+2019x-xy-2019y\)
\(=x\left(x+2019\right)-y\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-y\right)\)
c) \(x^2-9y^2-4x+4\)
\(=\left(x^2-4x+4\right)-9y^2\)
\(=\left(x-2\right)^2-\left(3y\right)^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
d) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Bài 2:
a) \(\left(6x^3y^3-27xy^2\right):\left(3x^2y\right)-2xy^2\)
\(=6x^3y^3:3x^2y-27xy^2:3x^2y-2xy^2\)
\(=2xy^2-\dfrac{9y}{x}-2xy^2\)
\(=-\dfrac{9y}{x}\)
b) \(\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}+\dfrac{3x+2}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}-\dfrac{3x+2}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(1-2x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)+\left(1-2x\right)\left(x-2\right)-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+4+x-2-2x^2+4x-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x}{x+2}\)
Bài 3:
a) \(3x\left(2x-3\right)-x\left(6x+4\right)=7-12x\)
\(\Rightarrow6x^2-9x-6x^2-4x=7-12x\)
\(\Rightarrow-13x=7-12x\)
\(\Rightarrow-13x+12x-7=0\)
\(\Rightarrow-x-7=0\)
\(\Rightarrow-x=7\)
\(\Rightarrow x=-7\)
b) \(3\left(x-5\right)-2x^2+10x=0\)
\(\Rightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 1
a) (6x4y2 - 3x3y3) : 3x3y2 = 6x4y2 : 3x3y2 - 3x3y3 : 3x3y2 = 2x - y
b) (2x - 1)(x2 - x + 3) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3
Bài 2
1) (x - 2)2 - (x - 3)2 = (x - 2 - x + 3)(x - 2 + x - 3) = 2x - 5>
2) 4x2 - 4xy + 2y2 + 1 = (4x2 - 4xy + y2) + y2 + 1 = (2x - y)2 + y2 + 1 > 0
vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)
\(x^2+y^2-xy-2x-2y+9=x^2+y^2+2xy-2x-2y+9-3xy\)
\(=\left(x+y\right)^2-2\left(x+y\right)+9-3xy=\left(x+y-2\right)\left(x+y\right)+9-3xy.\)
\(đếnđâytịt\)
b
c, =3 dễ
\(\frac{3x^2-6x+9}{x^2-2x+3}=\frac{3\left(x^2-2x+3\right)}{x^2-2x+3}=3\)
a, - \(\dfrac{1}{3}\).\(xy\).(3\(x^3\).y2 - 6\(x^2\) + y2)
= - \(x^4\).y3 + 2\(x^3\).y - \(\dfrac{1}{3}\).\(xy^3\)
b, (2\(x\) -3).(4\(x\)2 + 6\(x\) + 9)
= (2\(x\))3 - 33
= 8\(x^3\) - 27