Khôi Bùi

Đề phải là : \(\frac{9}{2\left(a+b+c\right)}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\left(x;y;z>0\right)\)

\(\Rightarrow\frac{a}{b^2}=\frac{y^2}{x};\frac{b}{c^2}=\frac{z^2}{y};\frac{c}{a^2}=\frac{x^2}{z};xyz=1\)

\(\frac{9}{2\left(a+b+c\right)}=\frac{9}{\frac{2\left(a+b+c\right)}{abc}}\left(abc=1\right)=\frac{9}{2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)}=\frac{9}{2\left(xy+yz+xz\right)}\)

Khi đó , ta có : \(P=\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}+\frac{9}{2\left(xy+yz+xz\right)}\)

\(=\frac{x^2}{z}+z+\frac{y^2}{x}+x+\frac{z^2}{y}+y+\frac{9}{2\left(xy+yz+xz\right)}-x-y-z\)

AD BĐT Cauchy , ta có :

\(P\ge2x+2y+2z+\frac{9}{\frac{2\left(x+y+z\right)^2}{3}}-\left(x+y+z\right)=x+y+z+\frac{27}{2\left(x+y+z\right)^2}\)

\(=\frac{x+y+z}{2}+\frac{x+y+z}{2}+\frac{27}{2\left(x+y+z\right)^2}\ge3.\sqrt[3]{\frac{27}{8}}=\frac{9}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

\(A=x^3+x^2y-2x^2-xy-y^2+3y+x-2019\)

\(=x^2\left(x+y-2\right)-y\left(x+y-2\right)+x+y-2-2017\)

\(=\left(x+y-2\right)\left(x^2-y+1\right)-2017\)

\(=0-2017=-2017\)

\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)

\(=\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left[\left(\sqrt{2}-1\right)^2-\left(\sqrt{3}-1\right)+1\right]\)

\(=\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)-2\left(\sqrt{2}+1\right)\left(\sqrt{6}+1\right)+\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\)

\(=\left(\sqrt{6}+1\right)\left(\sqrt{6}-\sqrt{3}+\sqrt{2}-1-2\sqrt{2}-2+\sqrt{6}+\sqrt{3}+\sqrt{2}+1\right)\)

\(=\left(\sqrt{6}+1\right)\left(2\sqrt{6}-2\right)\)

\(=2\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)=10\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\sqrt{4+\sqrt{15}}.\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}.\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}.\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)

\(\frac{x+y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{4}{\left(x+y\right)z}\ge\frac{4}{\frac{\left(x+y+z\right)^2}{4}}=\frac{4}{\frac{4^2}{4}}=1\)

\(\Rightarrow x+y\ge xyz\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=1;z=2\)

Min : Do x ; y không âm , \(x^2+y^2=1\) \(\Rightarrow\left|x\right|;\left|y\right|\le1\)

\(\Rightarrow0\le x;y\le1\)

\(\Rightarrow xy\ge0;x\ge x^2;y\ge y^2\)

\(P=\sqrt{1+2x}+\sqrt{1+2y}\)

\(\Rightarrow P^2=1+2x+1+2y+2\sqrt{1+2x+2y+2xy}\)

\(\ge2+2.1+2\sqrt{1+2.1+0}=4+2\sqrt{3}\)

\(\Rightarrow P\ge\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

Dấu " = " xảy ra \(\Leftrightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)

Max : Áp dụng BĐT phụ : \(\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\) , ta có :

\(P=\sqrt{1+2x}+\sqrt{1+2y}\le\sqrt{2\left(1+2x+1+2y\right)}\)

\(=\sqrt{2\left[2+2\left(x+y\right)\right]}\le\sqrt{2\left[2+2\sqrt{2\left(x^2+y^2\right)}\right]}=\sqrt{2\left(2+2\sqrt{2}\right)}=\sqrt{4+4\sqrt{2}}=2\sqrt{1+\sqrt{2}}\)

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

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\frac{\sqrt{2}.\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)

\(S=ab+2\left(a+b\right)\le\frac{a^2+b^2}{2}+2\sqrt{2\left(a^2+b^2\right)}=\frac{1}{2}+2\sqrt{2.1}=\frac{1}{2}+2\sqrt{2}=\sqrt{8}+\frac{1}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)

Đặt \(\sqrt{x^2+4}=a\) \(\Rightarrow a^2+1=x^2+5\)

\(x^2+4\ge4\forall x\Rightarrow a=\sqrt{x^2+4}\ge2\)

Đề bài đã cho trở thành : Cho \(a\ge2\) . Tìm min \(P=\frac{a^2+1}{a}\)

Giải : \(P=\frac{a^2+1}{a}=a+\frac{1}{a}=a+\frac{4}{a}-\frac{3}{a}\)

Áp dụng BĐT Cô - si cho 2 số ta có : \(P=a+\frac{4}{a}-\frac{3}{a}\ge4-\frac{3}{a}\)

Mà \(a\ge2\Rightarrow P\ge4-\frac{3}{2}=\frac{5}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=2\Leftrightarrow x=0\)

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ĐK : \(x\ge-1;y\ge0\)

Ta có HPT : \(\left\{{}\begin{matrix}\sqrt{x-1}+2\sqrt{y}=5\left(1\right)\\\frac{4x-9-y}{2\sqrt{x-1}+3\sqrt{y}}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}+4\sqrt{y}=10\\2\sqrt{x-1}-3\sqrt{y}=-4\end{matrix}\right.\)

\(\Rightarrow7\sqrt{y}=14\) \(\Leftrightarrow y=4\)

Thay y = 4 vào ( 1 ) , ta được : \(\sqrt{x-1}+2\sqrt{y}=5\)

\(\Leftrightarrow\sqrt{x-1}+4=5\) \(\Leftrightarrow x-1=1\Leftrightarrow x=2\)

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ĐK : \(x\ne-2\) , ta có :

\(\left\{{}\begin{matrix}\frac{1}{x+2}+\frac{2}{y-x}=3\\\frac{2}{x+2}-\frac{3}{y-x}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{3}{x+2}+\frac{6}{y-x}=9\\\frac{4}{x+2}-\frac{6}{y-x}=2\end{matrix}\right.\)

\(\Rightarrow\frac{7}{x+2}=11\Leftrightarrow x+2=\frac{7}{11}\Leftrightarrow x=\frac{-15}{11}\)

Thay \(x=-\frac{15}{11}\)vào pt : \(\frac{1}{x+2}+\frac{2}{y-x}=3\) được :

\(\frac{1}{\frac{-15}{11}+2}+\frac{2}{y+\frac{15}{11}}=3\) \(\Leftrightarrow\frac{11}{7}+\frac{2}{y+\frac{15}{11}}=3\)

Giải ra được : \(y=\frac{2}{55}\)

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1 ) Đặt \(x+y=S;xy=p\) , ta có :

\(\left\{{}\begin{matrix}S+p=7\\S^2-p=13\end{matrix}\right.\) \(\Rightarrow S^2+S=20\Leftrightarrow\left(S-4\right)\left(S+5\right)=0\Leftrightarrow\left[{}\begin{matrix}S=4\\S=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}p=3\\p=12\end{matrix}\right.\)

TH 1 : \(S=4;p=3\) . Giải pt : \(x^2-4x+3=0\)

TH 2 : S \(=-5;p=12\) . Giải pt : \(x^2+5x+12=0\)

( tự giải nha )

2 ) Ta có HPT :

\(\left\{{}\begin{matrix}x+y=7\\x^3+y^3=133\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=49\\\left(x+y\right)\left(x^2-xy+y^2\right)=133\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=49\\x^2-xy+y^2=19\end{matrix}\right.\)

\(\Rightarrow3xy=30\Leftrightarrow xy=10\)

\(\left\{{}\begin{matrix}x+y=7\\xy=10\end{matrix}\right.\) => pt : \(x^2-7x+10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=5\\y=2\end{matrix}\right.\)

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a ) \(\sqrt{\frac{49}{9}-\frac{4}{3}.\sqrt{5}}=\sqrt{5-2.\sqrt{5}.\frac{2}{3}+\frac{4}{9}}=\sqrt{\left(\sqrt{5}-\frac{2}{3}\right)^2}=\sqrt{5}-\frac{2}{3}\)

b ) \(\sqrt{\frac{64}{9}-\frac{2}{3}.\sqrt{7}}=\sqrt{7-2.\sqrt{7}.\frac{1}{3}+\frac{1}{9}}=\sqrt{\left(\sqrt{7}-\frac{1}{3}\right)^2}=\sqrt{7}-\frac{1}{3}\)

c ) \(\sqrt{\frac{79}{36}+\frac{2}{3}\sqrt{7}}=\sqrt{\frac{72}{36}+2.2.\frac{\sqrt{7}}{6}+\frac{7}{36}}=\sqrt{\left(2+\frac{\sqrt{7}}{6}\right)^2}=2+\frac{\sqrt{7}}{6}=\frac{12+\sqrt{7}}{6}\)

d ) \(\sqrt{\frac{45}{4}-\sqrt{11}}=\sqrt{\frac{44}{4}-\sqrt{11}+\frac{1}{4}}=\sqrt{11-\sqrt{11}+\frac{1}{4}}=\sqrt{\left(\sqrt{11}-\frac{1}{2}\right)^2}=\sqrt{11}-\frac{1}{2}\)

a ) Để ý thấy \(16\sqrt{3}=2.2\sqrt{3}.4=2.\sqrt{12}.4\) , như vậy , ta sẽ tách :

\(28=12+16\) \(\Rightarrow\sqrt{\sqrt{28+16\sqrt{3}}=\sqrt{\sqrt{12+16+16\sqrt{3}}}}=\sqrt{\sqrt{\left(\sqrt{12}+4\right)^2}}=\sqrt{\sqrt{12}+4}\)

\(=\sqrt{3+2.\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

b ) \(4\sqrt{3}=2.2\sqrt{3}\), tách \(7=4+3\)

c ) \(24\sqrt{5}=2.\sqrt{5}.12=2.\sqrt{5}.2.6=2.\sqrt{20}.6\) , tách : \(56=20+36\)

d ) \(2\sqrt{11}=2.11.1\) , tách : \(12=11+1\)

e ) \(4\sqrt{2}=2.\sqrt{2}.2.1=2.\sqrt{8}.1\) , tách : \(9=8+1\)

Ta có : \(\left\{{}\begin{matrix}x+3y-xy=3\\x^2+y^2+xy=3\end{matrix}\right.\)

Xét PT đầu : \(x+3y-xy=3\)

Chuyển vế được : \(\left(x-3\right)\left(1-y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

TH 1 : x = 3 \(;x^2+y^2+xy=3\)

\(\Rightarrow9+y^2+3y=3\) \(\Leftrightarrow y^2+3y+6=0\) ( ***** ) vì :

\(y^2+3y+6=y^2+3y+\frac{9}{4}+\frac{15}{4}=\left(y+\frac{3}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall y\)

TH 2 : \(y=1;x^2+y^2+xy=3\)

\(\Rightarrow x^2+x+1=3\) \(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\) ( t/m )

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ĐK : tự tìm nha

\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)

\(\Leftrightarrow\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}=x-\frac{4}{x}\)

Đặt \(\sqrt{x-\frac{1}{x}}=a;\sqrt{2x-\frac{5}{x}}=b\left(a,b\ge0\right)\Rightarrow x-\frac{1}{x}=a^2;2x-\frac{5}{x}=b^2\)

\(\Rightarrow b^2-a^2=x-\frac{4}{x}\)

Ta có : \(a-b=b^2-a^2\)

\(\Leftrightarrow a-b+a^2-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{x}-x=0\\a+b=-1\left(VL\right)\end{matrix}\right.\)

\(\Leftrightarrow4-x^2=0\Leftrightarrow x=2\) ( t/m )

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P/s : Làm siêu tắt

Ta có :

\(\left(1+\frac{a}{b}\right)^5+\left(1+\frac{b}{a}\right)^5\ge\left(1+\frac{a}{b}\right)\left(1+\frac{b}{a}\right)\left[\left(1+\frac{a}{b}\right)^3+\left(1+\frac{b}{a}\right)^3\right]\ge\left(1+\frac{a}{b}\right)^2\left(1+\frac{b}{a}\right)^2\left(2+\frac{a}{b}+\frac{b}{a}\right)=\frac{\left(a+b\right)^2.\left(a+b\right)^2}{a^2b^2}.\left(2+\frac{a}{b}+\frac{b}{a}\right)\ge\frac{4ab.4ab}{a^2b^2}.\left(2+2\right)=16.4=64\)

( AD BĐT phụ \(x^5+y^5\ge xy\left(x^3+y^3\right);x^3+y^3\ge xy\left(x+y\right)\) và BĐT Cô - si )

Dấu " = " xảy ra \(\Leftrightarrow a=b;a,b>0\)

Theo GT : \(xy+yz+xz=3xyz\Rightarrow\frac{xy+yz+xz}{xyz}=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

\(\frac{x^3}{x^2+z}=\frac{x\left(x^2+z\right)}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)

Tương tự , ta có : \(\frac{y^3}{y^2+x}\ge y-\frac{\sqrt{x}}{2}\) ; \(\frac{z^3}{z^2+y}\ge z-\frac{\sqrt{y}}{2}\)

\(\Rightarrow\frac{x^3}{x^2+z}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\)

Vì x ; y ; z dương , áp dụng BĐT Cô - si , ta có :

\(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};z+1\ge2\sqrt{z}\)

\(\Rightarrow x+y+z+3\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

=> \(\frac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\) => BĐT được c/m

Tiếp tục AD BĐT Cô - si , ta có :

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)

\(\Rightarrow x+y+z\ge\frac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9}{3}=3\) => BĐT được c/m

Có : \(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{2}\ge\frac{3.3-3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy ...

\(B=2a+3b+\frac{6}{a}+\frac{10}{b}=\frac{6}{a}+\frac{6}{b}+\left(2a+2b\right)+b+\frac{4}{b}=6\left(\frac{1}{a}+\frac{1}{b}\right)+2\left(a+b\right)+b+\frac{4}{b}\)

Áp dụng BĐT Cô - si cho các cặp số thực dương , ta có :

\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) => BĐT được c/m

\(b+\frac{4}{b}\ge4\)

\(\Rightarrow B\ge6.\frac{4}{4}+2.4+4=18\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=2\)

ĐK : \(x;y\ne0\)

Ta có : \(\left\{{}\begin{matrix}\frac{5}{x}+\frac{3}{y}=1\\\frac{2}{x}+\frac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{x}+\frac{3}{y}=1\\\frac{6}{x}+\frac{3}{y}=-3\end{matrix}\right.\)

\(\Rightarrow\frac{1}{x}=-3-1=-4\Leftrightarrow x=-\frac{1}{4}\)

Thay vào , ta có : \(y=\frac{1}{7}\)

Vậy ...