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\(M=\frac{2010}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+2010}\)
Hay: \(M=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\)
\(M=\frac{a\left(bc\right)}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a.1}{a\left(b+1+bc\right)}\)
\(M=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(M=\frac{bc+b+1}{bc+b+1}=1\)
Vậy M=1
\(2A=2+2^2+2^3+2^4+..+2^{60}\)
\(A=2A-A=2^{60}-1\)
\(B=\left(2^2\right)^{25}x2^{10}=2^{50}x2^{10}=2^{60}\)
=> A<B
a)x + 3/5 = 1/5 <=> x = 1/5 - 3/5 <=> x = -2/5
b)-1/2 - x = 1/3 - 1/-4 <=> -x = 1/3 + 1/4 + 1/2 <=> -x = 8/24 + 6/24 + 12/24 <=> -x = 26/24 <=> x = -26/24
c)x/14 = 1/7 + -3/14 <=> x/14 = 2/14 + -3/14<=> x/14 = -1/14 <=> x = -1
có j thiếu sót các bạn sửa và bình luận cho mình nha
a) \(x+\dfrac{3}{5}=\dfrac{1}{5}\Rightarrow x=\dfrac{1}{5}-\dfrac{3}{5}=\dfrac{-2}{5}\)
b) \(\dfrac{-1}{2}-x=\dfrac{1}{3}-\dfrac{1}{-4}\rightarrow\dfrac{-1}{2}-x=\dfrac{7}{12}\)\(\Rightarrow x=\dfrac{-1}{2}-\dfrac{7}{12}=\dfrac{-13}{12}\)
c) \(\dfrac{x}{14}=\dfrac{1}{7}+\dfrac{-3}{14}\rightarrow\dfrac{x}{14}=\dfrac{-1}{14}\Rightarrow x=-1\)
\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)
\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\left(abc=105\right)\)
\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)
\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}\)
\(=1\)
\(S=\frac{105}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+105}\)
\(=\frac{abc}{abc+ab+a}+\frac{b}{bc+b+1}+\frac{a}{ab+a+abc}\) \(\left(abc=105\right)\)
\(=\frac{abc}{a\left(bc+b+1\right)}+\frac{b}{bc+b+1}+\frac{a}{a\left(bc+b+1\right)}\)
\(=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}\)
\(=1\)
Ta có: M= abc/ ab+bc+ca
<=> 1/M = ab+ bc+ ca/ abc= 1/a+ 1/b+ 1/c (1)
Do: ab/ a+2b= 2/5 nên a+2b/ ab= 5/2
<=> 1/b+ 2/a= 5/2 (2)
Tương tự: bc/ b+2c= 3/4 nên b+2c/ bc= 4/3
<=> 1/c+2/b=4/3 (3)
ac/c+2a=3/5 <=> c+2a/ac=5/3
<=> 1/a+2/c=5/3 (4)
Cộng tổng của (2), (3), (4) ta đc:
( 1/b+2/a) + (1/c+2/b)+(1/a+2/c)= 5/2+4/3+5/3
<=> 3/a+3/b+3/c=5/2+3
<=> 3 x (1/a+1/b+1/c)=11/2 (5)
Thay (1) vào (5), ta có: 3 x 1/M = 11/2
<=> 1/M=11/6 <=>M=6/11
Vậy giá trị biểu thức M=6/11
Bài 1.
a) \(\dfrac{3}{14}.\dfrac{7}{20}+\dfrac{13}{20}=\dfrac{3}{40}+\dfrac{13}{20}=\dfrac{3}{40}+\dfrac{26}{40}=\dfrac{29}{40}\).
b) \(\left(2.3^{2010}+12.3^{2010}-3.3^{2010}\right):3^{2012}\)
\(=3^{2010}\left(2+12-3\right):3^{2012}\)
\(=3^{2010}.11:3^{2012}\)
\(=\left(3^{2010}:3^{2012}\right).11\)
\(=\dfrac{1}{9}.11\)
\(=\dfrac{11}{9}\).
Bài 2.
a) \(\left(5^{14}.25^{10}\right):125^3\)
\(=\left[5^{14}.\left(5^2\right)^{10}\right]:\left(5^3\right)^3\)
\(=\left[5^{14}.5^{20}\right]:5^9\)
\(=5^{34}:5^9\)
\(=5^{25}\).
b) \(\left(\dfrac{1}{2}\right)^5.\left(\dfrac{1}{64}\right)^9:\left(\dfrac{1}{16}\right)^5\)
\(=\dfrac{1}{2^5}.\dfrac{1}{64^9}:\dfrac{1}{16^5}\)
\(=\dfrac{1}{2^5}.\dfrac{1}{\left(2^6\right)^9}:\dfrac{1}{\left(2^4\right)^5}\)
\(=\dfrac{1}{2^5}.\dfrac{1}{2^{54}}:\dfrac{1}{2^{20}}\)
\(=\dfrac{1}{2^{59}}:\dfrac{1}{2^{20}}\)
\(=\dfrac{2^{20}}{2^{59}}\)
\(=\dfrac{1}{2^{39}}\).
1/140
2/16
3/A=B