Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
\(\text{ }\dfrac{\left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}.\left(x-c\right)+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}.\left(x-c\right)=1\)
\(\Leftrightarrow\left(x-c\right)\left(\dfrac{\left(x-a\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)}{\left(a-b\right)\left(a-c\right)}\right)=1\)
\(\Leftrightarrow\left(x-c\right)\dfrac{\left(a-x\right)\left(a-c\right)+\left(x-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\left(x-c\right)\left[\left(a^2-b^2\right)-x\left(a-b\right)-c\left(a-b\right)\right]=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a-b\right)\left(a+b-x-c\right)=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(\Leftrightarrow\left(x-c\right)\left(a+b-x-c\right)-\left(b-c\right)\left(a-c\right)=0\)
\(\Leftrightarrow ax+bx-x^2-xc-ac-bc+xc+c^2-ab+bc+ac-c^2=0\)
\(\Leftrightarrow x^2-ax-bx+ab=0\)
\(\Leftrightarrow x\left(x-a\right)+b\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=0\\x-b=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=a\\x=b\end{matrix}\right.\)
Vậy\(S=\left\{a,b\right\}\)
Mấy bài này đăng nhiều rồi bạn ;v
Bài 1: Nhân cả 2 vế cho a+b+c rồi rút gọn được đpcm
Bài 2: Thêm 1 rồi bớt 1 :v (x+y+xy+1-1)
\(\dfrac{ \left(x-a\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)}+\dfrac{\left(x-b\right)\left(x-c\right)}{\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-a\right)\left(x-c\right)+\left(x-b\right)\left(x-c\right)}{\left(b-a\right)\left(b-c\right)+\left(a-b\right)\left(a-c\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(x-a+x-b\right)}{\left(b-a\right)\left(b-c\right)-\left(b-a\right)\left(c-a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(b-c-c+a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-a-b\right)}{\left(b-a\right)\left(-2c+b+a\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)\left(2x-\left(a+b\right)\right)}{\left(b-a\right)\left(-2c+\left(a+b\right)\right)}=1\)
\(\Leftrightarrow\dfrac{\left(x-c\right)2x}{\left(b-a\right)\left(-2c\right)}=1\)
\(\Leftrightarrow\dfrac{2x^2-2xc}{-2cb+2ac}=1\)
a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)
b: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+1\right)\left(x+2\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)
2b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
<=> \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)
<=> (ab+bc+ca)(a+b+c)=abc
<=> (ab+bc+ca)(a+b+c)-abc=0
<=> (a+b)(b+c)(c+a) = 0
<=> a+b=0 hoặc b+c=0 hoặc c+a=0
<=> a=-b hoặc b=-c hoặc c = -a
sau đó thay vào cái cần c/m
ĐK: \(x\ne b;x\ne c\)
Phương trình tương đương:
\(\dfrac{2}{b-x}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=\dfrac{1}{c-x}\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
TH1: Nếu \(a=b\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}\Rightarrow\) pt tương đương \(0=0\) \(\Rightarrow\) đúng với mọi x
TH2: nếu \(a\ne b\), chia cả 2 vế cho \(\dfrac{1}{a}-\dfrac{1}{b}\) ta được:
\(\dfrac{2}{b-x}=\dfrac{1}{c-x}\Leftrightarrow2c-2x=b-x\Leftrightarrow x=2c-b\)
Khó vậy mày.