cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\) ( \(a\ne5;b\ne6\)). Chứng minh \(\dfrac{a}{b}=\dfrac{5}{6}\)
bạn nào biết làm ơn viết lời giải ra hộ mình với
Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\left(a\ne5,b\ne6\right)\). Như vậy tỉ số \(\dfrac{a}{b}\) bằng bao nhiêu?
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
=>(a+5)(b-6)=(a-5)(b+6)
=>ab-6a+5b-30=ab+6a-5b-30
=>-6a+5b=6a-5b
=>-12a=-10b
=>6a=5b
hay a/b=5/6
1.Tìm x, biết:
\(\dfrac{x}{3}=\dfrac{y}{5}\)và x+y =16
2.Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\left(a\ne5;b\ne6\right)\)
CMR: \(\dfrac{a}{b}=\dfrac{5}{6}\)
giúp tớ đi cc >_< !!!
1.Tìm x, biết:
x/3=y/5
Theo tính chất dãy tỉ số bằng nhau ta có:
x+y/3+5= 16/8=2
=>x=6; y=10
2.Cho a+5/a−5=b+6/b−6(a≠5;b≠6)
CMR: ab=56
Giải:
ta có a+5/a-5=b+6/b-6 =>a+5/b+6=a-5/b-6 (*)
=> a+5+a-5/b+6+b-6=2a/2b=a/b (1)
Lại có: (*)=a+5-a+5/b+6-b+6=10/12=5/6 (2)
Từ 1 và 2 suy ra a/b=5/6 (đpcm)
Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\) (a ≠ 5; b ≠ 6). Chứng minh rằng \(\dfrac{a}{b}=\dfrac{5}{6}\)
mọi người ơi giúp mik với, ai làm đc mik tick cho
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)
Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\) ( a khác 5 ; b khác 6 )
Chứng minh rằng \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
nhân ra ik ròi suy ra đpcm :D
cho\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)(a\(\ne\)5;b\(\ne\)6) chứng minh:\(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab+5b-6a-30=ab-5b+6a-30\)
\(\Rightarrow5b-6a=-5b+6a\)
\(\Rightarrow10b=12a\)
\(\Rightarrow5b=6a\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\left(đpcm\right)\)
Vậy \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\dfrac{a+5}{a-5}=\dfrac{a+6}{a-6}\)suy ra \(\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(a+6\right)\)
suy ra: \(6a=5b\)
suy ra: \(\dfrac{a}{b}=\dfrac{5}{6}\)
Một cách giải khác:
TH1: b = -6
VP = 0 => VT = 0 => a = -5
=> \(\dfrac{a}{b}=\dfrac{-5}{-6}=\dfrac{5}{6}\)
TH2: b \(\ne\) -6 nên:
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+5}{b+6}=\dfrac{a-5}{b-6}=\dfrac{a+5+a-5}{b+6+b-6}=\dfrac{a}{b}=\dfrac{a+5-a+5}{b+6-b+6}=\dfrac{10}{12}=\dfrac{5}{6}\)
\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\left(x\ne5\right)\Rightarrow x=?????\)
\(\Leftrightarrow7\cdot\left(x-1\right)=6\cdot\left(x+5\right)\)
\(\Leftrightarrow7x-7-6x-30=0\)
\(\Leftrightarrow x-37=0\)
\(\Leftrightarrow x=37\left(N\right)\)
Ta có: \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow7x-7-6x-30=0\)
\(\Leftrightarrow x=37\)
Tìm \(\dfrac{a}{b}\)
\(\dfrac{a}{b}x4+\dfrac{1}{6}=\dfrac{19}{6}\)
b) \(\dfrac{4}{5}:\dfrac{a}{b}x6=\dfrac{16}{5}\)
\(a)\)\(\dfrac{a}{b}\times4+\dfrac{1}{6}=\dfrac{19}{6}\)
\(\dfrac{a}{b}\times4=\dfrac{19}{6}-\dfrac{1}{6}\)
\(\dfrac{a}{b}\times4=\dfrac{18}{6}\)
\(\dfrac{a}{b}=\dfrac{18}{6}\div4\)
\(\dfrac{a}{b}=\dfrac{18}{6}\times\dfrac{1}{4}\)
\(\dfrac{a}{b}=\dfrac{18}{24}\)
\(Cho\frac{a+5}{a-5}=\frac{b+6}{b-6}\left(a\ne5,b\ne6\right).Cmr\frac{a}{b}=\frac{5}{6}\)
1. a) \(\dfrac{x-3}{5}=\dfrac{7}{x-1}\)
b) \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+11}=0\)
2. Cho \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)( a khác 5; b khác 6)
CMR: \(\dfrac{a}{b}=\dfrac{5}{6}\)
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
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 \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\RightarrowĐPCM\)
1.
a) \(\frac{x - 3}{5}=\frac{7}{x - 1} \)
(x - 3)(x - 1) = 35
x2 - x - 3x + 3 - 35 = 0
x2 - 4x - 32 = 0
x2 - 4x + 4 - 36 = 0
(x - 2)2 - 36 = 0
(x - 2 - 36)(x - 2 + 36) = 0
(x - 38)(x + 34) = 0
\(\Leftrightarrow \begin{bmatrix} x - 38 = 0 & & \\ x + 34 = 0 & & \end{bmatrix}\)
\(\Leftrightarrow \begin{bmatrix} x = 38 & & \\ x = - 34 & & \end{bmatrix}\)
pn bỏ dấu ngoặc bên phải nhé
Vậy x = 38 ; x = - 34
b) (x - 5)x + 1 - (x - 5)x + 11 = 0
\((x - 5)^{x + 1}\left [ 1- (x - 5)^{10} \right ]= 0\)
tới đây pn giải giống câu a