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21 tháng 7 2018

\(a,\dfrac{1}{x^2-x}+\dfrac{2x}{4x^3}-\dfrac{1}{x^2+x+1}\)

\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{2x^2}-\dfrac{1}{x^2+x+1}\)

\(=\dfrac{2x\left(x^2+x+1\right)+\left(x-1\right).\left(x^2+x+1\right)-2x^2.\left(x-1\right)}{2x^2.\left(x-1\right).\left(x^2+x+1\right)}\)

\(=\dfrac{2x^3+2x^2+2x+x^3-1-2x^3+2x^2}{2x^2.\left(x^3-1\right)}\)

\(=\dfrac{4x^2+2x+x^3-1}{2x^5-2x^2}\)

\(=\dfrac{x^3+4x^2+2x-1}{2x^5-2x^2}\)

21 tháng 7 2018

\(b,\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x+1\right).\left(x^2-x+1\right)}\)

\(=\dfrac{1}{x^2-x+1}+1-\dfrac{x^2+2}{\left(x^2-x+1\right)}\)

\(=\dfrac{x+1\left(x+1\right).\left(x^2-x+1\right)-\left(x^2+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x+1+x^3+1-x^2-2}{\left(x+1\right).\left(x^2-x+1\right)}\)

\(=\dfrac{x+0+x^3-x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x\left(1+x^2-x\right)}{\left(x+1\right).\left(x^2-x+1\right)}\)

\(=\dfrac{x}{x+1}\)

21 tháng 11 2017

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

22 tháng 11 2017

Cảm ơn, mình làm được rồi :>

28 tháng 11 2022

a: \(=\dfrac{1}{\left(x-y\right)\left(y-z\right)}-\dfrac{1}{\left(y-z\right)\left(x-z\right)}-\dfrac{1}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=0\)

b: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(x-y\right)\left(y-z\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(y^2-x^2\right)-z^2\left(y-x\right)-xy\left(y-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{\left(x-y\right)\left[-z\left(x+y\right)+z^2+xy\right]}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{-zx-zy+z^2+xy}{xyz\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{z\left(z-x\right)-y\left(z-x\right)}{xyz\left(y-z\right)\left(x-z\right)}=\dfrac{1}{xyz}\)

26 tháng 11 2022

a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)

\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

=0

c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)

\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(=\dfrac{1}{xyz}\)

 

26 tháng 12 2021

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

28 tháng 6 2017

Phép cộng các phân thức đại số

Phép cộng các phân thức đại số

8 tháng 11 2017

a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )

\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )

\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)

b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)

\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)

\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )

c) MTC = ( x+ 2)2(x - 2)2

Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)

\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)

\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)

8 tháng 11 2017

d) MTC = xyz( x - y)( y - z)( x - z)

Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

Cộng các phân thức lại ta có :

\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)

AH
Akai Haruma
Giáo viên
13 tháng 3 2021

Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

13 tháng 3 2021

Thầy ơi, nhưng câu này là đề thi huyện chỗ em á thầy, em cũng chả biết làm sao nữa, chả nhẽ đề thi huyện lại sai:"(

13 tháng 3 2017

Quy đồng tính bình thường.

\(A=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+2\left(\dfrac{1}{x-y}+\dfrac{1}{y-z}+\dfrac{1}{z-x}\right)\)\(=\dfrac{2x^2+2y^2+2z^2-2xy-2yz-2xz}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\dfrac{2yz+2xz+2xy-2x^2-2y^2-2z^2}{ }\)

=0