Cho a + b + c = 2001 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\)
Tính: \(s=\dfrac{a}{b+a}+\dfrac{b}{c+a}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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.
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:
\((a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})=\frac{a}{a+b}+\frac{a}{b+c}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{c}{b+c}+\frac{c}{a+c}\)
$\Leftrightarrow 2018.\frac{1}{2018}=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$
$\Leftrightarrow 1=1+1+1+S$
$S=1-1-1-1=-2$
Cho a + b + c = 2018 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\). Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Cho a + b + c = 2017
Và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
Tính S = \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}=\dfrac{1}{10}\)
\(\Rightarrow2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)=\dfrac{2017}{10}\)
\(\Rightarrow\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}=201,7\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{a+c}{a+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow3+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=201,7\)
\(\Rightarrow\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}=198,7\)
Ta có: \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{10}\)
\(=>2017\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2017}{10}\)
\(=>\dfrac{2017}{a+b}+\dfrac{2017}{b+c}+\dfrac{2017}{c+a}=201,7\)
Mà 2017 = a+b+c nên ta có:
\(=>\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=201,7\)
\(=>1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}=201,7\)
\(=>\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=201,7-3=198,7\)
CHÚC BẠN HỌC TỐT....
Cho các số a,b,c thoả mãn a+b+c = 126 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=16\)
Tính giá trị biểu thức \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Xét \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=126.16=2016\)
\(\Leftrightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=2016\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=2013\)
Vậy A = 2013
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 = \(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\)