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a, TK:
(x lẻ do \(2y^2-8y+3=2\left(y^2-4y\right)+3=x^2\) lẻ)
\(b,\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+4y+4\right)=9\\ \Leftrightarrow\left(x-2\right)^2+\left(y+2\right)^2=9\)
Vậy pt vô nghiệm do 9 ko phải tổng 2 số chính phương
a) ĐKXĐ: \(x\notin\left\{0;2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
Suy ra: \(x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-1}
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
=>x+2013=0
hay x=-2013
\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)
a) \(=-7\left(x^2-\frac{10}{7}x+\frac{2016}{7}\right)\)
\(=-7\left(x^2-2.\frac{5}{7}x+\frac{25}{49}+\frac{14087}{49}\right)\)
\(=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\)
ta có
\(\left(x-\frac{5}{7}\right)^2\ge0\)với mọi x
\(=>-7\left(x-\frac{5}{7}\right)^2\le0\)(nhân cả hai vế với -7)
\(=>-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)
trường hợp dấu "=" xảy ra khi và chỉ khi
\(\left(x-\frac{5}{7}\right)^2=0\)
\(=>x-\frac{5}{7}=0\)
\(=>x=\frac{5}{7}\)
vậy GTLN cảu biểu thức là \(-\frac{14087}{7}\) khi và chỉ khi x= \(\frac{5}{7}\)
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
a) \(-7x^2+10x-2016=-7\left(x^2-\frac{10x}{7}\right)-2016=-7\left(x^2-2.x.\frac{5}{7}+\frac{25}{49}\right)+\frac{25}{49}.7-2016=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)Vậy Max = \(-\frac{14087}{7}\Leftrightarrow x=\frac{5}{7}\)
b) \(\frac{x+5}{11}+\frac{x+2010}{6}\ge\frac{x-1}{2017}+\frac{x+6}{2010}\)
\(\Leftrightarrow\frac{x}{2011}+\frac{x}{6}+\frac{5}{2011}+335\ge\frac{x}{2017}+\frac{x}{2010}-\frac{1}{2017}+\frac{1}{335}\)
\(\Leftrightarrow x\left(\frac{1}{2011}+\frac{1}{6}-\frac{1}{2017}-\frac{1}{2010}\right)\ge\frac{1}{335}-\frac{1}{2017}-\frac{5}{2011}-335\)
\(\Leftrightarrow\frac{677389259}{4076467935}x\ge\frac{-455205582048}{1358822645}\) \(\Leftrightarrow x\ge-2016\)
Câu b) còn cách khác nữa bạn nhé. Mình làm cách này "xù" quá ^^
\(\frac{2-x}{2016}-1=\frac{1-x}{2017}+\frac{x}{2018}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-2018x}{4070306}+\frac{2017x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-2018x+2017x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1=\frac{1-x}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}-1+1=\frac{1-x}{4070306}+1\)
\(\Rightarrow\frac{2-x}{2016}=\frac{1-x+4070306}{4070306}\)
\(\Rightarrow\frac{2-x}{2016}=\frac{4070307-x}{4070306}\)
\(\Rightarrow4070306.\left(2-x\right)=2016.\left(4070307-x\right)\)
\(\Rightarrow8140612-4070306x=8205738912-2016x\)
\(\Rightarrow-4070306x+2016x=8205738912-8140612\)
\(\Rightarrow-4068290x=8197598300\)
\(\Rightarrow x=4,95\)
Vậy x=4,95
Chúc bn học tốt
Bạn kiểm tra lại đề nhé!
Nếu viết theo thứ tự trên thì 2 phân số cuối là: \(\frac{x-1}{2012}\)và \(\frac{x}{2013}\)