Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2\)
\(=\left(\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2=\left(\dfrac{b}{d}\right)^2\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}\)
\(=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có thêm tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) nx nha
![](https://rs.olm.vn/images/avt/0.png?1311)
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\left|x\right|=-x\)
nên x<=0
b: \(\left|x\right|>x\)
=>x<0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a}{b}=\dfrac{c}{d}\)
Lời giải:
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$
$\Rightarrow (a+b)(c-d)=(a-b)(c+d)$
$\Rightarrow ac-ad+bc-bd=ac+ad-bc-bd$
$\Rightarrow 2ad=2bc$
$\Rightarrow ad=bc$
$\Rightarrow \frac{a}{b}=\frac{c}{d}$ (đpcm)
Đặt \(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
Vì \(a,b,c\in Z\)*
\(\Rightarrow A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}\)
\(\Rightarrow A>\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow A>1\)
Vì \(a,b,c\in Z\)*
\(\Rightarrow A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{b+c+a}+\dfrac{c+b}{c+a+b}\)
\(\Rightarrow A< \dfrac{a+c+b+a+c+b}{a+b+c}=2\)
\(\Rightarrow A< 2\)
Vậy \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
Đặt \(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
Vì \(a,b,c\in Z\)*
\(\Rightarrow\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\)
Nên: \(\dfrac{b}{b+c}>\dfrac{b}{b+c+a};\dfrac{c}{c+a}>\dfrac{c}{c+a+b}\)
Vậy\(A=\)\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow A>1\)(1)
Lại có :\(\dfrac{a}{a+b}< \dfrac{a+b}{a+b+c}\)
Nên:\(\dfrac{b}{b+c}< \dfrac{b+c}{b+c+a};\dfrac{c}{c+a}< \dfrac{c+a}{c+a+b}\)
Vậy\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b+b+c+c+a}{a+b+c}\)
=\(\dfrac{2.\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow A< 2\)(2)
Từ (1) và (2) \(\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
\(\Rightarrow A\) không phải là số nguyên