\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\) ( a khác 5 ; b khác 6 )

Chứng minh rằng...">

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6 tháng 11 2017

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)

nhân ra ik ròi suy ra đpcm :D

14 tháng 4 2017

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)

\(\Rightarrow ab+5b-6a-30=ab-5b+6a-30\)

\(\Rightarrow5b-6a=-5b+6a\)

\(\Rightarrow10b=12a\)

\(\Rightarrow5b=6a\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\left(đpcm\right)\)

Vậy \(\dfrac{a}{b}=\dfrac{5}{6}\)

14 tháng 4 2017

\(\dfrac{a+5}{a-5}=\dfrac{a+6}{a-6}\)suy ra \(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(a+6\right)\)

suy ra: \(6a=5b\)

suy ra: \(\dfrac{a}{b}=\dfrac{5}{6}\)

AH
Akai Haruma
Giáo viên
7 tháng 2 2020

Bài 1:

$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:

\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)

$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)

Từ $(1);(2)$ suy ra đpcm.

AH
Akai Haruma
Giáo viên
7 tháng 2 2020

Bài 2:

Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:

$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)

5 tháng 11 2017

Sửa câu a:

(x - 2)2 - 36 = 0

(x - 2 - 6)(x - 2 + 6) = 0

(x - 8)(x + 4)= 0

\(\Leftrightarrow \begin{bmatrix} x - 8= 0 & & \\ x + 4 = 0 & & \end{bmatrix}\)

\(\Leftrightarrow \begin{bmatrix} x = 8 & & \\ x = - 4 & & \end{bmatrix}\)

pn bỏ dấu ngoặc bên phải nhé

Vậy x = 8; x = - 4

5 tháng 11 2017

2:

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{2a}{2b}=\dfrac{a}{b}\)

Từ đó suy ra \(\dfrac{a}{b}=\dfrac{5}{6}\)

\(\RightarrowĐPCM\)

27 tháng 7 2017

pn ơi hình như đề sai a+5/a-5 va b+6/b-6

27 tháng 7 2017

ta có : a+5/a-5=b+6/b-6
=> a+5/b+6=a-5/b-6
áp dụng dãy tỉ số bằng nhau ta được:
a+5/b+6=a-5/b-6 =(a+5+a-5)/(b+6+b-6)=(a+5-a+5)/(b+6-b+6)
=> 2a/2b = 10/12
=> a/b = 5/6

13 tháng 9 2017

\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)

\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)

ab-6a+5b-30=ab-5b+6a-30

12a=10b

\(\dfrac{a}{b}=\dfrac{5}{6}\)

13 tháng 9 2017

thanks

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

1/* Chứng minh rằng:

\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+...\dfrac{1}{49\times50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+..+\dfrac{1}{50}\)

2/* Cho:

A=\(\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+.....+\dfrac{1}{99\times100}\). Chứng minh rằng:\(\dfrac{7}{12}< A>\dfrac{5}{6}\)

Các bn giúp mk những bài này nha!

4
16 tháng 7 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\Rightarrow x=-2004\)

16 tháng 7 2017

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)

\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)

\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Rightarrow x+2004=0\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004\)

12 tháng 7 2017

2, a-b=ab => a=ab+b => a=b(a+1)

thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1

Theo bài ra ta có: a:b=a-b

=> a+1=a-b

=>-b=1

=> b=-1

Thay b=-1 vào a-b=ab ta có : a-(-1)=-a

=> a +1=-a

=>a=-1/2

Vậy a=-1/2. b=-1

19 tháng 7 2017

a,

\(a+b=-9\\ b+c=2\\ c+a=-3\\ \Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\\ 2a+2b+2c=-10\\ 2\left(a+b+c\right)=-10\\ a+b+c=-5\\ a+b=-9\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-9\right)+c=-5\Rightarrow c=4\\ b+c=2\\ \Rightarrow a+b+c=-5\Leftrightarrow a+2=-5\Rightarrow a=-7\\ c+a=-3\\ \Rightarrow a+b+c=-5\Leftrightarrow\left(-3\right)+b=-5\Rightarrow b=-2\)

Vậy \(a=-7;b=-2;c=5\)

b,

\(a+b=\dfrac{1}{2}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{-5}{6}\\ \Rightarrow a+b+b+c+c+a=\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{-5}{6}\\ 2a+2b+2c=\dfrac{6}{12}+\dfrac{9}{12}+\dfrac{-10}{12}\\ 2\left(a+b+c\right)=\dfrac{5}{12}\\ a+b+c=\dfrac{5}{24}\\ a+b=\dfrac{1}{2}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{2}+c=\dfrac{5}{24}\Rightarrow c=\dfrac{-7}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{5}{24}\Rightarrow a=\dfrac{-13}{24}\\ a+c=\dfrac{-5}{6}\\ \Rightarrow a+b+c=\dfrac{5}{24}\Leftrightarrow b+\dfrac{-5}{6}=\dfrac{5}{24}\Rightarrow b=\dfrac{25}{24}\)

Vậy \(a=\dfrac{-13}{24};b=\dfrac{25}{24};c=\dfrac{-7}{24}\)

c,

\(a+b=2\\ b+c=6\\ c+a=3\\ \Rightarrow a+b+b+c+c+a=2+6+3\\ 2a+2b+2c=11\\ 2\left(a+b+c\right)=11\\ a+b+c=5,5\\ a+b=2\\ \Rightarrow a+b+c=5,5\Leftrightarrow2+c=5,5\Rightarrow c=3,5\\ b+c=6\\ \Rightarrow a+b+c=5,5\Leftrightarrow a+6=5,5\Rightarrow a=-0,5\\ c+a=3\\ \Rightarrow a+b+c=5,5\Leftrightarrow b+3=5,5\Rightarrow b=2,5\)

Vậy \(a=-0,5;b=2,5;c=3,5\)

d,

\(a+b=\dfrac{5}{6}\\ b+c=\dfrac{3}{4}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+b+c+c+a=\dfrac{5}{6}+\dfrac{3}{4}+\dfrac{5}{3}\\ 2a+2b+2c=\dfrac{10}{12}+\dfrac{9}{12}+\dfrac{20}{12}\\ 2\left(a+b+c\right)=\dfrac{13}{4}\\ a+b+c=\dfrac{13}{8}\\ a+b=\dfrac{5}{6}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow\dfrac{5}{6}+c=\dfrac{13}{8}\Rightarrow c=\dfrac{19}{24}\\ b+c=\dfrac{3}{4}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow a+\dfrac{3}{4}=\dfrac{13}{8}\Rightarrow a=\dfrac{7}{8}\\ c+a=\dfrac{5}{3}\\ \Rightarrow a+b+c=\dfrac{13}{8}\Leftrightarrow b+\dfrac{5}{3}=\dfrac{13}{8}\Rightarrow b=\dfrac{-1}{24}\)

Vậy \(a=\dfrac{7}{8};b=\dfrac{-1}{24};c=\dfrac{19}{24}\)

19 tháng 7 2017

\(\left\{{}\begin{matrix}a+b=-9\\b+c=2\\c+a=-3\end{matrix}\right.\)

\(\Rightarrow a+b+b+c+c+a=\left(-9\right)+2+\left(-3\right)\)

\(\Rightarrow2a+2b+2c=-10\)

\(\Rightarrow2\left(a+b+c\right)=-10\)

\(\Rightarrow a+b+c=-5\)

\(\Rightarrow\left\{{}\begin{matrix}c=-5-9=-14\\a=-5-2=-7\\b=-5-\left(-3\right)=-2\end{matrix}\right.\)