Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\) =\(\frac{a+b+c+d}{b+c+d+a+c+d+a+b+d+a+b+c}\)
Vì a+b+c+d khác 0
=> b+c+d=a+c+d=a+b+d=a+b+c
=>a=b=c=d
Khi đó:
a + b = c+d
b+c= (a+d)
c+d=a+b
d+a=b+c
=>\(\frac{a+b}{c+d}=\frac{b+c}{a+d}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{b+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{b+a+b}{c}=\frac{a+b+c}{d}\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{b+a+b}{c}+1=\frac{a+b+c}{d}+1\)
\(=\frac{b+c+d}{a}+\frac{a}{a}=\frac{c+d+a}{b}+\frac{b}{b}=\frac{b+a+b}{c}+\frac{c}{c}=\frac{a+b+c}{d}+\frac{d}{d}\)
\(=\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
\(\Rightarrow a=b=c=d\)
Do đó \(\frac{a+b}{c+d}+\frac{b+c}{c+d}+\frac{c+d}{a+b}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1=3\)
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)
+ Ta có
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{a+b}{\left(a+b\right)+2\left(c+d\right)}=\frac{1}{3}\)
\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)
\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\Rightarrow a+b=c+d\)
Tương tự ta cũng c/m được
\(b+c=a+d\)
\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)
Ta có : \(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{d+a+b}=\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{c+d+a}+1=\frac{c}{d+a+b}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{c+d+a}=\frac{a+b+c+d}{d+a+b}=\frac{a+b+c+d}{a+b+c}\)
Nếu a + b + c + d = 0
=> a + b = - c - d
b + c = - a - d
c + d = - b - a
d + a = - b - c
Khi đó \(P=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{d+a}+\frac{-\left(b+a\right)}{b+a}=\frac{-\left(b+c\right)}{b+c}\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Nếu a + b + c + d \(\ne\)0
\(\Rightarrow\frac{1}{c+d}=\frac{1}{d+a}=\frac{1}{b+a}=\frac{1}{b+c}\)
\(\Rightarrow c+d=d+a=b+a=b+c\)
\(\Rightarrow a=b=c=d\)
Khi đó \(P=1+1+1+1=4\)
Vậy nếu a + b + c + d = 0 thì P = - 4
nếu a + b + c + d \(\ne\)0 thì P = 4
\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\)\(\frac{d}{a+b+c}\)
\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Mà: \(a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow A=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}+\frac{d+d}{d+d}\)
\(\Rightarrow A=1+1+1+1=4\)
số đo slaf
4
nhe sbn
bài dài
lắm mình
vhir tiện ghi
thế này thôi
\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
Vì \(a+b+c+d\ne0\) nên \(b+c+d=a+c+d=a+b+d=a+b+c\)
\(\Rightarrow a=b=c=d\)
\(\Rightarrow A=1+1+1+1=4\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)
suy ra: x/5 = 45 => x = 225
y/7 = 45 => y = 315
z/9 = 45 => z = 405
Đề đúng
\(\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+c+b}+1\)
\(\Leftrightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+c+b}\)
=> b+c+d = a+c+d = a+b+d = a+c+b
=> a=b=c=d
=> GTBT = 1+1+1+1 =4