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\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=> bc+ac+ab=0
ta có
\(bc+ac=-ab\)
<=> \(\left(bc+ac\right)^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)
tương tự
\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)
\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)
thay vào E ta đc
\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)
=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)
A = (2a + 2b +2c - 3c)^2 + (2b + 2c +2a - 3a)^2 + (2c + 2a +2b -3b)^2
Đặt a + b + c = x thì
A = (2x - 3c)^2 + (2x - 3a)^2 + (2x - 3b)^2
=4x^2 - 12cx + 9c^2 + 4x^2 - 12ax + 9x^2 + 4x^2 - 12bx + 9b^2
=12x^2 - 12x(a + b + c) + 9(a^2 + b^2 + c^2)
=12x^2 - 12x^2 + 9(a^2 + b^2 + c^2) =9(a^2 + b^2 + c^2) =9m
\(A=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2\)
\(=\left(4a^2+4b^2+c^2+8ab-4ac+4bc\right)+\left(4b^2+4c^2+a^2+8bc-4ba-4ac\right)\)\(+\left(4c^2+4a^2+b^2+8ac-4cb-4ab\right)\)
\(=9a^2+9b^2+9c^2\)
\(=9\left(a^2+b^2+c^2\right)\)
\(=9m\)
1. Phuc will look through a new English book tomorrow.
2. Which ethnic group has the largest population in Vietnam?
Hình như sai đề :
Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)
\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)
\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )
Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)
\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)
\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )
CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )
Thay ( * ) và ( * ') vào E , ta được :
\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)
\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)
\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)
\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)
\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)
\(=\dfrac{0}{2}=0\)
Vậy \(E=0\)
P = \(\frac{a^2c}{a^2c+c^2b+b^2a+}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)
P = \(\frac{a^2c+b^2a+c^2b}{a^2c+c^2b+b^2a}=1\)
\(P=\frac{\frac{a}{b}}{\frac{a}{b}+\frac{c}{a}+\frac{b}{c}}+\frac{\frac{b}{c}}{\frac{b}{c}+\frac{a}{b}+\frac{c}{a}}+\frac{\frac{c}{a}}{\frac{c}{a}+\frac{b}{c}+\frac{a}{b}}=\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}=1\)
= 4a^2+4b^2+c^2+8ab-4bc-4ac+4b^2+4c^2+a^2+8bc-4ac-4ab+4c^2+4a^2+b^2+8ac-4ab-4bc
= 7 ( a^2+b^2+c^2 ) = 7 .10 = 70
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