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Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{\left(a+b+c\right)^2}{\left(b+c+d\right)^2}\)
\(\Leftrightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{\left(a+b+c\right)^2}{\left(b+c+d\right)^2}\)
\(\Leftrightarrow\dfrac{a}{d}=\dfrac{\left(a+b+c\right)^2}{\left(b+c+d\right)^2}\left(đpcm\right)\)
Chúc bạn học tốt!
ai giải được ko
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\)
1.
giả sử điều đó đúng thì:
\(c\left(b+a\right)=a\left(c+d\right)\\ bc+ca=ac+ad\Rightarrow bc+ca=ca+bc\left(đúng\right)\)
\(\Rightarrow\dfrac{a}{b+a}=\dfrac{c}{d+c}\)
2.
\(\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\\ \dfrac{a-b}{b}-1=\dfrac{c-d}{d}-1\\ \dfrac{a-b}{b}=\dfrac{c-d}{d}\\ \left(a-b\right)d=\left(c-d\right)b\\ ad-bd=bc-bd\\ \Rightarrow ad-bd=ad-bd\left(đúng\right)\)
\(\Rightarrow\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\) cũng đúng
1)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
\(\dfrac{a}{b+a}=\dfrac{c}{c+d}\Leftrightarrow a\left(c+d\right)=c\left(b+a\right)\)
\(\Leftrightarrow ac+ad=bc+ac\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{a}{b+a}=\dfrac{c}{c+d}\)
2)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-2=\dfrac{c}{d}-2\)
\(\Leftrightarrow\dfrac{a}{b}-\dfrac{2b}{b}=\dfrac{c}{d}-\dfrac{2d}{d}\)
\(\Leftrightarrow\dfrac{a-2b}{b}=\dfrac{c-2d}{d}\rightarrowđpcm\)
a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)
= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)
= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(\dfrac{51}{2.50}=\dfrac{51}{100}\)
Lời giải:
a)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)
Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)
Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)
b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
a, Ta có: \(\dfrac{a}{a+b+c}< \dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\) (1)
\(\dfrac{b}{a+b+c}< \dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\) (1)
\(\dfrac{c}{a+b+c}< \dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\) (3)
Từ (1), (2), (3) \(\Rightarrow\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
Thầy mk hướng dẫn phần a như thế còn phần b mk ko bt lm, chúc p hk tốt 
Vì \(b\ne d;b+d\ne0\) nên áp dụng tính chất cảu dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
Vậy \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\) (đpcm)
Chúc bạn học tốt!!!
Ta có:Nếu
\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
thì \((a+c)(b-d)=(a-c)(b+d)\)
\(a(b-d)+c(b-d)=a(b+d)-c(b+d)\)
\(ab-ad+bc-cd=ab+ad-bc+cd\)
\(=\)\(ab-ab\)\(-ad+ad\)\(+bc-bc\)\(-cd+cd\)
\(=0\)
\(\Leftrightarrow\left(a+c\right)\left(b-d\right)\)\(=\left(a-c\right)\left(b+d\right)\)
\(\Leftrightarrow\dfrac{a+c}{b+d}\)\(=\dfrac{a-c}{b-d}\)
Ahhhhh!!! Bài gì khó dữ!!!
Ta có :
+) \(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\) \(\left(1\right)\)
+) \(c^2=b.d\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Đặt :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=k^3\)
Mặt khác :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=k^3\)
Áp dụng tính chất dãy tỉ lệ thức ta có :
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
\(\Leftrightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\rightarrowđpcm\)
Ta có: \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\left(b>0,d>0\right)\)
a) Giả sử: +) \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\) \(ad=bc\) (nhân chéo)
\(\Rightarrow\) nếu \(\dfrac{a}{b}< \dfrac{c}{d}\) thì \(ad< bc.\)
b) Giả sử \(ad=bc\) \(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\) nếu \(ad< bc\) thì \(\dfrac{a}{b}< \dfrac{c}{d}.\)
Ta có:\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)
\(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{a}{c}=\dfrac{b}{d}\)
Vậy \(\dfrac{a}{c}=\dfrac{b}{d}\left(\text{đ}pcm\right)\)
1.
- Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\left(a,b,c,d\ne0\right)\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
- Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}\) =\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)( đpcm).
2.
- Ta có:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
=> ( a+b ).(c-a) = (a-b).(c+a)
=> ac - a2 +bc-ba = ac +a2 -bc -ba
=> ac - a2 +bc-ba -(ac +a2 -bc -ba) =0
=> ac - a2 +bc-ba -ac -a2 +bc +ba = 0
=>ac - aa +bc-ba -ac -aa +bc +ba = 0
=> ( ac-ac) +( -aa-aa) +( bc+bc) + ( -ba+ba) =0
=> -2aa +2bc = 0
=> 2bc = 2aa
=> bc = aa
=> bc = a2
- Vậy nếu bc = a2 thì \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)( đpcm).