Cho \(a+b+c=2016\) và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{90}\). Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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Cho a+b+c+d=2000 và \(\dfrac{1}{a+b+c}+\dfrac{1}{b+c+d}+\dfrac{1}{c+d+a}+\dfrac{1}{d+a+b}=\dfrac{1}{40}\)
Tính S=\(\dfrac{a}{b+c+d}+\dfrac{b}{c+d+a}+\dfrac{c}{d+a+b}+\dfrac{d}{a+b+c}\)
a) Cho các số a, b, c thỏa mãn abc\(\ne\) 0 và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) =\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}\)=\(\dfrac{1}{3}\). Tính S= a + b + c + 2021.
Cho a+b+c=2007 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{90}\)
Tính S= \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Nhân cả hai vế của đẳng thức cho a+b+c ta được
\(\dfrac{a+b+c}{a+b}\)+\(\dfrac{a+b+c}{a+b}\)=\(\dfrac{a+b+c}{c+a}\)=\(\dfrac{a+b+c}{90}\)
=> a+ \(\dfrac{c}{a+b}\)+1+\(\dfrac{a}{b+c}\)+1+\(\dfrac{b}{c+a}\)=\(\dfrac{2007}{90}\)
=>\(\dfrac{a}{b+c}\)+\(\dfrac{b}{c+a}\)+\(\dfrac{c}{a+b}\)=\(\dfrac{2007}{90}\)-3= 22,3-3=19,3
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Cho a + b + c = 2007 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{90}\)
Tính S = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{90}\Leftrightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{a+b+c}{a+b}\)\(\Leftrightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1=\dfrac{2007}{90}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{193}{10}\)
\(\Rightarrow S=\dfrac{193}{10}\)
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Ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)
\(=2007.\frac{1}{90}-3\)
\(=22,3-3\)
\(=19,3\)
Vậy S = 19,3
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a) Cho \(a+b+c+d=2000\) và \(\dfrac{1}{a+b+c}+\dfrac{1}{b+c+d}+\dfrac{1}{c+d+a}+\dfrac{1}{d+a+b}=\dfrac{1}{40}\)
Tính giá trị của: \(S=\dfrac{a}{b+c+d}+\dfrac{b}{c+d+a}+\dfrac{c}{d+a+b}+\dfrac{d}{a+b+c}\)
b) Xác định tổng các hệ số của đa thức \(f\left(x\right)=\left(5-6x+x^2\right)^{2016}\cdot\left(5-6x+x^2\right)^{2017}\)
Cho a,b,c thỏa mãn a+b+c = 2021 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2021}\)
Tính Q = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
a+b+c=2018 và \(\dfrac{1}{a+b}\)+\(\dfrac{1}{b+c}\)+\(\dfrac{1}{a+c}\)=\(\dfrac{\text{1}}{\text{2018}}\)
S=\(\dfrac{a}{b+c}\)+\(\dfrac{b}{a+c}\)+\(\dfrac{c}{a+b}\)
Lời giải:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
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cho a+b+c=2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{4}\)
Tính \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
theo bài ra ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{`1}{4}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{a+b+c}{4}\)
\(\Rightarrow1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1=\dfrac{2016}{4}\)
\(\Rightarrow\left(1+1+1\right)+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)
\(\Rightarrow3+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=504\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=504-3\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)
vậy \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=501\)
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(a+b+c)(1/a+b+1/b+c+1/c+a)=(a+b+c)/4
(a+b+c)/(a+b)+(a+b+c)/(b+c)+(a+b+c)/(c+a)=(a+b+c)/4
=> 1+c/(a+b)+1+a/(b+c)+1+b/(c+a)=2016/4
<=>c/(a+b)+a/(b+c)+b/(c+a)+3=504
=> A=a/(b+c)+b/(c+a)+c/(a+b)=504-3=501
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tớ cần hỏi bài tập toán như sau:
cho a+b+c = 2023 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{2023}\)
tính giá trị biểu thức: Q = \(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\)