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Câu 1
Ta có : \(\frac{a}{b}=\frac{c}{d}=>\left(\frac{a}{b}+1\right)=\left(\frac{c}{d}+1\right)\left(=\right)\frac{a+b}{b}=\frac{c+d}{d}\)
=> ĐPCM
Câu 2
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\left(\frac{b}{a}+1\right)=\left(\frac{d}{c}+1\right)\left(=\right)\frac{b+a}{a}=\frac{d+c}{c}=>\frac{a}{b+a}=\frac{c}{d+c}\)
=> ĐPCM
Câu 3
Câu 3
Ta có \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(=) (a+b).(c-d)=(a-b).(c+d)(=)ac-ad+bc-bd=ac+ad-bc-bd(=)-ad+bc=ad-bc(=) bc+bc=ad+ad(=)2bc=2ad(=)bc=ad=> \(\frac{a}{b}=\frac{c}{d}\)
=> ĐPCM
Câu 4
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=>\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)
Lại có \(\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+c^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2) => ĐPCM
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{5a+4b}{5a-4b}=\frac{5\cdot bk+4b}{5\cdot bk-4b}=\frac{b\left(5k+4\right)}{b\left(5k-4\right)}=\frac{5k+4}{5k-4}\)
\(\frac{5c+4d}{5c-4d}=\frac{5\cdot dk+4d}{5\cdot dk-4d}=\frac{d\left(5k+4\right)}{d\left(5k-4\right)}=\frac{5k+4}{5k-4}\)
Do đó: \(\frac{5a+4b}{5a-4b}=\frac{5c+4d}{5c-4d}\)
Bài 2:
a: Đặt \(\frac{a}{b}=\frac{b}{c}=k\)
=>b=ck; a=bk=ck^2
\(\frac{a^2+b^2}{b^2+c^2}=\frac{\left(ck^2\right)^2+\left(ck\right)^2}{\left(ck\right)^2+c^2}=\frac{c^2k^2\left(k^2+1\right)}{c^2\left(k^2+1\right)}=k^2\)
\(\frac{a}{c}=\frac{ck^2}{c}=k^2\)
Do đó: \(\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)
b: Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
=>\(\begin{cases}c=dk\\ b=ck=dk\cdot k=dk^2\\ a=bk=dk^2\cdot k=dk^3\end{cases}\)
\(\left(\frac{a+b-c}{b+c-d}\right)^3=\left(\frac{dk^3+dk^2-dk}{dk^2+dk-d}\right)^3\)
\(=\left\lbrack\frac{dk\left(k^2+k-1\right)}{d\left(k^2+k-1\right)}\right\rbrack^3=k^3\)
\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)
Do đó: \(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\)
Ta có :
\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< ac\Leftrightarrow ab+ad< ab+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)\(\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)
\(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\)\(\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Ta có:2bd=c(b+d)
=>2bd=bc+cd
Mà a+c=2b (theo đề)
=>(a+c).d=bc+cd
=>ad+cd=bc+cd
=>ad=bc (cùng bớt đi cd)
=>a/b=c/d (đpcm)
a,
\(\dfrac{a}{b}=\dfrac{b}{d}\\ \Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{d^2}=\dfrac{ab}{bd}\\ \Rightarrow\dfrac{a^2+b^2}{b^2+d^2}=\dfrac{a}{d} \)