Cho: \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b},\right)\left(a,b,c\ne0,b\ne c\right)\) Chứng minh rằng: \(\frac{a}{b}=\frac{a-b}{c-b}\)
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cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0;b\ne c\right)\)) chứng minh rằng : \(\frac{a}{b}=\frac{a-c}{c-b}\)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(a,b,c\ne0,b\ne c\right)\).Chứng minh rằng\(\frac{a}{b}=\frac{a-c}{c-b}\)
1. Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)với a,b,c \(\ne0,b\ne c\). Chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-b}\)
theo bài ra ta có:
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{b}{ab}+\frac{a}{ab}\right)\\ \Rightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\\ \Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
=> 2ab = c(a + b)
=> ab + ab = ca + cb
=> ab - cb = ca - ab
=> b( a - c ) = a( c - b )
=> \(\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
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Cho 4 số dương a;b;c;d. Biết rằng \(b=\frac{a+c}{2};c=\frac{2bd}{b+d}\)
Chứng minh 4 số này lập thành 1 tỉ lệ thức
B2
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right);\left(a;b;c\ne0;b\ne c\right)\) . Chứng minh \(\frac{a}{b}=\frac{a-c}{c-b}\)
B1:
Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)
Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:
\(2b=a+\frac{2bd}{b+a}\)
\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)
\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)
\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)
\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)
\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)
B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)
Cho \(\frac{2}{a}=\frac{1}{b}+\frac{1}{c}\left(a,b,c\ne0,a\ne c\right)\)
Chứng minh rằng: \(\frac{b}{c}=\frac{b-a}{a-c}\)
\(\frac{2}{a}=\frac{1}{b}+\frac{1}{c}=\frac{b+c}{bc}\)
=> 2bc = a.(b + c)
=> bc + bc = ab + ac
=> bc - ac = ab - bc
=> c.(b - a) = b.(a - c)
\(\Rightarrow\frac{b-a}{a-c}=\frac{b}{c}\left(đpcm\right)\)
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Cho \(\frac{1}{c}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\)(với \(a,b,c\ne0;b\ne c\)) chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-d}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ca+bc\)
\(\Rightarrow ab-cb=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
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Trả lời :........................................................
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}......................\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Học sinh giỏi 6A
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cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\) với \(a,b,c\ne0,b\ne c\) chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-b}\)
Ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{c}\div\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{2}{c}=\frac{b+a}{ab}\)
\(\Rightarrow2ab=c\left(b+a\right)\)
\(\Rightarrow ab+ab=bc+ac\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)(đpcm)
cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right)\) ; \(abc\ne0\) và\(a\ne b\)
Chứng minh rằng: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c\)
1) Cho frac{a-left(c-bright)}{b-c}+frac{b-left(a-cright)}{c-a}+frac{c-left(b-aright)}{a-b}3CM frac{a}{left(b-cright)^2}+frac{b}{left(c-aright)^2}+frac{c}{left(a-bright)^2}02) Cho frac{1}{a}+frac{1}{c}frac{1}{b-c}-frac{1}{a-b}và acne0; ane b; bne cCM frac{a}{c}frac{a-c}{b-c}
Đọc tiếp
1) Cho \(\frac{a-\left(c-b\right)}{b-c}+\frac{b-\left(a-c\right)}{c-a}+\frac{c-\left(b-a\right)}{a-b}=3\)
CM \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
2) Cho \(\frac{1}{a}+\frac{1}{c}=\frac{1}{b-c}-\frac{1}{a-b}\)và \(ac\ne0\); \(a\ne b\); \(b\ne c\)
CM \(\frac{a}{c}=\frac{a-c}{b-c}\)