Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
Ta có:
Nếu:
\(\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\Leftrightarrow\left(2a+c\right)\left(b-d\right)=\left(a-c\right)\left(2b+d\right)\)
\(\Leftrightarrow2a\left(b-d\right)+c\left(b-d\right)=a\left(2b+d\right)-c\left(2b+d\right)\)
\(\Leftrightarrow2ab-2ad+bc-cd=2ab+ad-2bc+cd\)
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\dfrac{2a+c}{2b+d}=\dfrac{a-c}{b-d}\left(đpcm\right)\)
Đề có bị sao không vậy? \(S\) không thể bằng \(2\) Sửa đề:
Chứng minh rằng \(S\ge6\)
Giải:
Ta có:
\(S=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}\)
\(=\left(\dfrac{a}{c}+\dfrac{b}{c}\right)+\left(\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{c}{b}\right)\)
\(=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\)
\(\Rightarrow S\ge2+2+2=6\)
Vậy \(S\ge6\) (Đpcm)
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 \(x=14;y=26\)
Á p dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
\(\Leftrightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{a-b}{c-d}\right)^2\Leftrightarrow\dfrac{a}{c}.\dfrac{b}{d}=\left(\dfrac{a-b}{c-d}\right)^2\)
suy ra đpcm
Bài 2 : đề bài này chỉ cần a,b>0 , ko cần phải thuộc N* đâu
a, Áp dụng bất đẳng thức AM-GM cho 2 số lhoong âm a,b ta được :
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\) . Dấu "=" xảy ra khi a=b
b , Áp dụng BĐT AM-GM cho 2 số không âm ta được : \(a+b\ge2\sqrt{ab}\)
\(\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}=\dfrac{2}{\sqrt{ab}}\)
Nhân vế với vế ta được :
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2.2.\dfrac{\sqrt{ab}}{\sqrt{ab}}=4\left(đpcm\right)\)
Dấu "="xảy ra tại a=b
Bài 1.
Vì a, b, c, d \(\in\) N*, ta có:
\(\dfrac{a}{a+b+c+d}< \dfrac{a}{a+b+c}< \dfrac{a}{a+b}\)
\(\dfrac{b}{a+b+c+d}< \dfrac{b}{a+b+d}< \dfrac{b}{a+b}\)
\(\dfrac{c}{a+b+c+d}< \dfrac{c}{b+c+d}< \dfrac{c}{c+d}\)
\(\dfrac{d}{a+b+c+d}< \dfrac{d}{a+c+d}< \dfrac{d}{c+d}\)
Do đó \(\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}< M< \left(\dfrac{a}{a+b}+\dfrac{b}{a+b}\right)+\left(\dfrac{c}{c+d}+\dfrac{d}{c+d}\right)\)hay 1<M<2.
Vậy M không có giá trị là số nguyên.
Theo đề bài ta có \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\) ( tính chất dãy tỉ số = nhau )
=> \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-c}{b-d}\) ( tính chất dãy tỉ số = nhau )
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 - . -