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1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
\(3^6\) và \(8^2\)
\(8^2=\left(2^3\right)^2=2^6\)
\(\Rightarrow3^6>8^2\)
a) \(5^3\) và \(3^5\)
Vì : \(5^3=125\)
\(3^5=243\)
Vì 125 < 243 ⇔ \(5^3< 3^5\)
b) \(3^6\) và \(8^2\)
⇒ \(3^6=3^{3.2}=\left(3^3\right)^2=9^2\)
Vì \(9^2>8^2\) ⇔ \(3^6>8^2\)
c) \(16^{19}\) và \(8^{25}\)
⇒ \(16^{19}=\left(2^4\right)^{19}=2^{4.19}=2^{76}\)
⇒ \(8^{25}=\left(2^3\right)^{25}=2^{3.25}=2^{75}\)
Vì \(2^{76}>2^{75}\) ⇔ \(16^{19}>8^{25}\)
d) \(2^{12}\) và \(3^8\)
⇒ \(2^{12}=2^{3.4}=\left(2^3\right)^4=8^4\)
⇒ \(3^8=3^{2.4}=\left(3^2\right)^4=9^4\)
Vì \(8^4< 9^4\) ⇔ \(2^{12}< 3^8\)
e) \(27^{11}\) và \(81^8\)
⇒ \(27^{11}=\left(3^3\right)^{11}=3^{3.11}=3^{33}\)
⇒ \(81^8=\left(3^4\right)^8=3^{4.8}=3^{32}\)
Vì \(3^{33}>3^{32}\) ⇔ \(27^{11}>81^8\)
CHÚC BẠN HỌC TỐT !☕ ⚽ ⚡
1. 53 = 5.5.5 = 125
2. 27 = 2.2.2.2.2.2.2 = 128
3. 44 = 4.4.4.4 = 256
4. 73 = 7.7.7 = 343
6. 35 = 243
7. 26 = 64
8. 34 = 81
9. 83 = 512
11. 132 = 169
12. 112 = 121
13. 142 = 196
14. 152 = 225
16. 172 = 289
17. 182 = 324
18. 192 = 361
19. 202 = 400
21. 104 = 10000
22. 105 = 100000
23. 106 = 1000000
24. 107 = 10000000
câu 1 : điền dấu > , < , = thích hợp
\(a,0>\left(-25\right).\left(-19\right).\left(-1\right)^{2n}\)
\(b,\left(-3\right)^4.\left(-19\right)^2=3^4.19^2.\left(-1\right)^{100}\)
\(c,\left(-2006\right).9\left(-2007\right)>\left(-2008\right).2009\)
câu 2 : sắp xếp các số theo thứ tự tăng dần
- 37 ; 25 ; 0 ; dấu giá trị tuyệt đối nha / -18 / ; _ (-19 ) ; _ / - 39 / ; _ ( + 151 )
Có : \(-37;25;0;18;19;-39;-151\)
Thứ tự tăng dần : \(-151;-39;-37;25;19;18;0\)
câu 3 tính
\(\text{a ) -8 + 19}=11\)
\(\text{b ) ( -27 ) : ( -3 )}=9\)
c )\(4-\left(-13\right)=17\)
d )\(\text{ - 9 -13 -( -24 ) + 11=13}\)
\(e,323-6\left[3-7.\left(-9\right)\right]=-73\)
\(f,\left(-3\right)^5.\left(-3\right)^3-9\)\(=6552\)
\(g,9-8.16-13.8\)
\(=9-8.\left(16-13\right)\)
\(=9-8.4\)
\(=9-32\)
\(=-23\)
\(h,\left(-3\right)^2+\left\{-54:\left[\left(-2\right)^3+7.|-2|\right].\left(-2\right)^2\right\}\)
\(=9+\left\{-54:\left[\left(-8\right)+7.2\right].4\right\}\)
\(=9+\left\{-54:\left[\left(-8\right)+14\right].4\right\}\)
\(=9+\left\{-54:6.4\right\}\)
\(=9+\left\{-7.4\right\}\)
\(=9+\left(-28\right)\)
\(=-19\)
học tốt
a) \(9^{21}.9^{33}=9^{21+33}=9^{54}\)
b) \(19^{11}.19.19=19^{11+1+1}=19^{13}\)
c) \(25^2.5^2.125=5^4.5^2.5^3=5^{4+2+3}=5^9\)
d) \(t^{2021}.t^2.\left(t^2\right)^2=t^{2021}.t^2.t^4=t^{2021+2+4}=t^{2027}\)
e) \(123^{14}:123^{13}=123^{14-13}=123\)
f) \(64^2:8^3=\left(8^2\right)^2:8^3=8^4:8^3=8^{4-3}=8=2^3\)
g) \(6^{10}:6^3:36=6^{10}:6^3:6^2=6^{10-3-2}=6^5\)
h) \(m^{20}:m^{10}.m^{10}=m^{20-10+10}=m^{20}\)
\(4^8.2^{20}=2^{16}.2^{20}=2^{36}\)
\(9^{12}.27^5.81^4=3^{24}.3^{15}.3^{16}=3^{55}\)
mk chỉnh đề
\(64^3.4^5.16^2=4^9.4^5.4^4=4^{18}\)
\(25^{20}.125^4=5^{40}.5^{12}=5^{52}\)
\(x^7.x^4.x^3=x^{14}\)
a) \(\left(12x-4^3\right).8^3=4.8^4\)
\(12x-4^3=32\)
12x = 96
x = 8
b) \(\left(3x-2^4\right).7^3=2.7^4\)
3x - 24 = 14
3x = 30
x = 10
Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)