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Phùng Minh Quân

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Tổng: 5305 | Điểm tuần: 8 | Trả lời 7 ngày qua: 4 | Lượt trả lời trong tháng: 4

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Những câu trả lời của Phùng Minh Quân:

Vào lúc: 2020-06-05 15:30:07 Xem câu hỏi

\(y=\frac{\sqrt{2017\left(x-2015\right)}}{\sqrt{2017}\left(x+2\right)}+\frac{\sqrt{2016\left(x-2016\right)}}{\sqrt{2016}x}\le\frac{1}{2\sqrt{2017}}+\frac{1}{2\sqrt{2016}}\)

"=" \(\Leftrightarrow\)\(x=4032\)

Vào lúc: 2020-06-05 15:12:32 Xem câu hỏi

đề đúng: \(a,b,c>0\)

chuẩn hoá: \(a+b+c=3\)

\(\frac{1}{a^2+ab}+\frac{a}{2}+\frac{a+b}{4}\ge\frac{3}{2}\)\(\Leftrightarrow\)\(\frac{1}{a^2+ab}\ge\frac{3}{2}-\frac{3}{4}a-\frac{1}{4}b\)

tương tự \(\Rightarrow\)\(\Sigma\frac{1}{a^2+ab}\ge\frac{9}{2}-\left(a+b+c\right)=\frac{3}{2}=\frac{27}{2\left(a+b+c\right)^2}\)

dấu "=" xảy ra khi \(a=b=c=1\)

chưa học chuẩn hoá thì dùng cách này: 

gia su: \(a+b+c=3k>0\)

\(\frac{1}{a^2+ab}+\frac{a}{2k^3}+\frac{a+b}{4k^3}\ge\frac{3}{2k^2}\)\(\Leftrightarrow\)\(\frac{1}{a^2+ab}\ge\frac{3}{2k^2}-\frac{3}{4k^3}a-\frac{1}{4k^3}b\)

\(\Rightarrow\)\(\Sigma\frac{1}{a^2+ab}\ge\frac{9}{2k^2}-\frac{a+b+c}{4k^3}=\frac{3}{2k^2}=\frac{27}{2\left(a+b+c\right)^2}\)

dấu "=" xảy ra khi \(a=b=c=k\)

Vào lúc: 2020-06-03 16:13:40 Xem câu hỏi

đk: \(x+2y\ge0\)

\(x+2y=\sqrt{\frac{x^2+4y^2}{2}}+\sqrt{\frac{\left(x+y\right)^2}{3}+y^2}\ge\sqrt{\frac{\left(x+2y\right)^2}{4}}+\sqrt{\frac{\left(x+2y\right)^2}{4}}=x+2y\)

\(\Rightarrow\)\(x=2y\)\(\Rightarrow\)\(x=3-y=3-\frac{x}{2}\)\(\Rightarrow\)\(\hept{\begin{cases}x=2\\y=\frac{x}{2}=1\end{cases}}\)

Vào lúc: 2020-06-02 16:43:42 Xem câu hỏi

Chứng minh cái này đi: \(\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{2}{3}a+\frac{2}{3}\) ( gợi ý: bđt \(\Leftrightarrow\)\(\left(a-1\right)^2\left(a+1\right)\ge0\)

Tương tự với 2 ẩn kia \(\Rightarrow\)\(\Sigma\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{8}{27}\Pi\left(a+1\right)\ge\frac{64}{27}\sqrt{abc}\ge\frac{64}{27}\)

dấu "=" xảy ra khi \(a=b=c=1\)

Vào lúc: 2020-05-30 15:50:14 Xem câu hỏi

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

Vào lúc: 2020-05-16 18:34:39 Xem câu hỏi

\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)

Vào lúc: 2020-05-12 19:09:22 Xem câu hỏi

ta dễ chứng minh được \(x+y\ge\frac{2\sqrt{2}}{5}-\frac{2}{5}\)\(\Rightarrow\)\(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}>0\)

\(P=\frac{5\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\left(\frac{5}{2}\left(x+y-\left(\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\right)\left(\frac{5}{2}\left(x+y\right)+\sqrt{2}+1\right)-\frac{9}{4}\left(x-y\right)^2\right)}{\frac{5}{2}\left(x+y\right)+\sqrt{2}+1}\)

\(+\left(\frac{\frac{45}{2}\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)}{5\left(x+y\right)+\sqrt{2}+1}+\frac{9}{2}\right)\left(x-y\right)^2+6-4\sqrt{2}\ge6-4\sqrt{2}\)

Dấu "=" xảy ra khi \(x=y=\frac{\sqrt{2}-1}{5}\)

Vào lúc: 2020-05-10 18:41:07 Xem câu hỏi

Kaneki Ken

đk: \(x\ge0;y\ge0;x\ne-y\)

hpt \(\Leftrightarrow\)\(\hept{\begin{cases}2\sqrt{6x}\left(x+y+1\right)=4\sqrt{2}\left(x+y\right)\\\sqrt{7y}\left(x+y-1\right)=4\sqrt{2}\left(x+y\right)\end{cases}}\)

\(\Rightarrow\)\(2\sqrt{6x}\left(x+y+1\right)=\sqrt{7y}\left(x+y-1\right)\)

\(\Leftrightarrow\)\(\left(2\sqrt{6x}-\sqrt{7y}\right)\left(x+y+1\right)=0\)

... 

Vào lúc: 2020-05-06 19:01:55 Xem câu hỏi

\(abc\ge\frac{\left(252^3\left(63a+36b+28c+756\right)-252\right)\Sigma\left(28c+252\right)\left(63a-36b\right)^2}{63504\Pi\left(63a+252\right)\left(63a+36b+28c+756\right)}\)

\(+\frac{1}{63504}\Pi\left(63a+252\right)\left(\frac{63a+36b+28c-1512}{63a+36b+28c+756}\right)+\frac{508032.252}{63504}\ge2016\)

dau "=" xay ra khi \(\left(a;b;c\right)=\left(8;14;18\right)\)

Vào lúc: 2020-04-24 18:18:58 Xem câu hỏi

Có: \(\Delta=\left(m-2\right)^2\ge0\) => pt đã cho có nghiệm 

Vi-et: \(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=m-1\end{cases}}\)

\(C=\frac{2x_1x_2+3}{\left(x_1+x_2\right)^2+2}=\frac{2m+1}{m^2+2}\)

đến đây xét delta ra min max..

Vào lúc: 2020-04-24 18:01:46 Xem câu hỏi

hpt \(\Leftrightarrow\)\(\hept{\begin{cases}5\left(x+y\right)^2+\frac{2}{\left(x+y\right)^2}-12xy=\frac{251}{5}\\\frac{\left(x+y\right)^2+1}{x+y}=5-\left(x-y\right)\end{cases}}\) (*) 

đặt \(\left(a;b\right)=\left(x+y;x-y\right)\)\(\left(a\ne0\right)\)

hệ (*) \(\Leftrightarrow\)\(\hept{\begin{cases}5a^2+\frac{2}{a^2}-3\left(a^2-b^2\right)=\frac{251}{5}\\b=5-\frac{a^2+1}{a}\end{cases}}\Leftrightarrow\hept{\begin{cases}25a^4-150a^3+154a^2-150a+25=0\left(1\right)\\b=5-\frac{a^2+1}{a}\end{cases}}\)

pt (1) \(\Leftrightarrow\)\(\orbr{\begin{cases}a=\frac{1}{5}\Rightarrow b=\frac{-1}{5}\\a=5\Rightarrow b=\frac{-1}{5}\end{cases}}\)\(\Rightarrow\)\(\left(x;y\right)=\left\{\left(0;\frac{1}{5}\right);\left(\frac{12}{5};\frac{13}{5}\right)\right\}\)

Vào lúc: 2020-04-24 17:04:29 Xem câu hỏi

\(\Sigma\frac{a^3+1}{b^3+c^3+1}=(\frac{-\left(a+b\right)\left(c^3+1\right)}{ab\left(a+b+c\right)\left(a^3+b^3+1\right)}+\frac{\Sigma\left(a+b\right)^2}{3\left(a+b+c\right)}+2\left(a+b+c\right)\)

\(+\frac{\frac{1}{2}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\Sigma\frac{1}{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}\right)}{\left(a+b+c\right)^2+3\left(a+b+c\right)+9}+\frac{\Sigma\left(a-b\right)^2}{a+b+c})\left(a-b\right)^2+2\ge2\)

justforfun:) 

Vào lúc: 2020-04-22 19:32:52 Xem câu hỏi

\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)

Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)

Vào lúc: 2020-04-18 12:14:08 Xem câu hỏi

bài 2 cách 2 

Có: \(\Sigma\sqrt{a^2+b^2}\ge\sqrt{2}\left(a+b+c\right)\)

\(VT=\Sigma\frac{a^2+b^2}{b}-\left(a+b+c\right)\ge\frac{\left(\Sigma\sqrt{a^2+b^2}\right)^2}{a+b+c}-\left(a+b+c\right)\ge VP\)

Dấu "=" xảy ra khi a=b=c

Vào lúc: 2020-04-18 12:08:11 Xem câu hỏi

\(VT-VP=\frac{c\left(b\sqrt{c^2+a^2}-a\sqrt{a^2+b^2}\right)^2}{abc\left(a+b+c\right)}\)

\(+\frac{\Sigma\sqrt{a^2+b^2}\Sigma\frac{\left(a-b\right)^2}{\sqrt{a^2+b^2}+\frac{a+b}{\sqrt{2}}}}{2\left(a+b+c\right)}+\frac{1}{2\sqrt{2}}\Sigma\frac{\left(a-b\right)^2}{\sqrt{a^2+b^2}+\frac{a+b}{\sqrt{2}}}\ge0\)

Vào lúc: 2020-04-17 16:54:57 Xem câu hỏi

1) \(\left(a;b\right)=\left(\sqrt{3x+4y};\sqrt{8-x+y}\right)\) \(\left(a;b\ge0\right)\)

hpt \(\Leftrightarrow\)\(\hept{\begin{cases}4a+b=23\\3b-2\sqrt{-a^2-9b^2+110}=5\end{cases}}\Leftrightarrow\hept{\begin{cases}b=23-4a\\32-6a=\sqrt{-145a^2+1656a-4651}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=23-4a\\181a^2-2040a+5675=0\left(1\right)\end{cases}}\)

(1) \(\Leftrightarrow\)\(\orbr{\begin{cases}a=5\left(nhan\right)\Rightarrow b=3\left(nhan\right)\\a=\frac{1135}{181}\left(nhan\right)\Rightarrow b=\frac{-377}{181}\left(loai\right)\end{cases}}\)\(\Rightarrow\)\(a=5;b=3\)\(\Rightarrow\)\(x=3;y=4\)

Vào lúc: 2020-04-15 18:05:21 Xem câu hỏi

bđt \(\Leftrightarrow\)\(\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)\ge\left(\frac{10}{3}\right)^3abc\) (*) 

đặt \(\left(\sqrt{ab};\sqrt{bc};\sqrt{ca}\right)=\left(x;y;z\right)\)\(\Rightarrow\)\(xyz\le\frac{1}{27}\)

(*) \(\Leftrightarrow\)\(\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\ge\left(\frac{10}{3}\right)^3xyz\)

\(VT\ge\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\)

Có \(xy+1\ge10\sqrt[10]{\frac{xy}{9^9}}\)

Tương tự với \(yz+1\)\(;\)\(zx+1\)\(\Rightarrow\)\(VT\ge10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\)

Ta cần CM \(10^3\sqrt[10]{\frac{\left(xyz\right)^2}{9^{27}}}\ge\frac{10^3}{3^3}xyz\) đúng với \(xyz\le\frac{1}{27}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Vào lúc: 2020-04-15 11:38:52 Xem câu hỏi

\(\sqrt[4]{\frac{a^4+b^4}{2}}=\sqrt[12]{\frac{\left(\frac{a^6}{a^2}+\frac{b^6}{b^2}\right)^3}{8}}\ge\sqrt[12]{\frac{\left(a^3+b^3\right)^6}{8\left(a^2+b^2\right)^3}}\ge\sqrt[12]{\frac{\frac{\left(a^3+b^3\right)^4\left(a^2+b^2\right)^4}{\left(a+b\right)^2}}{8\left(a^2+b^2\right)^3}}\)

\(=\sqrt[12]{\frac{\left(a^3+b^3\right)^4\left(a^2+b^2\right)}{8\left(a+b\right)^2}}\ge\sqrt[12]{\frac{\frac{\left(a^3+b^3\right)^4\left(a+b\right)^2}{2}}{8\left(a+b\right)^2}}=\sqrt[3]{\frac{a^3+b^3}{2}}\)

dấu "=" xay ra khi a=b=c 

Vào lúc: 2020-02-05 16:22:48 Xem câu hỏi

Nyatmax can chung minh \(a^2-ab+bc+ca\ge0\)

Vào lúc: 2020-02-02 15:29:59 Xem câu hỏi

Có: \(4=\left(a+b\right)^2-\left(b-1\right)^2\le\left(a+b\right)^2\)\(\Rightarrow\)\(a+b\ge2\)

\(P=\frac{\frac{a^4}{a}+\frac{b^4}{b}}{ab}\ge\frac{\frac{\left(a^2+b^2\right)^2}{a+b}}{ab}\ge\frac{\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{a+b}}{ab}=\frac{\left(a+b\right)\left(a+b\right)^2}{4ab}\ge\frac{2\left(2\sqrt{ab}\right)^2}{4ab}=2\)

"=" \(\Leftrightarrow\)\(a=b=1\)

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