Chọn đáp án D

Giải

\(\hept{\begin{cases}x^2-xy+y-7=0\left(1\right)\\x^2+xy-2y=4\left(x-1\right)\left(2\right)\end{cases}}\)

Ta có: \(\left(2\right)\Leftrightarrow x^2-4x+4+xy-2y=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-2+y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\y=2-x\end{cases}}}\)

Thay x=2 vào phương trình (1) được:

\(4-2y+y-7=0\Leftrightarrow y=-3\)

Thay y=2-x vào phương trình (1) ta được:

\(x^2-x\left(2-x\right)+2-x-7=0\)

\(\Leftrightarrow2x^2-3x-5=0\)

\(\Leftrightarrow2x^2-5x+2x-5=0\)

\(\Leftrightarrow x\left(2x-5\right)+\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-1\end{cases}}}\)

Với x=-1 => y=3

Với x=\(\frac{5}{2}\)=> \(y=\frac{-1}{2}\)

Vậy tập nghiệm của hệ phương trình là (x;y)={(2;-3);(-1;3);\(\left(\frac{5}{2};\frac{-1}{2}\right)\)}

đang luyện Bu-nhi-a-cốp-ski :)) 

lời giải

Áp dụng BĐT Bu-nhi-a-cốp-ski,ta có :

\(\left(a^2+1\right)\left[1+\left(b+c\right)^2\right]\ge\left(a+b+c\right)^2\)

\(\Rightarrow\frac{3}{4}\left(a^2+1\right)\left[1+\left(b+c\right)^2\right]\ge\frac{3\left(a+b+c\right)^2}{4}\)

Cần chứng minh : \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{3}{4}\left(a^2+1\right)\left[1+\left(b+c\right)^2\right]\)

\(\Leftrightarrow4\left(b^2c^2+b^2+c^2+1\right)\ge3\left(b^2+c^2+2bc+1\right)\)

\(\Leftrightarrow\left(2bc-1\right)^2+\left(b-c\right)^2\ge0\)

Dấu"=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{a}{1}=\frac{1}{b+c}\\b=c\\2bc=1\end{cases}}\Leftrightarrow a=b=c=\pm\frac{1}{\sqrt{2}}\)

Ta có : \(\frac{1}{\sqrt{n}\left(n+1\right)}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Áp dụng vào bài toán, ta có :

\(VT< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

\(=2\left(1-\frac{1}{\sqrt{n+1}}\right)< 2\)

đề có thiếu không bạn ơi

Sửa đề thành : \(5x^2-x.4\left(m+1\right)+2=0\)

\(< =>5x^2-x\left(4m+4\right)+2=0\)

Ta có \(\Delta=\left[-\left(4m+4\right)\right]^2-4.5.1=4m^2+2.4m.4+4^2-20\)

\(=4m^2+32m-4=4\left(m^2+8m-1\right)\)

đến đây thì xin quỳ :))

Ta có: 

\(15\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=10\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+2014\)

\(\le10\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2014\)

=> \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le\frac{2014}{5}\)

\(P=\frac{1}{\sqrt{5x^2+2xy+2yz}}+\frac{1}{\sqrt{5y^2+2yz+2zx}}+\frac{1}{\sqrt{5z^2+2zx+2xy}}\)

=> \(P\sqrt{\frac{2014}{135}}=\frac{1}{\sqrt{5x^2+2xy+2yz}.\sqrt{\frac{135}{2014}}}\)

\(+\frac{1}{\sqrt{5y^2+2yz+2zx}\sqrt{\frac{135}{2014}}}+\frac{1}{\sqrt{\frac{135}{2014}}\sqrt{5z^2+2zx+2xy}}\)

\(\le\frac{1}{2}\left(\frac{1}{5x^2+2xy+2yz}+\frac{2014}{135}+\frac{1}{5y^2+2yz+2zx}+\frac{2024}{135}+\frac{1}{5z^2+2yz+2zx}+\frac{2014}{135}\right)\)

\(\le\frac{1}{2}\left[\frac{1}{81}\left(\frac{5}{x^2}+\frac{2}{xy}+\frac{2}{yz}\right)+\frac{1}{81}\left(\frac{5}{y^2}+\frac{2}{yz}+\frac{2}{zx}\right)+\frac{1}{81}\left(\frac{5}{z^2}+\frac{2}{zx}+\frac{2}{xy}\right)+\frac{2014}{45}\right]\)

\(=\frac{5}{162}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\frac{2}{81}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{1007}{45}\)

\(\le\frac{5}{162}.\frac{2014}{5}+\frac{2}{81}.\frac{2014}{5}+\frac{1007}{45}=\frac{2014}{45}\)

=> \(P\le\frac{2014}{45}:\sqrt{\frac{2014}{135}}=3\sqrt{\frac{2014}{135}}\)

Dấu "=" xảy ra <=> x = y = z = \(\sqrt{\frac{15}{2014}}\)

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