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\(a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow a+b+c-ab-bc-ca=0\)
\(\Leftrightarrow a+b+c-ab-bc-ca+abc-1=0\)
\(\Leftrightarrow\left(a-ac\right)+\left(b-bc\right)+\left(-ab+abc\right)+\left(c-1\right)=0\)
\(\Leftrightarrow-a\left(c-1\right)-b\left(c-1\right)+ab\left(c-1\right)+\left(c-1\right)=0\)
\(\Leftrightarrow\left(-a-b+ab+1\right)\left(c-1\right)=0\)
\(\Leftrightarrow\left[b\left(a-1\right)-\left(a-1\right)\right]\left(c-1\right)\)
\(\Leftrightarrow\left(b-1\right)\left(a-1\right)\left(c-1\right)=0\)
\(\Rightarrow\)\(\left[{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)(đpcm)
\(\left\{{}\begin{matrix}ab+bc+ca=abc\\a+b+c=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}abc-ab-bc-ca=0\\a+b+c-1=0\end{matrix}\right.\)
\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)
\(=abc-ab-ac+a-bc+b+c-1\)
\(=\left(abc-ab-bc-ca\right)+\left(a+b+c-1\right)\)
\(=0+0=0\) (ddpcm)
\(VT=\left(a-1\right)\left(b-1\right)\left(c-1\right)\\ =\left(ab-a-b+1\right)\left(c-1\right)\\ =abc-ab-ac+a-bc+b+c-1\\ =abc-\left(ab+bc+ca\right)+\left(a+b+c\right)-1\\ =abc-abc+1-1=0=VP\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2019}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{2019}\)
\(\Leftrightarrow2019\left(ab+bc+ac\right)=abc\)
\(\Leftrightarrow2019\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+b+c\right)-abc=0\)
\(\Leftrightarrow b\left(a+b+c\right)\left(a+c\right)+ca\left(a+c\right)=0\)
\(\Leftrightarrow\left(ab+b^2+bc+ac\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Suy ra a + b = 0 hoặc b + c = 0 hoặc a + c = 0
Mà a + b + c = 2019 nên phải có 1 trong ba số a,b,c bằng 2019 (đpcm)
1) Cho 3 số a,b,c thỏa mãn 0 < a <= b <= c. Chứng minh rằng:
a/b + b/c + c/a >= b/a + c/a + a/c
2) Giải phương trình:
( 2017 - x)^3 + ( 2019 - x)^3 + (2x - 4036)^3 = 0
3)
a) Rút gọn biểu thức : A = 1/1-x + 1/1+x + 2/1+x^2 + 4/1+x^4 + 8/1+x+8
b) Tìm x,y biết : x^2 + y^2 + 1/x^2 + 1/y^2 = 4
\(a+b+c=\frac{abc}{a}+\frac{abc}{b}+\frac{1}{c}\Leftrightarrow a+b+c=bc+ac+\frac{1}{c}\)
\(\Leftrightarrow c\left(a+b\right)-\left(a+b\right)+\frac{1}{c}-c=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-1\right)-\frac{c^2-1}{c}=0\)
\(\Leftrightarrow\left(a+b\right)\left(c-1\right)-\frac{\left(c+1\right)\left(c-1\right)}{c}=0\)
\(\Leftrightarrow\left(c-1\right)\left(a+b-\frac{c+1}{c}\right)=0\)
\(\Leftrightarrow\left(c-1\right)\left(a+b-\frac{c+abc}{c}\right)=0\)
\(\Leftrightarrow\left(c-1\right)\left(a+b-1-ab\right)=0\)
\(\Leftrightarrow\left(c-1\right)\left[a-1-b\left(a-1\right)\right]=0\)
\(\Leftrightarrow\left(c-1\right)\left(a-1\right)\left(1-b\right)=0\Rightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
`1/a+1/b+1/c=1/(a+b+c)`
`<=>(a+b)/(ab)+(a+b)/(c(a+b+c))=0`
`<=>(a+b)(ab+ac+bc+c^2)=0`
`<=>(a+b)(a+c)(b+c)=0`
`=>` $\left[ \begin{array}{l}a=-b\\b=-c\\c=-a\end{array} \right.$
`=>` PT luôn tồn tại 2 số đối nhau
\(abc=1\Rightarrow c=\frac{1}{ab}\)
\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Leftrightarrow a+b+\frac{1}{ab}=\frac{1}{a}+\frac{1}{b}+ab\)
\(\Leftrightarrow\left(ab-a-b+1\right)-\left(\frac{1}{ab}-\frac{1}{a}-\frac{1}{b}+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)-\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)-\frac{\left(a-1\right)\left(b-1\right)}{ab}=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(1-\frac{1}{ab}\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\)
\(\Leftrightarrow a=1\text{ hoặc }b=1\text{ hoặc }c=1\)
Cách khác: Nhân tung \(\left(a-1\right)\left(b-1\right)\left(c-1\right)\) ra, dựa vào giả thiết để suy ra no bằng 0.
thanks ban nhìu nha!