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x2 - 4x + 4
= x2 - 2.2x + 22
= x2 - 22
mà x2 - 22 = 0
=> x2 - 4 = 0
=> x2 = 4
=> x2 = 22
=> x = 2
x2 - 4x + 4
= x2 - 2.2x + 22
= x2 - 22
mà x2 - 22 = 0
=> x2 - 4 = 0
=> x2 = 4
=> x2 = 22
=> x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
(1/2x - 3/4)3 =(-1): (2/3)6
1/8x - 27/64 = (-1) : 64/729
1/8x- 27/64 = -729/64
1/8x = - 729/64 + 27/64
1/8x = -351/32
x =-351/32 : 1/8
x =-351/4
![](https://rs.olm.vn/images/avt/0.png?1311)
(x-5)2=(1-3x)2
=> x-5 = 1- 3x
=> 4x = 6
=> x = \(\frac{3}{2}\)
( x - 5 )2 = ( 1 - 3x ) 2
x - 5 = 1 - 3x
x = 1 - 3x + 5
x = 6 - 3x
x + 3x = 6
( 3 + 1 )x = 6
4x=6
=> x = 6 : 4
=> x = 1,5
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(A=\left|x+2\right|+\left|9-x\right|\ge\left|X+2+9-x\right|=11\)
Dấu "=' xảy ra khi \(\left(x+2\right)\left(9-x\right)\ge0\Leftrightarrow-2\le x\le9\)
Vậy MinA = 11 khi -2 =< x =< 9
b, Vì \(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow B=\frac{3}{4}-\left(x-1\right)^2\le\frac{3}{4}\)
Dấu "=" xảy ra khi x = 1
Vậy MaxB = 3/4 khi x=1
Ta có :\(A=\left|x+2\right|+\left|9-x\right|\ge\left|x+2+9-x\right|=11\)
Vậy \(A_{min}=11\) khi \(2\le x\le9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(-x5):x2=-x3
Học tốt!!!!!!!!!!!!!!
.........................................
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Bài giải
\(\left(-x^5\right)\text{ : }x^2=\left(-x^3\right)\left(-x^2\right)\text{ : }x^2\)
\(=\left(-x^3\right)\left(x^2\right)\text{ : }x^2\)
\(=\left(-x^3\right)\cdot1\)
\(=-x^3\)
Mình cũng không chắc lắm !