Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{a}{b}=\frac{-2,5}{4,5}\Rightarrow\frac{a}{-2,5}=\frac{b}{4,5}\)
áp dụng ...ta có
\(\frac{a}{-2,5}=\frac{b}{4,5}=\frac{a+b}{-2,5+4,5}=\frac{1,44}{2}=0,72\)
\(\frac{a}{-2,5}=0,72\Rightarrow a=0,72.\left(-2,5\right)=-1,8\)
\(\frac{b}{4,5}=0,72\Rightarrow b=0,72.4,5=3.24\)
a/b=-2,5/4,5=-5/9
=> a gồm -5 phần bằng nhau thì b gồm 9 phần bằng nhau
=> a+b gồm -5 + 9=4 phần bằng nhau => 1,44 gồm 4 phần bằng nhau
=> 1 phần bằng 1,44 : 4 = 0,36
=> a=0,36 x (-5)=-1,8
b=0,36 x 9 =-3,24
Lời giải:
$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$
$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$
$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$
$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$
$\Rightarrow a+b+c=0$ hoặc $a=b=c$
Nếu $a+b+c=0$ thì:
$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:
$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$
Lời giải:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow \frac{abc}{c(a+b)}=\frac{abc}{a(b+c)}=\frac{bca}{b(c+a)}\)
\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)
\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)
\(\Rightarrow a=b=c\) (do $a,b,c>0$)
$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow P=2+2+2=6\)
Áp dụng t/c dtsbn:
\(\dfrac{1}{a+b}=\dfrac{2}{b+c}=\dfrac{3}{c+a}=\dfrac{1+2+3}{2\left(a+b+c\right)}=\dfrac{6}{2\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)
\(\Rightarrow\left\{{}\begin{matrix}3a+3b=a+b+c\\3b+3c=2a+2b+2c\\3a+3c=3a+3b+3c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=2a\\b=0\end{matrix}\right.\)
\(Q=\dfrac{a+2021b+c}{a+2022b+c}=\dfrac{a+2a}{a+2a}=1\)
TH1: \(a+b+c+d\ne0\)
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)
\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)
\(\Rightarrow a=b=c=d\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=1+1+1+1\)
\(\Rightarrow P=4\)
TH2: \(a+b+c+d=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)
\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)
\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(\Rightarrow P=-4\)
bn mình nền của bn là nôb team trưởng team là t gaming
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{-2,5}{4,5}=\dfrac{-5}{9}\Rightarrow\dfrac{a}{-5}=\dfrac{b}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{-5}=\dfrac{b}{9}=\dfrac{a+b}{-5+9}=\dfrac{1,44}{4}=0,36\)
\(\Rightarrow\left\{{}\begin{matrix}a=-1,8\\b=3,24\end{matrix}\right.\)
Vậy...
Ta có: \(\dfrac{a}{b}=\dfrac{-2,5}{4,5}\Rightarrow\dfrac{a}{-2,5}=\dfrac{b}{4,5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{-2,5}=\dfrac{b}{4,5}=\dfrac{a+b}{-2,5+4,5}=\dfrac{1,44}{2}=0,72\)
( do \(a+b=1,44\))
\(\Rightarrow\left\{{}\begin{matrix}a=0,72.\left(-2,5\right)=-1,8\\b=0,72.\left(4,5\right)=3,24\end{matrix}\right.\)
Vậy.........
Chúc bạn học tốt!!!