M=\(\frac{2015\cdot a}{ab+2015\cdot a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\) biet abc=2015.Tinh M
cho abc=2015
tính M=\(\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)
\(M=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\) biết \(abc=2015\). Tính M.
Ta có:
\(M=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{abca}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{abca}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)
\(\Rightarrow M=\frac{ac+c+1}{ac+c+1}=1\)
Vậy M = 1
Thay 2015= abc vào M ta được:
M = \(\frac{abca}{ab+abca+abc}\) + \(\frac{b}{bc+b+abc}\) + \(\frac{c}{ac+c+1}\)
M = \(\frac{abca}{ab\left(1+ac+c\right)}\) + \(\frac{b}{b\left(c+1+ac\right)}\) + \(\frac{c}{ac+c+1}\)
M = \(\frac{ac}{1+ac+c}\) + \(\frac{1}{c+1+ac}\) + \(\frac{c}{ac+c+1}\)
M = \(\frac{1+ac+c}{1+ac+c}\) = 1
Vây M = 1
XONG !
Thay abc=2015 vào biểu thức M, ta có:
M=\(\frac{a^2bc}{ab+a^2bc+abc}\)+\(\frac{b}{bc+b+abc}\)+\(\frac{c}{ac+c+1}\)
=\(\frac{a^2bc}{ab\left(1+ac+c\right)}\)+\(\frac{b}{b\left(c+1+ac\right)}\)+\(\frac{c}{ac+c+1}\)
=\(\frac{ac}{ac+c+1}\)+\(\frac{1}{ac+c+1}\)+\(\frac{c}{ac+c+1}\)
=\(\frac{ac+c+1}{ac+c+1}\)
=1
Vậy M=1
CHÚC BẠN HỌC TỐT NHE
Cho abc= 2015
Tính M=\(\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)
\(M=\frac{abc.a}{ab+abc.a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+a}=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+a}=\frac{ac+c+1}{ac+c+1}=1\)
cho abc = 2015 , tính A=\(\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)
Ta có:
\(A=\frac{2015a}{ab+2015a+2015}+\frac{b}{bc+b+2015}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{abca}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{a^2bc}{ab.\left(1+ac+c\right)}+\frac{b}{b.\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}\)
\(\Rightarrow A=\frac{ac}{ac+1+c}+\frac{1}{ac+1+c}+\frac{c}{ac+1+c}\)
\(\Rightarrow A=\frac{ac+1+c}{ac+1+c}\)
\(\Rightarrow A=1.\)
Vậy \(A=1.\)
Chúc bạn học tốt!
Thay $abc=2015$ vào $A$ ta có:
\(\begin{array}{l} A = \dfrac{{{a^2}bc}}{{ab + {a^2}bc + abc}} + \dfrac{b}{{bc + b + abc}} + \dfrac{c}{{ac + c + 1}}\\ A = \dfrac{{{a^2}bc}}{{ab\left( {1 + ac + c} \right)}} + \dfrac{b}{{b\left( {c + 1 + ac} \right)}} + \dfrac{c}{{ac + c + 1}}\\ A = \dfrac{{ac}}{{ac + c + 1}} + \dfrac{1}{{ac + c + 1}} + \dfrac{c}{{ac + c + 1}}\\ A = \dfrac{{ac + c + 1}}{{ac + c + 1}} = 1 \end{array}\)
Cho a.b.c=2015.Tinh A= \(\frac{2015}{ab+a+2015}+\frac{2015}{bc+b+2015}+\frac{2015}{ca+c+2015}\)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR:
a,\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) b,\(\frac{2015\cdot a^2+2016\cdot b^2}{2015\cdot c^2+2016\cdot d^2}=\frac{ab}{cd}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
ÁP DỤNG TÍNH CHẤT DÃY TỈ SỐ BẰNG NHAU TA ĐƯỢC:
\(\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(2\right)\)
MẶT KHÁC: \(\frac{a}{c}.\frac{a}{c}=\frac{b}{d}.\frac{b}{d}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\left(1\right)\)
TỪ (1) VÀ (2) TA ĐƯỢC \(\frac{a^2+b^2}{c^2+d^2}==\frac{ab}{cd}\)
Cho abc =15
Tinh M = 2015.a / ab + 2015.a + 2015 + b / bc + b + 2015 + c / ac + c +1
https://olm.vn/hoi-dap/question/764972.html
cho 3 số a, b, c thuộc R t/m \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a^{^{ }}+b+c}\)
cmr: \(\frac{1}{a^{2015}}+\frac{1}{b^{2015}}+\frac{1}{c^{2015}}=\frac{1}{a^{2015}+b^{2015}+c^{2015}}\)
mong được mọi người giúp đỡ
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)
\(\Leftrightarrow\frac{a+b}{ab}=\frac{-a-b}{\left(a+b+c\right)c}\)
\(\Leftrightarrow\left(a+b\right)\left(a+b+c\right)c=-\left(a+b\right)ab\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Tự làm nốt
Cho 3 số dương a,b,c
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}\)
\(Tính:A=\frac{21ab^{2015}+12bc^{2015}+15ca^{2015}}{a^{2016}+b^{2016}+c^{2016}}\)