ĐK: a,b,c \(\ne\) 0

Theo tính chất dãy tỉ số bằng nhau ta có:

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

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

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

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

\(\Rightarrow\) \(\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Rightarrow\) \(\dfrac{b+c}{bc}=0\) \(\Rightarrow\) b + c = 0 (vì bc \(\ne\) 0 do a,b,c \(\ne\) 0)

\(\Rightarrow\) b = -c \(\Rightarrow\) b⁵ = (-c)⁵ \(\Rightarrow\) b⁵ + c⁵ = 0

Thay b⁵ + c⁵ = 0 vào M ta được:

M = (a¹⁹ + b¹⁹).(b⁵ + c⁵).(c²⁰⁰¹ + a²⁰⁰¹)

M = (a¹⁹ + b¹⁹).0.(c²⁰⁰¹ + a²⁰⁰¹)

M = 0 (đpcm)

Chúc bn học tốt!

a=\(\frac{1}{9}\)

c=\(\frac{3069}{500}\)

= 98/45-31/15

=1/9

\(=\frac{2001.\left(2000+2\right)+1000}{2001.\left(2000+3\right)-1001}\)

\(=\frac{2001.2000+2001.2+1000}{2001.2000+2001.3-1001}\)

\(=\frac{2001.2+1000}{2001.3-1001}\)

\(=\frac{2001.2+1000}{2001.2+2001-1001}\)

\(=\frac{2001.2+1000}{2001.2+1000}\)

\(=1\)

  2001 x 2002 + 1000 / 2001 x 2003 - 1001                                                                                                                                       = 2001 x 2002 + 1000 / 2001 x (2002 + 1) - 1001                                                                                                                             = 2001 x 2002 + 1000 / 2001 x 2002 + 2001 - 1001                                                                                                                           = 1000 / 2001 - 1001 = 1000 / 1000 = 1    

\(=\frac{2001\cdot\left(2003-1\right)+1000}{2001\cdot2003-1001}\)

\(=\frac{2001\cdot2002-2001-1000}{2001\cdot2003-1001}\)

\(=\frac{2001\cdot2002-1001}{2001\cdot2002-1001}\Rightarrow=1\)

phần a nhé

1/a+1/b+1/c=(a+b+c)(1/a+1/b+1/c)=3+(a/b+b/a)+(b/c+c/b)+(a/c+c/a)            do a+b+c=1

áp dụng bdt cosi cho các  so dương a/b,b/a,a/c,c/a,b/c,c/b

a/b+b/a >=2

b/c+c/b>=2

a/c+c/a>=2

cộng hết vào suy ra 1/a+1/b+1/c >=9       

C = 1 + (-3) + 5 + (-7) +...+ 2001 + (-2003)

C= (1 - 2003) + (2001 - 3) + (5 - 1999) + (1997 - 7) +...+ (1001 - 1003)

C= -2002 + 1998 - 1994 + 1990 +....-2

C= (-4) + (-4) +....+ (-4) - 2 (250 cặp (-4) )

C= 250 x (-4) - 2

C= -1000 - 2 = -1002

D = (-1001) + (-1000) + (-999) +...+ 1001 + 1002

D= (1001 - 1001) + (1000 - 1000) +...+ (1-1) + 0 + 1002

D= 0 + 0 +... + 0 + 0 + 1002

D= 1002

Ta có :

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Mà \(\dfrac{a^3}{b^3}=\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)

\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)

. là x á nha

=\(\frac{2006}{2008}.\frac{2001}{2004}.\frac{2008}{2002}.\frac{2004}{2006}.\frac{1001}{2001}\)

=\(\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)

=\(\frac{1001}{2002}\)

= \(\frac{2006\cdot2001\cdot2008\cdot2004\cdot1001}{2008\cdot2004\cdot2002\cdot2006\cdot2001}\)

= \(\frac{1\cdot1\cdot1\cdot1\cdot1001}{1\cdot1\cdot2002\cdot1\cdot1}\)

= \(\frac{1}{2}\)

\(\frac{2006}{2008}.\frac{2001}{2004}.\frac{2008}{2002}.\frac{2004}{2006}\frac{1001}{2001}\)

\(=\frac{2006.2001.2008.2004.1001}{2008.2004.2002.2006.2001}\)

\(=\frac{1001}{2002}\)

\(=\frac{1}{2}\)

\(A=2001+2001^2+...+2001^9\)

\(\Rightarrow2001A=2001^2+2001^3+...+2001^{10}\)

\(\Rightarrow2001A-A=\left(2001^2+2001^3+...+2001^{10}\right)-\left(2001+2001^2+...+2001^9\right)\)\(\Rightarrow2000A=2001^{10}-1\)

\(\Rightarrow A=\frac{2001^{10}-1}{2000}\)

\(\Rightarrow K=2000.\frac{2001^{10}-1}{2000}+1=2001^{10}-1+1=2001^{10}\)

Vậy K=2001¹⁰