Tìm max hoặc min:
a) ( 2-x) (x+4)
b) 4x2-6x+9
c) 3x2+15x-2
d) -2x(x-1)+10
e) 2x(3-x)+5
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3 tháng 10 2021
\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)
28 tháng 7 2021
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
28 tháng 7 2021
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
9 tháng 9 2021
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
9 tháng 9 2021
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
29 tháng 10 2021
Bài 1:
a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Bài 2:
a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)
\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)
N
1
29 tháng 8 2023
b: 4x^2-20x+25=(x-3)^2
=>(2x-5)^2=(x-3)^2
=>(2x-5)^2-(x-3)^2=0
=>(2x-5-x+3)(2x-5+x-3)=0
=>(3x-8)(x-2)=0
=>x=8/3 hoặc x=2
c: x+x^2-x^3-x^4=0
=>x(x+1)-x^3(x+1)=0
=>(x+1)(x-x^3)=0
=>(x^3-x)(x+1)=0
=>x(x-1)(x+1)^2=0
=>\(x\in\left\{0;1;-1\right\}\)
d: 2x^3+3x^2+2x+3=0
=>x^2(2x+3)+(2x+3)=0
=>(2x+3)(x^2+1)=0
=>2x+3=0
=>x=-3/2
a: =>x^2(5x-7)-3(5x-7)=0
=>(5x-7)(x^2-3)=0
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
20 tháng 6 2021
`a)x^2>4`
`<=>sqrtx^2>sqrt4`
`<=>|x|>2`
`<=>` \(\left[ \begin{array}{l}x>2\\x<-2\end{array} \right.\)
`b)x^2<9`
`<=>\sqrtx^2<sqrt9`
`<=>|x|<3`
`<=>-3<x<3`
`c)(x-1)^2>=4`
`<=>\sqrt{(x-1)^2}>=sqrt4`
`<=>|x-1|>=2`
`<=>` \(\left[ \begin{array}{l}x-1 \ge 2\\x-1 \le -2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x \ge 3\\x \le -1\end{array} \right.\)
`d)(1-2x)^2<=0,09`
`<=>\sqrt{(1-2x)^2}<=sqrt{0,09}`
`<=>|2x-1|<=0,3`
`<=>-0,3<=2x-1<=0,3`
`<=>0,7<=2x<=1,3`
`<=>0,35<=x<=0,65`
`e)x^2+6x-7>0`
`<=>x^2-x+7x-7>0`
`<=>x(x-1)+7(x-1)>0`
`<=>(x-1)(x+7)>0`
TH1:
\(\left[ \begin{array}{l}x-1>0\\x+7>0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x>1\\x>-7\end{array} \right.\)
`<=>x>1`
TH2"
\(\left[ \begin{array}{l}x-1<0\\x+7<0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x<1\\x<-7\end{array} \right.\)
`<=>x<-7`
`f)x^2-x<2`
`<=>x^2-x-2<0`
`<=>x^2-2x+x-2<0`
`<=>x(x-2)+x-2<0`
`<=>(x-2)(x+1)<0`
`<=>` \(\begin{cases}x-2<0\\x+1>0\\\end{cases}\)
`<=>` \(\begin{cases}x<2\\x>-1\\\end{cases}\)
`<=>-1<x<2`
20 tháng 6 2021
a) x2 > 4
<=> \(\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\)
b) \(x^2< 9\)
<=> \(-3< x< 3\)
c) \(\left(x-1\right)^2\ge4\)
<=> \(\left[{}\begin{matrix}x-1\ge2< =>x\ge3\\x-1\le-2< =>x\le-1\end{matrix}\right.\)
d) \(\left(1-2x\right)^2\le0,09\)
<=> \(-0,3\le1-2x\le0,3\)
<=> \(1,3\ge2x\ge0,7\)
<=> \(0,65\ge x\ge0,35\)
e) \(x^2+6x-7>0\)
<=> \(\left(x+7\right)\left(x-1\right)>0\)
<=> \(\left[{}\begin{matrix}x-1>0< =>x>1\\x+7< 0< =>x< -7\end{matrix}\right.\)
f) \(x^2-x< 2\)
<=> \(x^2-x-2< 0\)
<=> \(\left(x-2\right)\left(x+1\right)< 0\)
<=> \(\left\{{}\begin{matrix}x+1>0< =>x>-1\\x-2< 0< =>x< 2\end{matrix}\right.\)
<=> -1 < x < 2
g) \(4x^2-12x\le\dfrac{-135}{16}\)
<=> \(64x^2-192x+135\le0\)
<=> (8x - 15)(8x - 9) \(\le0\)
<=> \(\left\{{}\begin{matrix}8x-15\le0< =>x\le\dfrac{15}{8}\\8x-9\ge0< =>x\ge\dfrac{9}{8}\end{matrix}\right.\)
<=> \(\dfrac{9}{8}\le x\le\dfrac{15}{8}\)
25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5