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Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
Ta có: a+b-c/c = b+c-a/a = c+a-b/b = a+b-c+b+c-a+c+a-b/c+a+b
= a+b+c/a+b+c = 1 (Áp dụng tính chất dãy tỉ số bằng nhau)
Trường hợp 1 : Nếu a+b+c = 0 => a=0; b=0 ; c=0 => P =1
Trường hợp 2: Nếu a+b+c khác 0 => a+b+c = 1
=> a+b = 1-c => b+c = 1-a
=> a+c = 1-b
Ta lại có:
1-c-c/c =1 => 1- 2c/c =1 => 1-2c = c => 1 = 3c=> c= 1/3
1-a-c/a = 1 => 1- 2a/a=1 => 1-2a =a => 1 = 3a => a= 1/3
1-b-b/b = 1 => 1-2b/b = 1 => 1-2b = b => 1= 3b => b= 1/3
=> P= (1+ 1/3 : 1/3). (1+ 1/3 : 1/3). ( 1+ 1/3 :1/3)
= 2 . 2. 2 =8
Vậy P = 1 hoặc = 8
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
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) \(\Rightarrowđpcm\)
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)
a.Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
=>\(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)(đpcm)
b.Vì\(\dfrac{a}{b}=\dfrac{c}{d}\)
=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
=>\(\dfrac{a}{c}-1=\dfrac{b}{d}-1\)
=>\(\dfrac{a-c}{c}=\dfrac{b-d}{d}\)(đpcm)
a)\(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
=>\(\dfrac{a}{b}\) -1= \(\dfrac{c}{d}\) -1
=> \(\dfrac{a}{b}\) - \(\dfrac{b}{b}\) = \(\dfrac{c}{d}\) - \(\dfrac{d}{d}\)
=> \(\dfrac{a-b}{b}\) = \(\dfrac{c-d}{d}\)
a: Đặt a/b=b/c=c/d=k
=>a=bk; b=ck; c=dk
=>a=bk; b=dk^2; c=dk
=>a=dk^3; b=dk^2; c=dk
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{dk^3+dk^2+dk}{dk^2+dk+d}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
=>\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
c: Đặt a/2003=b/2004=c/2005=k
=>a=2003k; b=2004k; c=2005k
4(a-b)(b-c)=(c-a)^2
=>4(2004k-2003k)(2005k-2004k)=(2005k-2003k)^2
=>4*k*k=(2k)^2(luôn đúng)
=>ĐPCM
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)