Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1
Ta có : \(\frac{a}{b}=\frac{c}{d}=>\left(\frac{a}{b}+1\right)=\left(\frac{c}{d}+1\right)\left(=\right)\frac{a+b}{b}=\frac{c+d}{d}\)
=> ĐPCM
Câu 2
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\left(\frac{b}{a}+1\right)=\left(\frac{d}{c}+1\right)\left(=\right)\frac{b+a}{a}=\frac{d+c}{c}=>\frac{a}{b+a}=\frac{c}{d+c}\)
=> ĐPCM
Câu 3
Câu 3
Ta có \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(=) (a+b).(c-d)=(a-b).(c+d)(=)ac-ad+bc-bd=ac+ad-bc-bd(=)-ad+bc=ad-bc(=) bc+bc=ad+ad(=)2bc=2ad(=)bc=ad=> \(\frac{a}{b}=\frac{c}{d}\)
=> ĐPCM
Câu 4
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=>\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)
Lại có \(\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+c^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2) => ĐPCM
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{a^2}{b^2};\frac{a}{b}.\frac{c}{d}=\frac{c}{d}.\frac{c}{d}=\frac{c^2}{d^2}\\ \Rightarrow\frac{a}{b}.\frac{c}{d}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
a) Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2.\left(k^2-1\right)}{d^2.\left(k^2-1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
b) Giải:
Để \(P\in Z\Rightarrow2x-3⋮x+1\)
Ta có:
\(2x-3⋮x+1\)
\(\Rightarrow\left(2x+2\right)-5⋮x+1\)
\(\Rightarrow5⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;5;-5\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=5\Rightarrow x=4\)
+) \(x+1=-5\Rightarrow x=-6\)
Vậy \(x\in\left\{0;-2;4;-6\right\}\)
\(\Rightarrow5⋮x+1\)
1)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(tính chất dãy tỉ số bằng nhau)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
2)\(P=\frac{2x-3}{x+1}=\frac{2x+2-5}{x+1}=\frac{2\left(x+1\right)-5}{x+1}=2-\frac{5}{x+1}\)
\(\Rightarrow P\in Z\Leftrightarrow2-\frac{5}{x+1}\in Z\Leftrightarrow\frac{5}{x+1}\in Z\Leftrightarrow5⋮x+1\Leftrightarrow x+1\inƯ\left(5\right)\)
\(\Rightarrow x+1\in\left\{-1;-5;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;-6;0;4\right\}\)
Đẳng thức đầu tiên sai:
Ví dụ: \(a=1;b=2;c=3;d=6\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Nhưng \(\dfrac{a.d}{c.d}\ne\dfrac{a^2-b^2}{b^2-d^2}\)
Với đẳng thức thứ 2:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a}{c}.\frac{b}{d}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)và \(\frac{a^2+b^2}{c^2+d^2}=\left(\frac{a+b}{c+d}\right)^2\)