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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Bài 1
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\rightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
Bài 2
a) \(25^3\div5^2=\left(5^2\right)^3\div5^2=5^6\div5^2=5^4\)
b) \(\left(\frac{3}{7}\right)^{21}\div\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}\div\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}\div\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
c) \(3-\left(\frac{-6}{7}\right)^0+\left(\frac{1}{2}\right)^2\div2=3-1+\frac{1}{4}\times\frac{1}{2}=2+\frac{1}{8}=\frac{17}{8}\)
Bài 3
a) \(9\times3^3\times\frac{1}{81}\times3^2=3^2\times3^3\times\frac{1}{3^4}\times3^2=3^3\)
b) \(4\times2^5\div\left(2^3\times\frac{1}{16}\right)=2^2\times2^5\div\left(2^3\times\frac{1}{2^4}\right)=2^7\div\frac{1}{2}=2^6\)
c) \(3^2\times2^5\times\left(\frac{2}{3}\right)^2=3^2\times2^5\times\frac{2^2}{3^2}=3^2\times\frac{2^7}{3^2}=2^7\)
d) \(\left(\frac{1}{3}\right)^2\times\frac{1}{3}\times9^2=\left(\frac{1}{3}\right)^3\times3^4=\frac{1}{3^3}\times3^4=3^1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A) \(\left(\frac{1}{3}\right)^{^2}.\frac{1}{3}.9^2=3=3^1\)(viết dưới dạng lũy thừa)
B)\(8< 2^n< 2.16\)
\(2^3< 2^n< 2.2^4\)
\(2^3< 2^n< 2^5\)
\(\Rightarrow3< n< 5\)
mà n là số tự nhiên => n = 4
C) |-x| = 1 => |x| = 1 => x = -1 hoặc x = 1.
|2x| = 6.7 + (-3,3) - 0.4 = 42 - 3,3 - 0 = 42 - 3,3 = 38,7
=> 2x = 38,7 hoặc 2x = -38,7
=> x = 19,35 hoặc x = -19,35
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1 a)x+1/3=1/4
x= -1/12
b) x-3=1
x=4
c)3x-1=-4
3x=-3
x=-1
Câu 2:
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(3^4\cdot3^5:\dfrac{1}{27}==3^9\cdot3^3=3^{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(9.3^3.\frac{1}{81}.3^2=3^2.3^3.\frac{1}{3^4}.3^2=3^7.\frac{1}{3^4}=3^3\)
b) \(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{16}=2^7:2^3.16=2^4.2^4=2^8\)
c) \(3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=2^5.2^2=2^7\)
d) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^3.\left(3^2\right)^2=\frac{1^3}{3^3}.3^4=1^3.3=3^1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^{\left(-4\right)}.3^2=3^{2+3-4+2}=3^3.\)
\(b,4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.2^{-4}\right)=2^{2+5}:2^{3-4}=2^7:2^{-1}=2^{7-\left(-1\right)}=2^8.\)
\(c,3^2.2^5.\left(\frac{2}{3}\right)^2=3^2.2^5.\frac{2^2}{3^2}=\left(\frac{3^2}{3^2}\right).\left(2^5.2^2\right)=1.2^{5+2}=2^7\)
\(d,\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^2.\frac{1}{3}.\left(3^2\right)^2=\left(\frac{1}{3}\right)^{2+1}.3^4=\left(\frac{1}{3}\right)^3.\left(\frac{1}{3}\right)^{-4}=\left(\frac{1}{3}\right)^{3-4}=\left(\frac{1}{3}\right)^{-1}=3^1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) bài 1
để \(x\in Z\)thì \(3x-1⋮x-1\)
mà \(x-1⋮x-1\)
\(\Rightarrow3\left(x-1\right)⋮x-1\)
\(\Rightarrow\left(3x-1\right)-\left[3x-3\right]⋮x-1\)
\(\Rightarrow2⋮x-1\)
\(\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
ta có bảng
x-1 | 1 | -1 | 2 | -2 |
x | 2 | 0 | 3 | -1 |
vậy \(x\in\left\{2;0;3;-1\right\}\)