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\(\dfrac{3a-b}{a+b}=\dfrac{3}{4}\)
\(\Rightarrow3\left(a+b\right)=4\left(3a-b\right)\)
\(\Rightarrow3\left(a+b\right)-4\left(3a-b\right)=0\)
\(\Rightarrow\left(3a+3b\right)-\left(12a-4b\right)=0\)
\(\Rightarrow3a+3b-12a+4b=0\)
\(\Rightarrow-9a+7b=0\)
\(\Rightarrow-9a=-7b\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{-7}{-9}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{7}{9}\)
Vậy giá trị tỉ số \(\dfrac{a}{b}=\dfrac{7}{9}\)
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/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 \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\)
\(\Rightarrow ab\left(b+c\right)=bc\left(a+b\right)\)
\(\Rightarrow ab^2+abc=abc+b^2c\)
\(\Rightarrow ab^2=b^2c\)
\(\Rightarrow a=c\)
Đến đây ko còn manh mối :v
1)
a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)
\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)
\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)
\(\Rightarrow x=-\dfrac{55}{6}\)
b) 4,25 : 8 = -3,5 : x
\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)
\(x=\dfrac{-3,5.8}{4,25}\)
\(x=\dfrac{-28}{4,25}\)
2.
\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)
\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
Ta có thể lặp đc các tỉ lệ thức sau:
\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)
\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)
\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)
\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)
Từ TLT: \(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{\left(a+c\right)}{\left(b+d\right)}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{ac}{bd}\left(dpcm\right)\)
Bài 1:
\(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)
\(=2007.\dfrac{1}{90}-3\)
\(=19,3\)
Vậy S = 19,3
5b)\(S=1+3+3^2+...+3^{2013}\)
\(\Rightarrow3S=3+3^2+3^3+...+3^{2014}\)
\(\Rightarrow3S-S=3^{2014}-1\)
\(\Rightarrow S=\dfrac{3^{2014}-1}{2}\)
C : 4
Ta có: \(\dfrac{2}{5}=\dfrac{a}{10}\) \(\Leftrightarrow\dfrac{4}{10}=\dfrac{a}{10}\) \(\Rightarrow a=4\)
-> C: 4