Xuất phát từ giả thiết , ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

=> \(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)

=> \(\left(a+b+c\right)\left(ab+bc+ac\right)=abc\)

=> \(\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)

=> \(a\left(ab+bc+ac\right)+b\left(ab+bc+ac\right)+c\left(ab+bc+ac\right)-abc=0\)=> a²b + abc + a²c + ab² + b²c + abc + abc + bc² + ac² - abc = 0

=> ab(a + b) + ac( a + c) + bc( b + c) + 2abc = 0

=> ab( a + b + c) + ac( a + b + c ) + bc( b + c) = 0

=> ( a + b + c)a( b + c) + bc( b + c) = 0

=> ( b + c)( a² + ab + ac + bc) = 0

=> ( b + c)( a + b)( c + a) = 0

Suy ra :

* b = -c

*a = -b

* c = -a

TH1 :Với b = -c

\(VT=\dfrac{1}{a^{1995}}+\dfrac{1}{\left(-c\right)^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{a^{1995}}\)

\(VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{a^{1995}+\left(-c\right)^{1995}+c^{1995}}=\dfrac{1}{a^{1995}}=VT\)

TH2 : với a = -b

\(VT=\dfrac{1}{\left(-b\right)^{1995}}+\dfrac{1}{b^{1995}}+\dfrac{1}{c^{1995}}=\dfrac{1}{c^{1995}}\)

\(VP=\dfrac{1}{a^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{\left(-b\right)^{1995}+b^{1995}+c^{1995}}=\dfrac{1}{c^{1995}}=VT\)

TH3 . c = -a , Tương tự

Vậy , đẳng thức được Chứng minh

giúp mik đi nhé 

mik đang cần gấp nèeee

a,-7/12

b,5/14

c,-3/8

d,-1

,

,,,,,,,,,ư

Cô làm rồi em nhé:

https://olm.vn/cau-hoi/giup-em-voiii.8161766187032

a: =3990-(463*72-39^2):15

=3990-2121=1869

b: =1995-963-172=860

c: =33/11-7/11=26/11

d: =3/2+5=13/2

`@` `\text {Answer}`

`\downarrow`

`a,`

`3990 - (463 \times 72 - 39 \times 39) \div 15`

`= 3990 - (33336 - 1521) \div 15`

`= 3990 - 31815 \div 15`

`= 3990 - 2121`

`= 1869`

`b,`

`1995 - 321 \times 3 - 6020 \div 35`

`= 1995 - 963 - 172`

`= 1032 - 172`

`= 860`

`c,`

`3 - 7/11`

`= 33/11 - 7/11`

`= 26/11`

`d,`

`1/4 \div 1/6 + 5`

`= 1/4 \times 6 + 5`

`= 3/2 + 5`

`= 13/2`

a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

b) \(\dfrac{6\cdot9-2\cdot17}{63\cdot3-119}\)

\(=\dfrac{2\left(3\cdot9-17\right)}{7\cdot\left(3\cdot9-17\right)}\)

\(=\dfrac{2}{7}\)

=>(x+1/1998+1)+(x+2/1997+1)=(x+3/1996+1)+(x+4/1995+1)

=>x+1999=0

=>x=-1999