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\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\\ \Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\\ \Leftrightarrow12a=10b\\ \Leftrightarrow6a=5b\Leftrightarrow\dfrac{a}{b}=\dfrac{5}{6}\)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab+5b-6a-30=ab-5b+6a-30\)
\(\Rightarrow5b-6a=-5b+6a\)
\(\Rightarrow10b=12a\)
\(\Rightarrow5b=6a\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\left(đpcm\right)\)
Vậy \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
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.
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)
Có \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
ab-6a+5b-30=ab-5b+6a-30
12a=10b
\(\dfrac{a}{b}=\dfrac{5}{6}\)
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 \(\dfrac{a}{b}=\dfrac{5}{6}\)
\(\RightarrowĐPCM\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
nhân ra ik ròi suy ra đpcm :D