Cho \(\frac{1}{c}\)=\(\frac{1}{2}\)(\(\frac{1}{a}\)+\(\frac{1}{b}\)) (với a,b,c\(\ne\)0 ; b\(\ne\)c) chứng minh rằng \(\frac{a}{b}\)=\(\frac{a-c}{c-b}\)
Cho a+b+c=0; a,b,c≠0. Chứng minh \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\)
\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
cho \(\frac{1}{c}\)=\(\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)(với a, b, c \(\ne\)0; b\(\ne\)0)
chứng minh rằng: \(\frac{a}{b}=\frac{a-c}{c-b}\)
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c.\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Cho a+b+c=0; a,b,c≠0. Chứng minh :
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\) ( do \(a+b+c=0\) )
\(=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
\(=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) ( đpcm )
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)(Với a,b,c \(\ne\)0 ; b\(\ne\)c)
CMR \(\frac{a}{b}=\frac{a-c}{c-b}\)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\) ( với a, b, c \(\ne\)0 và b\(\ne\)c). CMR : \(\frac{a}{b}=\frac{a-c}{c-b}\)
mong mn giúp đỡ
Áp dụng tỉ dãy số bằng nhau. Ta có:
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Leftrightarrow\frac{1+1+1}{a+b+c}=1\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{a}{b}\Leftrightarrow1-1\Leftrightarrow0\)
\(\Rightarrow PT=\frac{a-c}{c-b}=\frac{\left(a-c\right)^0}{\left(c-b\right)^0}=0\)
Vậy dấu = xảy ra khi a - c = a , c - b = b
Ta có ĐPCM
Ps: Chả biết đúng hay không nữa
như này mới đúng nè
ta có\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)
\(\Rightarrow\frac{b}{ab}+\frac{a}{ba}=\frac{2}{c}\)
\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)
\(\Rightarrow\left(b+a\right)c=2ab\)
\(\Rightarrow cb+ca=ab+ab\)
\(\Rightarrow ca-ab=ab-cb\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a-c}{c-b}=\frac{a}{b}\)
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) với a, b, c ≠ 0 và M = \(\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}\)
CMR M = 3abc
C/Minh đẳng thức:
a) \(\left(\frac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{2}{a-1}\) (với a>0, b>0, a≠b)
b)\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\) (với a>0, b>0,a≠b)
c) \(\frac{2\sqrt{a}+3\sqrt{b}}{\sqrt{ab}+2\sqrt{a}-3\sqrt{b}-6}-\frac{6-\sqrt{ab}}{\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6}=\frac{a+9}{a-9}\) (với a≥0, b≥0,a≠9)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\) ( với a, b, c \(\ne\)0 và b \(\ne\)c ). Chứng minh rằng : \(\frac{a}{b}=\frac{a-c}{c-b}\)
mn giúp tôi đc ko???
cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)vơi a,b,c \(\ne\) 0; b\(\ne\) c chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-b}\)
ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)
\(=\frac{2}{c}=\frac{b+a}{ab}\)
= \(c\left(b+a\right)=ab\times2\)
= cb +ca = ab+ab
= ab - cb = ac-ab
\(=b\left(a-c\right)=a\left(c-b\right)\)
= \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)
\(\frac{1}{c}=\frac{a+b}{2ab}\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ac+bc\)
\(ab-bc=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)
Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a;b;c khác 0 và \(b\ne c\)
CMR: \(\frac{a}{b}=\frac{a-c}{c-b}\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(dpcm\right)\)