Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Xét đề bài , ta thấy :
\(\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\)
\(\frac{b}{b+c+d}>\frac{b}{a+b+c+d}\)
\(\frac{c}{c+d+a}>\frac{c}{a+b+c+d}\)
\(\frac{d}{d+a+b}>\frac{d}{a+b+c+d}\)
Vậy , \(\frac{a}{b+c+d}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>1\)
mặt khác , ta lại có :
\(\frac{a}{b+c+d}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)
\(=\left(\frac{a}{d+b+c}+\frac{c}{c+d+a}\right)+\left(\frac{b}{b+c+d}+\frac{d}{d+a+b}\right)\)
Mà \(\frac{a}{b+c+d}+\frac{c}{c+d+a}< \frac{a}{a+c}+\frac{c}{c+a}=1\)
\(\frac{b}{b+c+d}+\frac{d}{d+a+c}< \frac{b}{b+d}+\frac{d}{d+b}=1\)
=> \(\frac{a}{b+c+d}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2\)
Vậy . . .
a) Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)
\(\Rightarrow ad< bc\) ( đpcm. )
b) Vì \(b>0;d>0\) \(\Rightarrow b+d>0\)
Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Leftrightarrow ad< bc\) (*)
Thêm \(ab\) vào \(2\) vế (*), ta có:
\(ab+ad< ba+bc\)
\(a.\left(b+d\right)< b.\left(a+c\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)
Thêm \(cd\) vào \(2\) vế (*), ta được:
\(ad+cd< cb+cd\)
\(\left(a+c\right).d< c.\left(b+d\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\) ( đpcm )
a)ta có \(\dfrac{a}{b}\)<\(\dfrac{c}{d}\)\(\Rightarrow\)\(\dfrac{a\times d}{b\times d}\)=\(\dfrac{c\times b}{d\times b}\)\(\Rightarrow\)a\(\times\)d=c\(\times\)d\(\Rightarrow\)ad=bc
b)theo câu a ta có \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad=bc\)(1)
Thêm ab vào 2 vế của (1):ad+ab=bc+ab
a(b+d)<b(a+c)\(\Rightarrow\)\(\dfrac{a}{b}< \dfrac{a+c}{b+d}\)(2)
Thêm cd vào 2 vế của (1):ad+cd<bc+cd
d(a+c)<c(b+d)\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\)(3)
Từ(2)và(3)\(\Rightarrow\)\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) ⇒ a=bk, c=dk
a) Ta có: ✽ \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)
✽\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
nên \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a-c}{c}=\dfrac{bk-dk}{dk}=\dfrac{k\left(b-d\right)}{dk}=\dfrac{b-d}{d}\)
Vậy \(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)
\(a,\)
Xét \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
\(b,\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) (Chứng minh câu a)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a}{b}\)
\(c,\)
Xét \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(d,\)
\(\dfrac{a}{c}=\dfrac{b}{d}\) (Chứng minh câu c)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(e,\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2a+b}{2c+d}\)
\(\Rightarrow\dfrac{2a+b}{2c+d}=\dfrac{a}{c}\)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)(đpcm)
b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{a}{c}-1=\dfrac{b}{d}-1\)
=>\(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)(đpcm)
a)\(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
=>\(\dfrac{a}{b}\) -1= \(\dfrac{c}{d}\) -1
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\left(1\right)\\ \dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}\left(2\right)\\ \dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}\left(3\right)\\ \dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}\left(4\right)\)
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
1 < B < 2 => KL