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Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt : \(P=\frac{48^2\cdot8^5\cdot100^9}{12^2\cdot2^{15}\cdot4^2}\)
\(=\frac{\left(2^4\cdot3\right)^2\cdot\left(2^3\right)^5\cdot\left(2^2\cdot5^2\right)^9}{\left(2^2\cdot3\right)^2\cdot2^{15}\cdot\left(2^2\right)^2}\)
\(=\frac{2^8\cdot3^2\cdot2^{15}\cdot2^{18}\cdot5^{18}}{2^4\cdot3^2\cdot2^{15}\cdot2^4}\)
\(=\frac{2^{41}\cdot3^2\cdot5^{18}}{2^{23}\cdot3^2}=2^{18}\cdot5^{18}=\left(2\cdot5\right)^{18}=10^{18}\)
Vậy : \(P=10^{18}\)
![](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)
Đặt A = 1+2+2^2+2^3+....+2^60
2A = 2+2^2+2^3+2^4+.....+2^61
2A-A= ( 2+2^2+2^3+....+2^61)-(1+2+2^2+.....+2^60)
A = 2^61-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x\right|=7\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Vậy \(x\in\left\{\pm7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đầu bài là như này đúng không hả bạn
\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
Ta có :\(\frac{1}{2}+\frac{2}{3}:\left(x-1\right)\)\(=\frac{3}{4}\)
\(\frac{2}{3}:\left(x-1\right)\)\(=\frac{1}{4}\)
\(\left(x-1\right)\)\(=\frac{8}{3}\)
\(x=\frac{11}{3}\)
= 5 - 1 + 1/4 : 2
= 4 + 1/8
= 33/8
K CHO MÌNH NHA