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Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm
Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:
\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm
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.
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Câu 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2}{c^2}=\dfrac{b^2k^2}{d^2k^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2a^2+3b^2}{2c^2+3d^2}=\dfrac{2b^2k^2+3b^2}{2d^2k^2+3d^2}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{2a^2+3b^2}{2c^2+3d^2}\)
b: \(\dfrac{2a-3c}{c}=\dfrac{2bk-3dk}{dk}=\dfrac{2b-3d}{d}\)
a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
a) Ta co: a/b = c/d= k
=> a=bk
c=dk
Ta co: a-b/a+b = bk-b/bk+b = b(k-1)/b(k+1) = k-1/k+1 (1)
Ta co: c-d/c+d = dk-d/dk+d = d(k-1)/d(k+1) = k-1/k+1 (2)
Tu (1) va (2)
=> a-b/a+b=c-d/c+d
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\) (1)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b) Từ (*) ta có:
\(\dfrac{7a-4b}{3a+5b}=\dfrac{7bk-4b}{3bk+5b}=\dfrac{b\left(7k-4\right)}{b\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (3)
\(\dfrac{7c-4d}{3c+5d}=\dfrac{7dk-4d}{3dk+5d}=\dfrac{d\left(7k-4\right)}{d\left(3k+5\right)}=\dfrac{7k-4}{3k+5}\) (4)
Từ (3) và (4) suy ra \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (5)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (6)
\(\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}=\dfrac{\left[\left(dk\right)-\left(bk\right)\right]^2}{\left(d-b\right)^2}=\dfrac{\left[k\left(d-b\right)\right]^2}{\left(d-b\right)^2}=k^2\) (7)
Từ (5), (6) và (7) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
Bài 2:
a)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=> a = b = c
b)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
=> x = y = z (theo a)
Thay x = y = z vào biểu thức, ta có:
\(M=\dfrac{x^{333}.x^{666}}{x^{999}}=1\)
c)
\(ac=b^2\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(ab=c^2\Rightarrow\dfrac{b}{c}=\dfrac{c}{a}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Rightarrow a=b=c\)
Thay a = b = c vào biểu thức, ta có:
\(M=\dfrac{a^{333}}{a^{111}.a^{222}}=1\)
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{ac}{c^2}\)=\(\dfrac{bd}{d^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{d^2}{c^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)= \(\dfrac{2c^2-ac}{2c^2-bd}\)
=> \(\dfrac{a}{b}\)=\(\dfrac{2c^2-ac}{2c^2-bd}\)=>\(\dfrac{a^2}{b^2}\)=\(\dfrac{2c^2-ac}{2d^2-bd}\)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)= \(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)=\(\dfrac{ma+nb}{mc+nd}\)=\(\dfrac{ma-nb}{mc-nd}\)
=> \(\dfrac{ma+nb}{ma-nb}\)=\(\dfrac{mc+nd}{mc-nd}\)
c) 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:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^3}{c^3}\)=\(\dfrac{b^3}{d^3}\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^3\)(2)
Từ (1) và (2) suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^3\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)