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Bài làm
1. Rút gọn
\(\frac{2^5.7+2^6}{2^5.5^2-2^5.7}=\frac{2^5\left(7+2\right)}{2^5\left(25-7\right)}=\frac{9}{18}=\frac{1}{2}\)
2. So sánh
Ta có: \(\frac{2.7+6.21+9.28}{4.9+8.27+12.36}\)
\(=\frac{14+126+252}{36+216+432}=\frac{392}{684}=\frac{98}{171}\)
Mà ta thấy, mẫu số của phân số 98/171 lớn hơn mẫu số của phân sốc 49/84 ( 171 > 84 )
=> \(\frac{98}{171}< \frac{49}{84}\)
Vậy \(\frac{2.7+6.21+9.28}{4.9+8.27+12.36}< \frac{49}{84}\)
3. Tìm x
a) ( 2x + 14 ) : 22 - 3 = 1
( 2x + 14 ) : 4 - 3 = 1
( 2x + 14 ) : 4 = 1 + 4
( 2x + 14 ) : 4 = 5
2x + 14 = 5 x 4
2x + 14 = 20
2x = 6
x = 3
Vậy x = 3
b) 32x - 5 + 2 = 29
9x - 5 + 2 = 29
9x - 5 = 27
9x = 32
x = 32 : 9 = \(\frac{32}{9}\)
Vậy x = \(\frac{32}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(6⋮\left(x-1\right)\left(đkxđ:x\ne1;x\inℕ\right)\)
\(\Rightarrow x-1\in U\left(6\right)=\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{2;3;4;7\right\}\)
b) \(14⋮\left(2x+3\right)\left(đkxđ:x\ne-\dfrac{3}{2};x\inℕ\right)\)
\(\Rightarrow2x+3\in U\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow x\in\left\{-1;-\dfrac{1}{2};2;\dfrac{9}{2}\right\}\)
\(\Rightarrow x\in\left\{-2\right\}\)
\(a,6⋮\left(x-1\right)\\ \Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\ Ta.có:x-1=-6\Rightarrow x=-5\left(loại\right)\\ x-1=-3\Rightarrow x=-2\left(loại\right)\\ x-1=-2\Rightarrow x=-1\left(loại\right)\\ x-1=-1\Rightarrow x=0\left(nhận\right)\\ x-1=1\Rightarrow x=2\left(nhận\right)\\ x-1=2\Rightarrow x=3\left(nhận\right)\\ x-1=3\Rightarrow x=4\left(nhận\right)\\ x-1=6\Rightarrow x=7\left(nhận\right)\\ Vậy:x\in\left\{0;2;3;4;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3/4-2x| 1|=7/8
7.5-3|5-2x|=-4.5
|x +4/15|+ 1/2=3.5
Ai giúp mình với, Mình đang cần gấp
(Ai làm nhanh mình k nha)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
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5x + 2 chia hết cho 9 - 2x
=> 2(5x + 2) = 10x + 4 chia hết cho 9 - 2x
=> 10x + 4 + 5(9 - 2x) = 10x + 4 + 45 - 10x = 49 chia hết cho 9 - 2x
=> 9 - 2x thuộc Ư(49) = {1, 7, 49}
=> 2x thuộc {8, 2, -40}
=> x thuộc {1, 4, -20}
Vậy x thuộc {1, 4, -20}
Học tốt nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
( 2x - 1 )3 = 53
=> 2x - 1 = 5
=> 2x = 6
=> x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left|2x+1\right|=3\)
=> 2x + 1 = 3 hoặc 2x + 1 = -3
=> 2x = 3 - 1 hoặc 2x = -3 - 1
=> 2x = 2 hoặc 2x = -4
=> x = 1 hoặc x = -2
b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)
=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)
=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)
=> \(\frac{5}{2}x=\frac{16-9}{6}\)
=> \(\frac{5}{2}x=\frac{7}{6}\)
=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)
c, \(3\cdot5^{x-3}+1=16\)
=> 3 . 5x-3 = 16 - 1
=> 3 . 5x-3 = 15
=> 5x-3 = 15 : 3
=> 5x-3 = 5
=> x - 3 = 5 : 5
=> x - 3 = 1
=> x = 1 + 3 = 4
d, \((x-1)^2=25\)
=> \((x-1)^2=5^2\)
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)
=> 4 + |3x + 1| = -196
=> |3x + 1| = -196 - 4 = -200
=> |3x + 1| = -200
Không thỏa mãn điều kiện
4554:18 ra may rua