Ta có: \(M=\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}\)

Thế: abc = 2010 ta được:

\(M=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)

\(\Leftrightarrow\frac{a^2bc}{ab\left(1+ac+c\right)}+\frac{ab}{ab\left(c+1+ac\right)}+\frac{abc}{ab\left(ac+c+1\right)}\)

\(\Leftrightarrow\frac{a^2bc+ab+abc}{ab\left(1+ac+c\right)}=\frac{ab\left(ac+1+c\right)}{ab\left(1+ac+c\right)}=1\)

Vậy \(M=1\)

nhóm như sau:

                   (1+3+5+....+2009)                ---             ( 2+4+6+.....+2010)

=   {{ ((2009 -1)/2 +1) x (2009 +1) } / 2 }}    ---    {{ (( 2010 - 2) /2+1) x (2010+2))  / 2 }}

=                          1010025                     ---                        1011030

=                                                       -1005

giá trị rút gọn là sao pạn?mình ko pit mình chỉ pit kết quả:

C=1-2+3-4+....+2007-2008+2009-2010

=(1-2)+(3-4)+...+(2007-2008)+(2009-2010)

=-1+-1+....+-1+-1

=-1.(-2010-1+1):2

=1.(-2010):2

=1.(-1005)

=-1005

e moi co lop 6 nen k giai duoc

phungtuantu  thek thì bl lm j hả bạn 

Ta có: \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}.\)

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

\(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=\frac{a-b}{2009-2010}=\frac{b-c}{2010-2011}=\frac{c-a}{2011-2009}.\)

\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)

\(\Rightarrow\frac{a-b}{-1}.\frac{b-c}{-1}=\left(\frac{c-a}{2}\right)^2\)

\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{2^2}\)

\(\Rightarrow\frac{\left(a-b\right).\left(b-c\right)}{1}=\frac{\left(c-a\right)^2}{4}.\)

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2.1\)

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)=\left(c-a\right)^2\)

\(\Rightarrow4.\left(a-b\right).\left(b-c\right)-\left(c-a\right)^2=0.\)

Hay \(M=0.\)

Vậy \(M=0.\)

Chúc bạn học tốt!

Đặt \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}=k\)

\(=>\hept{\begin{cases}a=2009k\\b=2010k\\c=2011k\end{cases}}\)

Ta có : \(4\left(a-b\right)\left(b-c\right)=4\left(2009k-2010k\right)\left(2010k-2011k\right)\)

\(=4\left(-k\right)\left(-k\right)=4k^2\)

Lại có : \(\left(c-a\right)^2=\left(2011k-2009k\right)^2=\left(2k\right)^2=4k^2\)

Suy ra \(M=4k^2-4k^2=0\)

a,

A = 2010²⁰¹⁰.[7¹⁰:7⁸-3.16-2²⁰¹⁰:2²⁰¹⁰] 

= 2010²⁰¹⁰.[7²-48-1]

= 2010²⁰¹⁰.0 = 0

b,

B = 1

\(A=2010^{2010}.\left[7^{10}:7^8-3.16-2^{2010}:2^{2010}\right]\)

\(A=2010^{2010}.\left[7^2-48-1\right]\)

\(A=2010^{2010}.0\)

\(Vay\)\(A=0\)

A= 2010²⁰¹⁰(7² - 48 - 1)

A=2010²⁰¹⁰(49-48-1)

A=2010²⁰¹⁰.0

A=0