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\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(M=\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(\Leftrightarrow M=\left(\frac{2x}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1}{x+1}\)
\(\Leftrightarrow M=\frac{2x-x-1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\cdot\frac{x+1}{x+\sqrt{x}+1}\)
\(\Leftrightarrow M=\frac{x-1}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)
\(\Leftrightarrow M=\frac{x-1}{x\sqrt{x}+1}\)
\(\hept{\begin{cases}\frac{4x}{x+1}+\frac{x}{y}=4\left(1\right)\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\left(2\right)\end{cases}}\)
ĐK: \(x\ne-1;0\) và \(y\ne0\)
(2) <=> \(\frac{2x}{x+1}+4=\frac{3x}{y}\)<=> \(2\frac{x}{x+1}-3\frac{x}{y}=-4\)
Đặt: \(\frac{x}{x+1}=u;\frac{x}{y}=v\)
Ta có hệ phương trình:
\(\hept{\begin{cases}4u+v=4\\2u-3v=-4\end{cases}}\)<=> \(\hept{\begin{cases}u=\frac{4}{7}\\v=\frac{12}{7}\end{cases}}\)
Khi đó ta có: \(\frac{x}{x+1}=\frac{4}{7};\frac{x}{y}=\frac{12}{7}\)
<=> \(x=\frac{4}{3};y=\frac{7}{9}\)
\(\hept{\begin{cases}\frac{4x}{x+1}+\frac{x}{y}=4\left(1\right)\\\frac{2}{x+1}+\frac{4}{x}=\frac{3}{y}\end{cases}\left(x\ne-1;y\ne0\right)}\)
Chia (1) cho x ta có: \(\hept{\begin{cases}\frac{4}{x+1}+\frac{1}{y}=\frac{4}{x}\left(3\right)\\\frac{2}{x+1}-\frac{3}{y}=\frac{-4}{x}\left(4\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{6}{x+1}-\frac{2}{y}=0\left(\left(3\right)+\left(4\right)\right)\\\frac{2}{x+1}-\frac{3}{y}=\frac{-4}{x}\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x+1=3y\\\frac{2}{3y}-\frac{3}{y}=\frac{-4}{x}\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3y-1\\\frac{-7}{3y}=\frac{-4}{x}\end{cases}\Leftrightarrow}\hept{\begin{cases}x-3y=1\\7x-12=0\end{cases}\Leftrightarrow}\hept{\begin{cases}7x-21y=-7\\7x-12y=0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}9y=7\\x=3y-1\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{7}{9}\\x=\frac{4}{3}\end{cases}}}\)
Vậy \(\left(x;y\right)=\left(\frac{4}{3};\frac{7}{9}\right)\)
Ta có:
\(P=\frac{x^3}{1+y}+\frac{y^3}{1+x}=\frac{x^3}{xy+y}+\frac{y^3}{xy+x}=\frac{x^4}{x+1}+\frac{y^4}{y+1}\)
\(=\frac{x^4-1}{x+1}+\frac{y^4-1}{y+1}+\frac{1}{x+1}+\frac{1}{y+1}\)
\(=\left(x^2+1\right)\left(x-1\right)+\left(y^2+1\right)\left(y-1\right)+\frac{1}{x+1}+\frac{1}{y+1}\)
\(=\left(x^2+1\right)\left(x-1\right)+\left(\frac{1}{x^2}+1\right)\left(\frac{1}{x}-1\right)+\frac{1}{x+1}+\frac{1}{\frac{1}{x}+1}\)
\(=\left(x^2+1\right)\left(x-1\right)-\frac{\left(x^2+1\right)\left(x-1\right)}{x^3}+\frac{1}{x+1}+\frac{x}{x+1}\)
\(=\frac{\left(x^2+1\right)\left(x-1\right)\left(x^3-1\right)}{x^3}+\frac{x+1}{x+1}\)
\(\ge\frac{2x\left(x-1\right)^2\left(x^2+x+1\right)}{x^3}+1\)
\(=\frac{2\left(x-1\right)^2.3\sqrt[3]{x^2.x.1}}{x^2}+1=\frac{6\left(x-1\right)^2}{x}+1\)
\(=6\frac{x^2-2x+1}{x}+1=6.\frac{x^2+1}{x}-11\ge12-11=1\)
Dấu "=" xảy ra <=> x = y = 1
Vậy min P = 1 tại x = y = 1.
Đề thi tuyển sinh chuyên Khoa học tự nhiên-Đại Học quốc gia Hà Nội năm học 2017-2018
ta có: \(ab+bc+ca+abc=2\)
\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)=\left(1+a\right)+\left(1+b\right)+\left(1+c\right)\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)}+\frac{1}{\left(1+b\right)\left(1+c\right)}+\frac{1}{\left(1+c\right)\left(1+a\right)}=1\)
đặt \(x=\frac{1}{1+a};y=\frac{1}{1+b};z=\frac{1}{1+c}\Rightarrow xy+yz+xz=1\)
ta có \(P=\frac{a+1}{\left(a+1\right)^2+1}+\frac{b+1}{\left(b+1\right)^2+1}+\frac{c+1}{\left(c+1\right)^2+1}\)
\(=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}+\frac{\frac{1}{y}}{\frac{1}{y^2}+1}+\frac{\frac{1}{z}}{\frac{1}{z^2}+1}=\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)
\(=\frac{x}{\left(x+y\right)\left(y+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+y\right)\left(z+x\right)}\)
\(=\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{2}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
mà \(9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+z+zx\right)\)
\(\Leftrightarrow x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\ge6xyz\)(đúng vì theo BĐT Cosi)
\(\Rightarrow P\le\frac{2}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)}=\frac{9}{4\left(x+y+z\right)}\le\frac{9}{4\sqrt{3}}=\frac{3\sqrt{3}}{4}\)
(vì \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3\))
Vậy \(P_{max}=\frac{3\sqrt{3}}{4}\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow a=b=c=\sqrt{3}-1\)
gọi G là giao của tia đối tia CD với AM (ta giả sử cung AC < cung BC)
ý c: từ b suy ra tam giác CDE đồng dạng CFD
=> \(\widehat{ECD}=\widehat{FCD}\)
ta có: \(\widehat{ECD}+\widehat{GCE}=180^o\)
\(\widehat{FCD}+\widehat{GCF}=180^o\)
\(\widehat{GCE}=\widehat{GCF}\)suy ra đccm
ý d: CM IK//AB
Ta có: \(\widehat{FDB}=\widehat{FCB}\)(BDCF nôi tiếp đường tròn)
\(\hept{\begin{cases}\widehat{FCB}+\widehat{FBC}=90^o\\\widehat{DCA}+\widehat{CAD}=90^o\end{cases}}\)
mà \(\widehat{CAD}=\widehat{FBC}\)(cùng chắn cung BC)
\(\Rightarrow\widehat{FCB}=\widehat{DCA}\Rightarrow\widehat{FDB}=\widehat{DCA}\)(1)
Tương tự:
\(\hept{\begin{cases}\widehat{ECA}+\widehat{EAC}=90^o\\\widehat{DCB}+\widehat{DBC}=90^o\end{cases}}\)
mà \(\widehat{EAC}=\widehat{DBC}\)(cùng chắn cung AC)
\(\Rightarrow\widehat{ECA}=\widehat{DCB}\). mà \(\widehat{ECA}=\widehat{EDA}\)(tứ giác ECDA nội tiếp nên 2 góc kia cùng chắn cung AE)
\(\Rightarrow\widehat{DCB}=\widehat{EDA}\)(2)
(1)+(2) => \(\widehat{ACD}+\widehat{BCD}=\widehat{FDB}+\widehat{EDA}\)
\(\Rightarrow\widehat{ICK}=\widehat{FDB}+\widehat{EDA}\)\(\Rightarrow\widehat{ICK}+\widehat{IDK}=\widehat{FDB}+\widehat{EDA}+\widehat{IDK}=180^o\)
suy ra tứ giác IDKC nội tiếp \(\Rightarrow\widehat{CKI}=\widehat{CDI}=\widehat{CAE}=\widehat{CBA}\)
mà góc CKI và góc CBA ở vị trí đồng vị suy ra IK//AB. ta đc đccm.