Bỏ dấu ngoặc rồi viết các biểu thức sau
a) (m-n) (m+n)
b) x(a-b)-x(a+b)
c) (a2-ax+by)-(by-a2-ax)
d) (a-b) (a+b)-(b-a)b
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AH
Akai Haruma
Giáo viên
12 tháng 3 2020
Lời giải:
a)
$x(a-b)-x(a+b)=xa-xb-xa-xb=-2xb$
b)
$(a^2-ax+by)-(by-a^2-ax)=a^2-ax+by-by+a^2+ax=2a^2$
c)
$(a-b)(a+b)-(b-a)b=a^2-b^2-(b^2-ab)=a^2-b^2-b^2+ab=a^2-2b^2+ab$
DN
12 tháng 3 2020
\(a,x\left(a-b\right)-x\left(a+b\right)\)
\(=ax-bx-ax-bx=-2bx\)
\(b,\left(a^2-ax+by\right)-\left(by-a^2-ax\right)\)
\(=a^2-ax+by-by+a^2+ax=2a^2\)
\(c,\left(a-b\right)\left(a+b\right)-\left(b-a\right).b\)
\(=a^2-b^2-b^2+ab\)
\(=\left(a-b\right)^2-b^2=\left(a-b-b\right)\left(a-b+b\right)\)
\(=\left(a-2b\right).a\)
FR
1
3 tháng 1 2017
giúp mk vs các bn ui, mai mk nộp bài rùi, mk cần gấp lắm lắm,...giúp mk nha....
7 tháng 1 2018
\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)19 tháng 2 2018
tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
7 tháng 8 2018
\(a.a\left(b+c\right)+3b+3c=a\left(b+c\right)+3\left(b+c\right)=\left(b+c\right)\left(a+3\right)\)
\(b.a\left(c-d\right)+c-d=\left(c-d\right)\left(a+1\right)\)
\(c.b\left(a-c\right)+5a-5c=b\left(a-c\right)+5\left(a-c\right)=\left(a-c\right)\left(b+5\right)\)
\(d.a\left(m-n\right)+m-n=\left(m-n\right)\left(a+1\right)\)
\(e.mx+my+5x+5y=m\left(x+y\right)+5\left(x+y\right)=\left(x+y\right)\left(m+5\right)\)
\(f.ma+mb-a-b=m\left(a+b\right)-\left(a+b\right)=\left(a+b\right)\left(m-1\right)\)
\(g.4x+by+4y+bx=4x+bx+by+4y=x\left(b+4\right)+y\left(b+4\right)=\left(b+4\right)\left(x+y\right)\)
\(h.1-ax-x+a=\left(a+1\right)-x\left(a+1\right)=\left(a+1\right)\left(1-x\right)\)
\(k.x^{m+2}-x^m=x^m\left(x^2-1\right)=x^m\left(x-1\right)\left(x+1\right)\)
\(m.\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)=\left(b-a\right)^2-\left(b-a\right)\left(a+b\right)=\left(b-a\right)\left(b-a-a-b\right)=-2a\left(b-a\right)\)
\(n.a\left(a-b\right)\left(a+b\right)-\left(a^2-2ab+b^2\right)=a\left(a-b\right)\left(a+b\right)-\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab-a+b\right)\)
28 tháng 6 2016
a) \(M=ax+bx+ay+by=x\cdot\left(a+b\right)+y\cdot\left(a+b\right)=\left(a+b\right)\cdot\left(x+y\right)=2\cdot17=34.\)
b) \(N=ax-by+bx-ay=a\left(x-y\right)+b\left(x-y\right)=\left(a+b\right)\left(x-y\right)=7\cdot1=7\)
10 tháng 7 2021
Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+b^2y^2+2axby\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay=bx\)
hay \(\dfrac{a}{x}=\dfrac{b}{y}\)
10 tháng 7 2021
Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2abxy+b^2y^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\)
\(\Leftrightarrow ay-bx=0\)
\(\Leftrightarrow ay=bx\Leftrightarrow\dfrac{a}{b}=\dfrac{x}{y}\)
a) (m - n) (m + n) = m2 + mn - mn + n2 = m2 + n2
b) x(a - b) - x(a + b) = ax - bx - ax - bx = -2bx
c) (a2 - ax + by) - (by - a2 - ax) = a2 - ax + by - by + a2 + ax = 2a2
d) (a - b) (a + b) - (b - a)b = a2 + ab - ab - b2 - b2 + ab = a2 - 2b2 + ab
cam on ban