cho a+b+c=2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{4}\)
Tính \(A=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Bài 1:
a) Cho a(y+z) = b(z+c) = c(x+y) Tính: \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-c}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
b) \(Cho\dfrac{a}{2014}=\dfrac{b}{2015}=\dfrac{c}{2016}cm:4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
c) \(\dfrac{a}{a'}+\dfrac{b'}{b}=1\) và \(\dfrac{b}{b'}+\dfrac{c'}{c}=1\)
cm: abc+a'b'c'=0
bài 4:
a) \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\) Tính: \(\dfrac{x}{y}\)
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) Tính P = \(\dfrac{xy+yz+xz}{x^2+y^2-z^2}\)
c) \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
Tính : P = \(\dfrac{a+b}{c+d}+\dfrac{c+b}{a+d}=\dfrac{c+d}{a+b}=\dfrac{a+d}{c+b}\)
d) \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\) Tính: \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
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 = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
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.
\(M=\dfrac{3^6.45^4-15^4.9^4}{27^4.25^3+7.45^6}\)
Cho a+b+c=2010 và \(\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{c+a}=\dfrac{1}{201}TínhS=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(M=\dfrac{3^{14}\cdot5^4-3^{12}\cdot5^4}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)
Cho \(a+b+c=2016\) và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{90}\). Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(a+b+c=2016\Rightarrow\left\{{}\begin{matrix}a=2016-\left(b+c\right)\\b=2016-\left(c+a\right)\\c=2016-\left(a+b\right)\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{2016-\left(b+c\right)}{b+c}+\dfrac{2016-\left(c+a\right)}{c+a}+\dfrac{2016-\left(a+b\right)}{a+b}\)\(\Rightarrow S=2016\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(\Rightarrow S=2016.\dfrac{1}{90}-3\)
\(\Rightarrow S=\dfrac{97}{2}\)
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 = \(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\)
Cho \(abc\ne0\) và \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}.\) Tính \(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
Giúp ik
Lời giải:
Bạn tham khảo cách làm tương tự tại đây:
https://hoc24.vn/cau-hoi/cho-dfracab-2017ccdfracbc-2017aadfracca-2017bbvoi-a-b-c-ne0-tinhp-left1dfracabrightleft1dfracb.161494910584
Kết quả $P=8$ hoặc $P=-1$