Thực hiện phép tính
\(\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
Những câu hỏi liên quan
Thực hiện phép tính :
\(\dfrac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\dfrac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\dfrac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
30 tháng 1 2018 lúc 21:13
Thực hiện phép tính:
1) Adfrac{1}{left(a-bright)left(a-cright)}+dfrac{1}{left(b-aright)left(b-cright)}+dfrac{1}{left(c-aright)left(c-bright)}
2) Bdfrac{1}{aleft(a-bright)left(a-cright)}+dfrac{1}{bleft(b-aright)left(b-cright)}+dfrac{1}{cleft(c-aright)left(c-bright)}
3, Cdfrac{bc}{left(a-bright)left(a-cright)}+dfrac{ac}{left(b-aright)left(b-cright)}+dfrac{ab}{left(c-aright)left(c-bright)}
4) Ddfrac{a^2}{left(a-bright)left(a-cright)}+dfrac{b^2}{left(b-aright)left(b-cright)}+dfrac{c^2}{left(c-ar...
Đọc tiếp
Thực hiện phép tính:
1) \(A=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
2) \(B=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
3, \(C=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ac}{\left(b-a\right)\left(b-c\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)
4) \(D=\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Đúng 0
Bình luận (0)
Thực hiện phép tính :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
Ta có :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}\)
\(=\dfrac{b\left(b-c\right)-a\left(a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b^2-bc-a^2+ac}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b^2-a^2\right)-\left(bc-ac\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(b-a\right)\left(b+a-c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(-a-b+c\right)}{ab\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}\)
Như vậy :
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{-a-b+c}{ab\left(a-c\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{c\left(-a-b+c\right)+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{-ac-bc+c^2+ab}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(-ac+ab\right)-\left(bc-c^2\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(b-c\right)}{abc\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{1}{abc}\)
Đúng 0
Bình luận (0)
1. Thực hiện phép tính:
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)
\(=\dfrac{1}{a\left(a-b\right)\left(a-c\right)}-\dfrac{1}{b\left(a-b\right)\left(b-c\right)}+\dfrac{1}{c\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{bc\left(b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\dfrac{ac\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\dfrac{ab}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{b^2c-bc^2-a^{ 2}c+ac^2+a^2b-ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(b-c\right)\left(b+c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(b-c\right)\left(a-b\right)\left(a+b-b-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{1}{abc}\)
Đúng 0
Bình luận (0)
từ giả thiết, ta có dfrac{1}{xy}+dfrac{1}{yz}+dfrac{1}{zx}1
đặt left(dfrac{1}{xy};dfrac{1}{yz};dfrac{1}{zx}right)left(a;b;cright)Rightarrow a+b+c1 left(dfrac{ac}{b};dfrac{ab}{c};dfrac{bc}{a}right)left(dfrac{1}{x^2};dfrac{1}{y^2};dfrac{1}{z^2}right)
ta có VTdfrac{1}{sqrt{1+dfrac{1}{x^2}}}+dfrac{1}{sqrt{1+dfrac{1}{y^2}}}+dfrac{1}{sqrt{1+dfrac{1}{z^1}}}sqrt{dfrac{1}{1+dfrac{ac}{b}}}+sqrt{dfrac{1}{1+dfrac{ab}{c}}}+sqrt{dfrac{1}{1+dfrac{bc}{a}}}
dfrac{1}{sqrt{dfrac{b+ac}{b}}}+dfrac{1}{sqrt{dfrac...
Đọc tiếp
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
Sửa đề: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
Ta có: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\dfrac{b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Đúng 1
Bình luận (0)
Thực hiên phép tính:
a) dfrac{1}{left(a-bright)left(b-cright)}+dfrac{1}{left(b-cright)left(c-aright)}+dfrac{1}{left(c-aright)left(a-bright)}
b) dfrac{left[a^2-left(b+cright)^2right]left(a+b-cright)}{left(a+b+cright)left(a^2+c^2-2ac-b^2right)}
c) dfrac{x-1}{x^3}-dfrac{x+1}{x^3-x^2}+dfrac{3}{x^3-2x^2+x}
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}
Đọc tiếp
Thực hiên phép tính:
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
b) \(\dfrac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
d) \(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right]:\dfrac{x-y}{x}\)
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) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
Đúng 0
Bình luận (1)
Thực hiện phép tính:
a) \(3x.\left(2x^2-3x+4\right)\)
b) \(\left(x+3\right)^2+\left(3x-2\right)\left(x+4\right)\)
c) \(\dfrac{2x-4}{x-1}+\dfrac{2x+2}{x^2-1}\)
`a)3x(2x^2-3x+4)`
`=6x^3-9x^2+12x`
______________________________________________
`b)(x+3)^2+(3x-2)(x+4)`
`=x^2+6x+9+3x^2+12x-2x-8`
`=4x^2+16x+1`
______________________________________________
`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]` `ĐK: x \ne +-1`
`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`
`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`
`=[2x^2-2]/[x^2-1]`
`=2`
Đúng 0
Bình luận (2)
14 tháng 3 2022 lúc 22:02
Cho các số thực dương a,b,c có abc=1 chứng minh rằng:
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}\ge\dfrac{1}{4}\)