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Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{5k+3}{5k-3}\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{5k+3}{5k-3}\)
Do đó: \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7b^2k^2+8\cdot bk\cdot b}{11\cdot b^2\cdot k^2-8b^2}=\dfrac{7k^2+8k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+8\cdot dk\cdot d}{11\cdot d^2\cdot k^2-8d^2}=\dfrac{7k^2+8k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2+8ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) thay \(a=bk;c=dk\) ta có
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)
từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) thay \(a=bk;c=dk\) ta có
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)
\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)
\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)
từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
a)Đặt \(\dfrac{a}{b}=\dfrac{c}{b}=k\left(k\ne0\right)\)
=> a=bk; c=dk
+) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(1\right)\)
+) \(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) cũng đặt và cm tương tự
Mk chỉ làm 1 câu thôi mấy câu sau tương tự theo cách đó nhoa:v
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\dfrac{b^4}{d^4}\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{bk^4+b^4}{dk^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)
\(\Rightarrow\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\Rightarrowđpcm\)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^4}{c^4}\)=\(\dfrac{b^4}{d^4}\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^4\)(2)
Từ (1) và (2)suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^4\)=\(\dfrac{a^4+b^4}{c^4+d^4}\)(đpcm)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5c}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a+3b}{5c+3d}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{5a}{5b}\)=\(\dfrac{3b}{3d}\)=\(\dfrac{5a-3b}{5c-3d}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)=\(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\) (đpcm)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Do đó: \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{a}{c}\right)^2\)và \(\dfrac{a}{c}\).\(\dfrac{b}{d}\)=\(\left(\dfrac{b}{d}\right)^2\)
=>\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\) và \(\dfrac{ab}{cd}\)=\(\dfrac{b^2}{d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ab}{cd}\)=\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{7a^2}{7c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{3ab}{3cd}\)=\(\dfrac{7a^2+3ab}{7c^2+3cd}\)(1)
Ta có: \(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=> \(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a^2}{c^2}\)=\(\dfrac{b^2}{d^2}\)=\(\dfrac{11a^2}{11c^2}\)=\(\dfrac{8b^2}{8d^2}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)(2)
Từ (1) và (2) suy ra:
\(\dfrac{7a^2+3ab}{7c^2+3cd}\)=\(\dfrac{11a^2-8b^2}{11c^2-8d^2}\)=\(\dfrac{7a^2+3ab}{11a^2-8b^2}\)=\(\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
a) Ta có:
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}\)
\(\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
b)
\(\frac{2a-b}{2a+b}=\frac{2bk-b}{2bk+b}=\frac{b(2k-1)}{bb(2k+1)}=\frac{2k-1}{2k+1}\)
\(\frac{2c-d}{2c+d}=\frac{2dk-d}{2dk+d}=\frac{d(2k-1)}{d(2k+1)}=\frac{2k-1}{2k+1}\)
\(\Rightarrow \frac{2a-b}{2a+b}=\frac{2c-d}{2c+d}\) (đpcm)
từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{5a+3b}{7a-2b}=\dfrac{5.ck+3.dk}{7.ck-2.dk}=\dfrac{k.\left(5c+3d\right)}{k.\left(7c-2d\right)}=\dfrac{5c+3d}{7c-2d}\)Vậy \(\dfrac{5a+3b}{7a-2b}=\dfrac{5c+3d}{7c-2d}\left(đpcm\right)\)
b) từ \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=k=>\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
ta có:\(\dfrac{7a^2+3ab}{11a^2+8.b^2}=\dfrac{7.c^2.k^2+3.c.d.k^2}{11.c^2.k^2+8.d^2.k^2}=\dfrac{k^2.\left(7.c^2+3.c.d\right)}{k^{2.}\left(11.c^2+8.d^2\right)}\) vậy .......
c)\(từ\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\dfrac{a+b}{c+d}\right)^2\)(1)
Mặt khác:\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1).(2)=>......
1,
Giải:
a, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\) (1)
\(\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
\(\Rightarrowđpcm\)
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a-b}{a}=\dfrac{bk-b}{bk}=\dfrac{b\left(k-1\right)}{bk}=\dfrac{k-1}{k}\)
\(\Rightarrow\dfrac{c-d}{c}=\dfrac{dk-d}{dk}=\dfrac{d\left(k-1\right)}{dk}=\dfrac{k-1}{k}\)
\(\dfrac{k-1}{k}=\dfrac{k-1}{k}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\rightarrowđpcm\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+2c}{3b+2d}=\dfrac{-5a+3c}{-5b+3d}\)
Vậy..............(đpcm)
Chúc bạn học tốt!!!
Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-5a}{-5b}=\dfrac{3c}{3d}=\dfrac{-5a+3c}{-5b+3d}\)
Vậy \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{2a+2c}{2b+2d}=\dfrac{-5a+3c}{-5b+3d}\Leftrightarrowđpcm\)
*a/b=c/d=k=>a=bk;c=dk
Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3
Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3
=>2a+3b/2a-3b=2c+3d/2c-3d
*a/b=c/d=>a/c=b/d=k
=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)
k^2=a/c.b/d=ab/cd (2)
Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2
*a/b=c/d=>a/c=b/d=k=a+b/c+d
=>k^2=(a+b/c+d)^2
k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2
=>(a+b/c+d)^2=a^2+b^2/c^2+d^2
Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)
a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)
Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)
Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
bạn cứ đặt công thức gốc là k sau đó thay vào các câu là được thui
`a/b = c/d`
`=> a/c = b/d = (11a)/(11c) = (11b)/(11d) = (3b)/(3d) = (3a)/(3c)`
Áp dụng t/chất dãy tỉ số bằng nhau:
`=> (11a + 3b)/(11c + 3d) = (3a - 11b)/(3c - 11d)` (đpcm)
Trường hợp a=0 => c=0 . Đẳng thức luôn đúng
Trường hợp a khác 0 => c khác 0
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{11a}{11c}=\dfrac{3b}{3d}=\dfrac{11a+3b}{11c+3d}\) ; (1)
tương tự
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a}{3c}=\dfrac{11b}{11d}=\dfrac{3a-11b}{3c-11d}\) ; (2)
từ (1) và (2) suy ra
\(\dfrac{11a+3b}{11c+3d}=\dfrac{3a-11b}{3c-11d}\)