Tìm GTNN của \(\left(2x-1\right)^2+\left(5x+1\right)^2\)
K
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Những câu hỏi liên quan
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
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`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
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`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
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`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
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`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 4 2021
\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)
\(A=\left|2021-x\right|+\left|2020-x\right|\)
\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)
\(A_{min}=1\) khi \(2020\le x\le2021\)
10 tháng 7 2023
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
10 tháng 7 2023
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
25 tháng 10 2021
\(A=\dfrac{1}{2}+\left|2x-1\right|\ge\dfrac{1}{2}\forall x\)
\(minA=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{\left|x\right|+2007}{2008}\ge\dfrac{0+2007}{2008}=\dfrac{2007}{2008}\)
\(minB=\dfrac{2007}{2008}\Leftrightarrow x=0\)
KN
27 tháng 10 2019
a) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-3\right|\ge\left|\left(x-1\right)+\left(3-x\right)\right|=2\)
Vậy\(A_{min}=2\Leftrightarrow\left(x-1\right)\left(3-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-1\ge0\\3-x\ge0\end{cases}}\Leftrightarrow1\le x\le3\)
\(TH1:\hept{\begin{cases}x-1\le0\\3-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge3\end{cases}}\left(L\right)\)
Vậy \(A_{min}=2\Leftrightarrow1\le x\le3\)
3 tháng 12 2023
a: \(f\left(x\right)=2x^2-7x+9\)
=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)
Đặt f'(x)=0
=>\(4x-7=0\)
=>\(x=\dfrac{7}{4}\)
\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)
\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)
\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)
Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)
nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)
b: \(f\left(x\right)=x^2+5x+3\)
=>\(f'\left(x\right)=2x+5\)
f'(x)=0
=>2x+5=0
=>2x=-5
=>\(x=-\dfrac{5}{2}\)
\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)
\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)
\(f\left(6\right)=6^2+5\cdot6+3=69\)
Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)
HP
7 tháng 3 2017
đặt |3x-5|= y ,ĐK : y >/ 0
F=y2-6y+10 đến đây đơn giản
ý sau khai triển tử của I rồi rút gọn được I=10x+40/x+41 >/ 2.20+41=81 (áp dụng bđt AM-GM)
Đặt
\(A=4x^2-4x+1+25x^2+10x+1\)
\(A=29x^2+6x+2\)
\(A=29\left(x^2-\frac{6}{29}x+\frac{9}{841}\right)+\frac{49}{29}\)
\(A=29\left(x-\frac{3}{9}\right)^2+\frac{49}{29}\ge\frac{49}{29}\)
\(A_{min}=\frac{49}{29}\) khi \(x=\frac{3}{29}\)
Chúc bạn học tốt !!!