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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(=>\hept{\begin{cases}a=b.k\\c=d.k\end{cases}}\)
\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{b.k-b}{d.k-d}\right)^2=\left(\frac{b.\left(k-1\right)}{d.\left(k-1\right)}\right)^2\)\(=\frac{\left(b^2.\left(k-1\right)^2\right)}{\left(d^2.\left(k-1\right)^2\right)}=\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}=\frac{b^2}{d^2}\)\(\left(1\right)\)
\(\frac{ab}{cd}=\frac{b.k.b}{d.k.d}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\)
Đặt \(\frac{a}{b}\)= \(\frac{c}{d}\)= k => a= bk ; c = dk
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)= \(\frac{b^2.\left(k-1\right)^2}{d^2.\left(k-1\right)^2}\)= \(\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}\)= \(\frac{bk.b}{dk.d}\)= \(\frac{b^2}{d^2}\) (2)
Từ (1) và (2) ->> \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) = \(\frac{ab}{cd}\)
Với \(a,b,c,d\ne0\) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{b}{d}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\Rightarrow\frac{a-b}{c-d}=\frac{b}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}\)\(=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
\(\Rightarrow\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
=>Đpcm
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\left(\frac{a+b}{c+d}\right)^3\left(1\right)\)
\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}\left(2\right)\)
Từ (1); (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\left(đpcm\right)\)
a, Ta co : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\)\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(1)
Xet :\(\frac{a}{a+b}=\frac{c}{c+d}\Rightarrow\frac{a}{c}=\frac{a+b}{c+d}\)(2)
Tu (1) va (2) \(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
b
1)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ac-ad=ac-bc\Leftrightarrow a\left(c-d\right)=c\left(a-b\right)\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
2) Gọi độ dài các cạnh của tam giác đó là a,b,c thì a : b : c = 3 : 4 : 5 ; a + b + c = 36
\(\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{36}{12}=3\Rightarrow\hept{\begin{cases}a=3.3=9\\b=3.4=12\\c=3.5=15\end{cases}}\).Vậy tam giác đó có 3 cạnh là 9 cm ; 12 cm ; 15 cm
3)\(\hept{\begin{cases}a:b:c:d=3:4:5:6\\a+b+c+d=3,6\end{cases}\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{d}{6}=\frac{a+b+c+d}{3+4+5+6}=\frac{3,6}{18}=0,2}\)
=> a = 0,2.3 = 0,6 ; b = 0,2.4 = 0,8 ; c = 0,2.5 = 1 ; d = 0,2.6 = 1,2
4)\(\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}:5=\frac{y}{2}:5\Leftrightarrow\frac{x}{15}=\frac{y}{10}\)
\(\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{5}:2=\frac{z}{7}:2\Leftrightarrow\frac{y}{10}=\frac{z}{14}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{14}=\frac{x+y+z}{15+10+14}=\frac{184}{39}=4\frac{28}{39}\Rightarrow\hept{\begin{cases}x=4\frac{28}{39}.15=70\frac{10}{13}\\y=4\frac{28}{39}.10=47\frac{7}{39}\\z=4\frac{28}{39}.14=66\frac{2}{39}\end{cases}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)
\(\frac{a+b^3}{c+d^3}=\frac{bk+b^3}{dk+d^3}\)
Đề bài sai nhé bạn