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BÀI 1:
\(\dfrac{a}{k}=\dfrac{x}{a}\Rightarrow a^2=kx\)
\(\dfrac{b}{k}=\dfrac{y}{b}\Rightarrow b^2\)=ky
Vay \(\dfrac{a^2}{b^2}=\dfrac{kx}{ky}=\dfrac{x}{y}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=3(a+b+c+d)/a+b+c+d=3
suy ra k=3
taco:\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}+\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)=>\(\dfrac{b+c+d}{a}+1=\dfrac{c+d+a}{b}+1=\dfrac{a+b+d}{c}+1=\dfrac{a+b+c}{d}+1=k+1\)=>\(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}=k+1=\dfrac{a+b+c+d+a+b+c+d+a+b+c+d}{a+b+c+d}=\dfrac{4.\left(a+b+c+d\right)}{a+b+c+d}=4\)
=>k+1=4
=>k=3
Theo tính chất dãy tỉ số bằng nhau ,ta có :
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)
\(=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
=> k = 3
sửa: \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)
giải:
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}\\ =\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\\ =\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3=k\)
vậy k=3
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\left(1\right)\\ \dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}\left(2\right)\\ \dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}\left(3\right)\\ \dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}\left(4\right)\)
Từ (1) (2) (3) (4) => \(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>\dfrac{a+b+c+d}{a+b+c+d}\\ \Rightarrow\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>1\left(4\right)\)
Mặt khác
\(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}=\left(\dfrac{a}{a+b+c}+\dfrac{c}{c+d+a}\right)+\left(\dfrac{b}{b+c+d}+\dfrac{d}{d+a+b}\right)\)
mà \(\dfrac{a}{a+b+c}+\dfrac{c}{c+d+a}< \dfrac{a}{a+c}+\dfrac{c}{c+a}\) ; \(\dfrac{b}{b+c+d}+\dfrac{d}{d+a+b}< \dfrac{b}{b+d}+\dfrac{d}{b+d}\)
=>\(\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \left(\dfrac{a}{a+c}+\dfrac{c}{a+c}\right)+\left(\dfrac{b}{b+d}+\dfrac{b}{b+d}\right)=2\)(5)
Từ (4) (5) => \(1< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\)
Vậy B không phải là số nguyên
Ta có:
+) \(\dfrac{a}{k}=\dfrac{b}{k}\Rightarrow a=b\)
+) \(\dfrac{x}{a}=\dfrac{y}{b}\)mà a=b \(\Rightarrow x=y\)
Ta lại có:
+)a=b \(\Rightarrow\) \(\dfrac{a^2}{b^2}=\left(\dfrac{a}{b}\right)^2=1^2=1\)(1)
+)x=y \(\Rightarrow\dfrac{x}{y}=1\)(2)
* Từ (1) và (2) \(\Rightarrow\)\(\dfrac{a^2}{b^2}=\dfrac{x}{y}\)
Vậy \(\dfrac{a^2}{b^2}=\dfrac{x}{y}\)
CHÚC BẠN HỌC TỐT!
theo bài ra ta có:
\(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=k\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1=\frac{a+b+c}{d}+1=k+1\) \(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=k+1\)
vì a + b + c + d khác 0 => a = b = c = d
ta có:
\(\Rightarrow\frac{4a}{a}=\frac{4b}{b}=\frac{4c}{c}=\frac{4d}{d}=k+1\)
=> 4 = 4 = 4 = 4 = k + 1
=> k + 1 = 4
=> k = 3
vật k = 3
theo đầu bài
=>\(\dfrac{b+c+d}{a}\)=\(\dfrac{c+d+a}{b}\)=\(\dfrac{d+a+b}{c}\)=\(\dfrac{a+b+c}{d}\)=\(\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)=\(\dfrac{3\left[a+b+c+d\right]}{a+b+c+d}\)=>=3
=>k=3
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}=\dfrac{\left(a+a+a\right)+\left(b+b+b\right)+\left(c+c+c\right)+\left(d+d+d\right)}{a+b+c+d}=\dfrac{3a+3b+3c+3d}{a+b+c+d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
Vậy \(k=3\)
áp dụng tính chất dãy tỉ số băng nhau ta có
\(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{a}\)=\(\dfrac{a+b+c}{b+c+a}\)=1
=>k=1