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Bạn chỉ cần để ý điều này thôi: \(\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=x^2-2+\frac{1}{x^2}\)
Do đó giả thiết viết lại thành:
\(\left(a^2-2+\frac{1}{a^2}\right)+\left(b^2-2+\frac{1}{b^2}\right)+\left(c^2-2+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-\frac{1}{a}=0\\b-\frac{1}{b}=0\\c-\frac{1}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{a}\\b=\frac{1}{b}\\c=\frac{1}{c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b^2=1\\c^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2\right)^{1010}=1^{1010}\\\left(b^2\right)^{1010}=1^{1010}\\\left(c^2\right)^{1010}=1^{1010}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{2020}=1\\b^{2020}=1\\c^{2010}=1\end{matrix}\right.\) \(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}=3\)
Câu 1:
a: Để M là số nguyên thì \(2x^3-6x^2+x-3-5⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)
b: Để N là số nguyên thì \(3x^2+2x-3x-2+5⋮3x+2\)
\(\Leftrightarrow3x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-\dfrac{1}{3};-1;1;-\dfrac{7}{3}\right\}\)
a ) Để \(\dfrac{3}{-x^2+2x+4}\) đạt GTlN thì :
\(-x^2+2x+4\) phải đạt GTNN ( chắc ai cũng biết )
Ta có :
\(-x^2+2x+4\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2-5\)
Tới đây chắc bạn hỉu rồi nhỉ ?
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
Bài làm:
Ta có:
(a-b)2+(b-c)2+(c-a)2=(a+b-2c)2+(b+c-2a)2+(c+a-2b)2
<=> a2-2ab+b2+b2-2bc+c2+c2-2ca+a2=6a2+6b2+6c2-6(ab+bc+ca)
<=> \(4a^2+4b^2+4c^2-4ab-4bc-4ca=0\)
<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
<=> \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=\left(a+b-2c\right)^2+\left(b+c-2a\right)^2+\left(c+a-2b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4ab-4bc-4ca=\left(a+b\right)^2\)
\(+\left(b+c\right)^2+\left(c+a\right)^2-4\left(b+c\right)a+4a^2-4\left(c+a\right)b+4b^2-4\left(a+b\right)c+4c^2\)
\(\Leftrightarrow-4ab-4bc-4ca=-4\left(b+c\right)a+4a^2-4\left(c+a\right)b+4b^2-4\left(a+b\right)c+4c^2\)
\(\Leftrightarrow ab-\left(a+b\right)c+c^2+bc-\left(b+c\right)a+a^2+ca-\left(c+a\right)b+b^2=0\)
\(\Leftrightarrow ab-ac-bc+c^2+bc-ba-ca+a^2+ca-cb-ab+b^2=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
Ta có:
\(Q=\dfrac{2a-b}{3a-b}+\dfrac{5b-a}{3a+b}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)}{\left(3a-b\right)\left(3a+b\right)}+\dfrac{\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a-b\right)\left(3a+b\right)}\)
\(Q=\dfrac{3a^2+15ab-6b^2}{9a^2-b^2}\)
Ta lại có:
\(6a^2-15ab+5b^2=0\)
\(\Rightarrow3a^2+15ab-6b^2=9a^2-b^2\left(1\right)\)
Thay (1) vào Q
=> Q = 1
Bài 1: Ta có: \(B=\dfrac{4+2\left|4-2x\right|}{5}\)
Do \(\left|4-2x\right|\ge0\left(\forall x\right)\Rightarrow2\left|4-2x\right|\ge0\left(\forall x\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left|4-2x\right|=0\Leftrightarrow x=2\)
\(\Rightarrow MinB=\dfrac{4+2.0}{5}=\dfrac{4}{5}\)
Vậy GTNN của \(B=\dfrac{4}{5}\Leftrightarrow x=2\)
Bài 2:a, \(A=\dfrac{12}{3+\left|5x+1\right|+\left|2y-1\right|}\)
Do \(\left|5x+1\right|\ge0\left(\forall x\right);\left|2y-1\right|\ge0\left(\forall y\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{5};y=\dfrac{1}{2}\)
\(\Rightarrow\left|5x+1\right|+\left|2y-1\right|\ge0\left(\forall x;y\right)\)
\(\Rightarrow3+\left|5x+1\right|+\left|2y-1\right|\ge3\left(\forall x;y\right)\)
\(\Rightarrow\dfrac{1}{3+\left|5x+1\right|+\left|2y-1\right|}\le\dfrac{1}{3}\left(\forall x;y\right)\)
\(\Rightarrow A=\dfrac{12}{3+\left|5x+1\right|+\left|2y-1\right|}\le4\left(\forall x;y\right)\)
Vậy Max A = 4 \(\Leftrightarrow x=-\dfrac{1}{5};y=\dfrac{1}{2}\)
b, \(B=\dfrac{5}{\left(4x^2+4x+1\right)+\left(y^2+2y+1\right)+1}=\dfrac{5}{\left(2x+1\right)^2+\left(y+1\right)^2+1}\)Bn tự cm: \(\left(2x+1\right)^2+\left(y+1\right)^2+1\ge1\left(\forall x;y\right)\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2};y=-1\)
Vậy ta cx dễ dàng tìm được: Max\(B=\dfrac{5}{0+0+1}=5\) \(\Leftrightarrow x=-\dfrac{1}{2};y=-1\)
\(4=2a^2+\dfrac{1}{a^2}+\dfrac{b^2}{4}=\left(a^2+\dfrac{1}{a^2}-2\right)+\left(a^2+\dfrac{b^2}{4}+ab\right)-ab+2\)
\(\Rightarrow4=\left(a-\dfrac{1}{a}\right)^2+\left(a+\dfrac{b}{2}\right)^2-ab+2\)
\(\Rightarrow ab=\left(a-\dfrac{1}{a}\right)^2+\left(a+\dfrac{b}{2}\right)^2-2\ge-2\)
\(M_{min}=-2\) khi \(\left\{{}\begin{matrix}a-\dfrac{1}{a}=0\\a+\dfrac{b}{2}=0\end{matrix}\right.\) \(\Rightarrow\left(a;b\right)=\left(1;-2\right);\left(-1;2\right)\)