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Ta đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(a=b\times k\) ; \(c=d\times k\)
a) Ta có: \(\dfrac{a}{b}=\dfrac{b\times k}{d\times k}=\dfrac{b}{d}\) (1)
=> \(\dfrac{a+b}{c+d}=\dfrac{b\times k+b}{d\times k+d}=\dfrac{b\times\left(k+1\right)}{d\times\left(k+1\right)}=\dfrac{b}{d}\) (2)
Từ (1),(2) => đpcm
b)
\(\dfrac{a+b}{a}=\dfrac{b\times k+b}{b\times k}=\dfrac{b\times\left(k+1\right)}{b\times k}=\dfrac{k+1}{k}\) (1)
\(\dfrac{c+d}{c}=\dfrac{d\times k+d}{d\times k}=\dfrac{d\times\left(k+1\right)}{d\times k}=\dfrac{k+1}{k}\) (2)
Từ (1),(2) => đpcm
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}+1=\dfrac{c}{d}+1=>\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{b}-1=\dfrac{c}{d}-1=>\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=>ad=cb=>ad+ac=cb+ac\)
\(=>a\left(c+d\right)=c\left(a+b\right)=>\dfrac{a}{c}=\dfrac{a+b}{c+d}=>\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
a) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có \(\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\)\( \Rightarrow d(a + b) = b(c + d)\)\( \Rightarrow ad + bd = bc + bd\)
\( \Rightarrow ad = bc\) (luôn đúng)
\( \Rightarrow \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}\)
b) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có: \(\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}\)
\(\begin{array}{l} \Rightarrow d(a - b) = b(c - d)\\ \Leftrightarrow ad - bd = bc - bd\\ \Leftrightarrow ad = bc\end{array}\) ( luôn đúng)
Vậy \(\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}\)
c) Vì \(\dfrac{a}{b} = \dfrac{c}{d}\) nên \(ad = bc\)
Ta có: \(\dfrac{a}{{a + b}} = \dfrac{c}{{c + d}}\)
\(\begin{array}{l} \Rightarrow a(c + d) = c(a + b)\\ \Leftrightarrow ac + ad = ac + bc\\ \Leftrightarrow ad = bc\end{array}\) (luôn đúng)
Vậy \(\dfrac{a}{{a + b}} = \dfrac{c}{{c + d}}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}=k^2\)
\(\dfrac{ac}{bd}=k^2\)
Do đó: \(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
a) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\) và \(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)
\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b) \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-b}{b}=\dfrac{b\left(k-1\right)}{b}=k-1\\\dfrac{c-d}{d}=\dfrac{d\left(k-1\right)}{d}=k-1\end{matrix}\right.\)\(\Rightarrow\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
c) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
d) \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a}{c}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
Câu 1: D
Câu 2: C
Câu 3:
\(\sqrt{\dfrac{9}{49}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\dfrac{3}{7}\) là số hữu tỉ
\(\sqrt{49}=7\) là số hữu tỉ
\(\sqrt{0,9}=\sqrt{\dfrac{9}{10}}=\dfrac{3}{\sqrt{10}}\) là số vô tỉ
\(\sqrt{0,03}\) là số vô tỉ
=>Trong các số này có hai số là số vô tỉ đó là \(\sqrt{0,9};\sqrt{0,03}\)