1--Tìm GTNN:
a) \(A=\left|2x-2\right|+\left|2x-2017\right|\)
b) \(B=\left|x-2\right|+\left|x-8\right|\)
2--Tìm x:
\(\left|x+3\right|+\left|x+7\right|=4x\)
✰✰✰✰
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NT
Nguyễn thành Đạt
CTVHS
8 tháng 9 2023
Bạn xem lại đề nhé.
a) \(A=x^2+5y^2+2xy-4x-8y+2015\)
\(A=x^2-4x+4-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2\right)^2-2y\left(x-2\right)+y^2+2011+4y^2\)
\(A=\left(x-2-y\right)^2+4y^2+2011\)
Vì \(\left(x-y-2\right)^2\ge0;4y^2\ge0\)
\(\Rightarrow A_{min}=2011\)
Dấu bằng xảy ra : \(\Leftrightarrow\left\{{}\begin{matrix}x-y-2=0\\4y^2=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
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`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`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`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
2 tháng 1 2018
\(A=\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+2002\)
\(\Leftrightarrow A=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2002\)
Đặt \(x^2-9x+14=y\)
\(\Rightarrow A=\left(y-6\right)\left(y+6\right)+2002\)
\(\Leftrightarrow A=y^2-36+2002\)
\(\Leftrightarrow A=y^2+1966\ge1966\)
Dấu "=" xảy ra khi
\(x^2-9x+14=0\)
\(\Leftrightarrow x=2,7\)
CW
18 tháng 7 2016
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
24 tháng 5 2018
a) \(\left|x+2\right|+\left|x-3\right|=7\)
Lập bảng xét dấu:
| x | -2 3 |
| x + 2 | - 0 + \(|\) + |
| x - 3 | - \(|\) - 0 + |
* Nếu \(x< -2\) thì pttt:
\(-x-2-x+3=7\)
\(\Leftrightarrow-2x+1=7\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\left(tm\right)\)
* Nếu \(-2\le x\le3\) thì pttt:
\(x+2-x+3=7\)
\(\Leftrightarrow5=7\) ( vô lí )
* Nếu \(x>3\) thì pttt:
\(x+2+x-3=7\)
\(\Leftrightarrow2x-1=7\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(tm\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-3;4\right\}\)
24 tháng 5 2018
b) \(\left|x+2\right|-6x=1\)
* Nếu \(x+2>0\Leftrightarrow x>2\) thì pttt:
\(x+2-6x=1\)
\(\Leftrightarrow-6x=-1\)
\(\Leftrightarrow x=1\left(ktm\right)\)
* Nếu \(x+2< 0\Leftrightarrow x< 2\) thì pttt:
\(-x-2-6x=1\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=-\dfrac{3}{7}\left(tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{-3}{7}\right\}\)
a, Ta có :
\(A=\left|2x-2\right|+\left|2x-2017\right|=\left|2x-2\right|+\left|2017-2x\right|\ge\left|2x-2+2017-2x\right|=2015\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2x-2\right)\left(2017-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-2\ge0\\2017-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-2\le0\\2017-2x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x\ge2\\2017\ge2x\end{matrix}\right.\\\left\{{}\begin{matrix}2x\le2\\2017\le2x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge1\\\dfrac{2017}{2}\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le1\\\dfrac{2017}{2}\le x\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1\le x\le\dfrac{2017}{2}\\x\in\varnothing\end{matrix}\right.\)
Vậy ...
b, Tương tự
c, \(\left|x+3\right|+\left|x+7\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+3\right|\ge0\\\left|x+7\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x-3\right|+\left|x+7\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x\ge0\)
Với \(x\ge0\) ta có :
+) \(\left|x+3\right|=x+3\)
\(\left|x+7\right|=x+7\)
\(\Leftrightarrow\left|x+3\right|+\left|x+7\right|=x+3+x+7=4x\)
\(\Leftrightarrow2x+10=4x\)
\(\Leftrightarrow10=2x\)
\(\Leftrightarrow x=5\)
Vậy ..
B1b)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(B=\left|x-2\right|+\left|x-8\right|\)
\(B\ge\left|x-2\right|+\left|8-x\right|=6\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(8-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\le0\\8-x\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\ge0\\8-x\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le2\\x\ge8\end{matrix}\right.\left(C\right)}\\\left\{{}\begin{matrix}x\ge2\\x\le8\end{matrix}\right.\left(L\right)}\end{matrix}\right.\)
TH1: chọn, TH2: loại.
Vậy \(MIN_B=6\Leftrightarrow2\le x\le8\)