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Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2-b^2}{ab}=\dfrac{b^2k^2-b^2}{bk\cdot b}=\dfrac{b^2\left(k^2-1\right)}{b^2k}=\dfrac{k^2-1}{k}\)
\(\dfrac{c^2-d^2}{cd}=\dfrac{d^2k^2-d^2}{dk\cdot d}=\dfrac{d^2\left(k^2-1\right)}{d^2\cdot k}=\dfrac{k^2-1}{k}\)
Do đó: \(\dfrac{a^2-b^2}{ab}=\dfrac{c^2-d^2}{cd}\)
b: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(bk+b\right)^2}{b^2k^2+b^2}=\dfrac{b^2\cdot\left(k+1\right)^2}{b^2\left(k^2+1\right)}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
\(\dfrac{\left(c+d\right)^2}{c^2+d^2}=\dfrac{\left(dk+d\right)^2}{d^2k^2+d^2}=\dfrac{\left(k+1\right)^2}{k^2+1}\)
Do đó: \(\dfrac{\left(a+b\right)^2}{a^2+b^2}=\dfrac{\left(c+d\right)^2}{c^2+d^2}\)
A)\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)=\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\) (đpcm)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+b}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a.b}{c.d}=\dfrac{a+b}{c+d}.\dfrac{a+b}{c+d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2c}{2d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\left(đpcm\right)\)
Vậy...
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\) (!)
Thay (!) vào đề bài:
VT = \(c\left(k+2\right).d\left(k+1\right)\left(1\right)\)
\(VP=c\left(k+1\right).d\left(k+2\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow VT=VP\)
hay \(\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\).
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
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) \(\Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\left(\dfrac{b}{d}\right)^2\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}=\left(\dfrac{b}{d}\right)^2\) (2)
Từ (1) và (2)=> \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\) (đpcm)
Dặt a/b = c/d = k
=> a = kb; c = kd
Sau đó thay vào biểu thức sẽ ra keets quả bằng nhau.
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) chứng minh \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{a}{b}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
mà cần chứng minh: \(\left(\dfrac{a+b+c}{b+c+d}\right)=\dfrac{a}{d}\left(2\right)\)
từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\) \(\dfrac{a^3}{b^3}=\dfrac{a}{d}\Rightarrow a^3.d=b^3.a\)
\(\Rightarrow a^2.d=b^3\)
vì \(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow a.c=b^2\)
\(\Rightarrow a.b.c=b.c\left(3\right)\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a.d=b.c\left(4\right)\)
từ \(\left(3\right)\) và \(\left(4\right)\) \(\Rightarrow a.a.d=b^3\)
\(\Rightarrow a^2.d=b^3\left(đpcm\right)\)
vậy \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)