Cho a, b, c ≠ 0 thoả mãn \(\left\{{}\begin{matrix}a+b+c=2021\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2021}\end{matrix}\right.\) . Chứng minh: \(\frac{1}{a^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{a^{2021}+b^{2021}+c^{2021}}\)
Những câu hỏi liên quan
cho a,b,c khác 0; a+b+c khác 0 thỏa mãn: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
CMR: \(\frac{1}{a^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{a^{2021}+b^{2021}+c^{2021}}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow ab^2+a^2b+ac^2+a^2c+bc^2+b^2c+2abc=0\)
\(\Leftrightarrow ab^2+a^2b+ac^2+bc^2+a^2c+abc+b^2c+abc=0\)
\(\Leftrightarrow\left(a+b\right)ab+c^2\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(c^2+ab+bc+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Vậy ta có các trường hợp: \(a=-b,c=0\)hoặc \(b=-c,a=0\)hoăc \(a=-c,b=0\).
Với từng trường hợp ta đều có đpcm.
Cho a,b,c ,(a+b+c) là các số thực khác 0 thỏa mãn điều kiện: \(\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\\a^3+b^3+c^3=2^9\end{cases}}\)
Tính \(A=a^{2021}+b^{2021}+c^{2021}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0
=> Hoặc a=-b hoặc b=-c hoặc c=-a
Ko mất tổng quát, g/s a=-b
a) Ta có: vì a=-b thay vào ta được:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)
=> đpcm
b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
cho \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). chứng minh:\(2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\)
Cho a,b,c ,(a+b+c) là các số thực khác 0 thỏa mãn điều kiện: \(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\\a^3+b^3+c^3=2^9\end{matrix}\right.\)
Tính \(A=a^{2021}+b^{2021}+c^{2021}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0
=> Hoặc a=-b hoặc b=-c hoặc c=-a
Ko mất tổng quát, g/s a=-b
a) Ta có: vì a=-b thay vào ta được:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)
=> đpcm
b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
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Cho các số thực a,b,c thỏa mãn a+b+c=2021 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2021}\). Tính \(Q=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}\)
\(=1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}\)
\(=3+Q\)
Suy ra \(3+Q=1\Leftrightarrow Q=-2\).
Cho b^2=ac (b+c khác 0)
Chứng minh: $\frac{ (a+b)^{2021} }{ (b+c)^{2021} }$=$\frac{ a^{2021}+ b^{2021} }{b^{2021}+c^{2021}}$
11 tháng 2 2020 lúc 21:04
1. left{{}begin{matrix}a,b,c0a+b+c1end{matrix}right.. Cmr: frac{ab}{sqrt{left(1-cright)^2left(1+cright)}}+frac{bc}{sqrt{left(1-aright)^2left(1+aright)}}+frac{ca}{sqrt{left(1-bright)^3left(1+bright)}}lefrac{3sqrt{2}}{8}
2. left{{}begin{matrix}a,b,c0a+b+cle1end{matrix}right.. Cmr: frac{1}{a^2+b^2+c^2}+frac{1}{ableft(a+bright)}+frac{1}{bcleft(b+cright)}+frac{1}{acleft(a+cright)}gefrac{87}{2}
3. left{{}begin{matrix}a,b,c0ab+bc+ca2abcend{matrix}right.. Cmr: frac{1}{aleft(2a-1right)^2}+frac{1}{bleft...
Đọc tiếp
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\zy+yz+zx=27\end{matrix}\right.\)
a.
\(\frac{3x-36}{12}=\frac{5y-45}{15}=\frac{z-1}{1}=\frac{3x+5y-z-50}{26}=\frac{-48}{26}\)
\(\Rightarrow\frac{x-12}{4}=\frac{-48}{26}\Rightarrow x=...\)
Tương tự với y, z, nhưng chắc bạn nhầm đề, nếu pt bên dưới là -2 thì nó ra \(\frac{-52}{26}=-2\) kết quả đẹp hơn nhiều
b. Không rõ đề
c.
\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)
Cho \(\left\{{}\begin{matrix}a,b,c>0\\\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=3\end{matrix}\right.\)
Tìm MAX A \(=\frac{1}{\left(2a+b+c\right)^2}+\frac{1}{\left(2b+a+c\right)^2}+\frac{1}{\left(2c+a+b\right)^2}\)
Áp dụng BĐT Cô - si ta có :
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\ge\frac{2}{\frac{x+y}{2}}=\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\left(1\right)\)
Áp dụng BĐT trên ta được :
\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\Rightarrow\left(\frac{1}{2a+b+c}\right)^2\le\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2\)
Chứng minh tương tự rồi cộng các vế lại cho nhau ta được :
\(A\le\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\frac{1}{16}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2\)
\(\Rightarrow16A\le\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\left(\frac{1}{a+c}+\frac{1}{b+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2\)
\(=\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(c+a\right)^2}+\frac{2}{\left(a+b\right)\left(a+c\right)}+\frac{2}{\left(b+c\right)\left(a+b\right)}+\frac{2}{\left(a+c\right)\left(b+c\right)}\)
Đặt \(\left(\frac{1}{a+b};\frac{1}{b+c};\frac{1}{c+a}\right)\rightarrow\left(x;y;z\right)\)
Khi đó \(16A\le2x^2+2y^2+2z^2+2xy+2yz+2zx\)
Ta có BĐT phụ sau :
\(xy+yz+zx\le x^2+y^2+z^2\) ( tự chứng minh ) (2)
Áp dụng ta được :
\(16A\le4x^2+4y^2+4z^2=\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)
\(\Rightarrow4A\le\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\)
Từ (1) \(\Rightarrow\frac{1}{\left(x+y\right)^2}\le\frac{1}{16}\left(\frac{1}{x}++\frac{1}{y}\right)^2\)( bình phương 2 vế lên )
Áp dụng BĐT này ta được :
\(4A\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{16}\left(\frac{1}{b}+\frac{1}{c}\right)^2+\frac{1}{16}\left(\frac{1}{c}+\frac{1}{a}\right)^2\)
\(\Rightarrow64A\le\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{2}{bc}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{ac}+\frac{1}{a^2}\)
\(\Rightarrow32A\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
Áp dụng BĐT (2) ta được :
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Rightarrow32A\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=3+3=6\)
\(\Rightarrow A\le\frac{6}{32}=\frac{3}{16}\)
Dấu " = " xảy ra khi a=b=c=1
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