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Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
Do đó: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
Biết \(\dfrac{a^2 + b^2}{c^2 + d^2}=\dfrac{ab}{cd}\) với a,b,c,d khác 0. Chứng minh rằng:
\(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc\(\dfrac{a}{b}=\dfrac{d}{c}\) cái \(\dfrac{a}{b}=\dfrac{c}{d}\)thì mình chứng minh được rồi còn cái\(\dfrac{a}{b}=\dfrac{d}{c}\)thì chưa mong các bạn giúp ạ
Bài 1:
a) \(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Leftrightarrow x^2.25=6.24\)
\(\Leftrightarrow x^2.25=144\)
\(\Leftrightarrow x^2=144:25\)
\(\Leftrightarrow x^2=5,76\)
\(\Leftrightarrow x=2,4\)
b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(x+5\right)\)
\(\Leftrightarrow7x-7=6x+30\)
\(\Leftrightarrow7x=6x+30+7\)
\(\Leftrightarrow7x=6x+37\)
\(\Leftrightarrow7x-6x=37\)
\(\Leftrightarrow x=37\)
c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x-2\right).x+\left(x-2\right).7=\left(x+4\right).x-\left(x+4\right)\)
\(\Leftrightarrow x^2-2x+7x-14=x^2+4x-x-4\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow x^2+5x-14+4-3x-x^2=0\)
\(\Leftrightarrow\left(x^2-x^2\right)+\left(5x-3x\right)-\left(14-4\right)=0\)
\(\Leftrightarrow2x-10=0\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=10:2=5\)
Bài 2:
\(\dfrac{x}{7}=\dfrac{y}{13}\) và \(x+y=40\)
Ta có: \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)
Do đó \(\left\{{}\begin{matrix}\dfrac{x}{7}=2\Rightarrow x=14\\\dfrac{y}{13}=2\Rightarrow y=26\end{matrix}\right.\)
Vậy \(x=14;y=26\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
Ta có:
Nếu:
\(\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\Leftrightarrow\left(2a+c\right)\left(b-d\right)=\left(a-c\right)\left(2b+d\right)\)
\(\Leftrightarrow2a\left(b-d\right)+c\left(b-d\right)=a\left(2b+d\right)-c\left(2b+d\right)\)
\(\Leftrightarrow2ab-2ad+bc-cd=2ab+ad-2bc+cd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\left(đpcm\right)\)
Gọi d là ƯCLN(\(\dfrac{a+b}{2};\dfrac{b+c}{2};\dfrac{c+a}{2}\))(\(d\ne0,d⋮2\))
Ta có:\(\dfrac{a+b}{2}⋮d,\dfrac{b+c}{2}⋮d,\dfrac{c+a}{2}⋮d\)
\(\Rightarrow\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}⋮d\)
\(\Rightarrow\dfrac{a+b+b+c+c+a}{2}⋮d\)
\(\Rightarrow a+b+c⋮d\)
\(\Rightarrow a,b,c⋮d\)
\(\Rightarrow\)ƯCLN(a,b,c)=ƯCLN(\(\dfrac{a+b}{2};\dfrac{b+c}{2};\dfrac{c+a}{2}\))
P/S không chắc đâu nhất là 2 bước cuối
b)B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
B<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
B<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
B<\(1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+...+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)-\dfrac{1}{9}\)
B<1-\(\dfrac{1}{9}\)
B<\(\dfrac{8}{9}\)(1)
ta có:
B>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{10}\)
B>\(\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)...+\left(\dfrac{1}{9}+\dfrac{1}{9}\right)-\dfrac{1}{10}\)
B>\(\dfrac{1}{2}-\dfrac{1}{10}\)
B>\(\dfrac{2}{5}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{b}=k\)
\(\Rightarrow a=c.k;c=b.k\)
Suy ra:
\(\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{\left(c.k\right)^2+\left(b.k\right)^2}{b^2+\left(b.k\right)^2}=\dfrac{k^2.\left(c^2+b^2\right)}{b^2.\left(k^2+1\right)}\)
\(=\dfrac{k^2.\left[\left(b.k\right)^2+b^2\right]}{b^2.\left(k^2+1\right)}=\dfrac{k^2.\left[b^2.\left(k^2+1\right)\right]}{b^2.\left(k^2+1\right)}=k^2\) (1)
\(\dfrac{a}{b}=\dfrac{c.k}{b}=\dfrac{b.k^2}{b}=k^2\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2+c^2}{b^2+c^2}=\dfrac{a}{b}\)
Chúc học tốt!!
đề sai òi - . -