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\(\left(x+1\right)^3=27\)
\(\left(x+1\right)^3=3^3\)
\(\Rightarrow x+1=3\)
\(x=2\)
\(\left(x+1\right)^3=27\)
\(< =>\left(x+1\right)^3=3.3.3=3^3\)
\(< =>x+1=3< =>x=3-1=2\)
\(\left(2x+3\right)^3=9.81\)
\(< =>\left(2x+3\right)^3=9.9.9\)
\(< =>\left(2x+3\right)^3=9^3\)
\(< =>2x+3=9< =>2x=6\)
\(< =>x=\frac{6}{2}=3\)
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.
Giải:
a) Đặt:
\(A=1+2^2+2^3+2^4+...+2^{2018}\)
\(\Leftrightarrow2A=2+2^3+2^4+2^5+...+2^{2019}\)
\(\Leftrightarrow2A-A=\left(2+2^{2019}\right)-\left(1+2^2\right)\)
\(\Leftrightarrow A=2+2^{2019}-1-2^2\)
\(\Leftrightarrow A=2+2^{2019}-5\)
\(\Leftrightarrow A=2^{2019}-3\)
Vậy \(A=2^{2019}-3\).
b) Đặt:
\(B=1+5+5^2+5^3+...+5^{2017}\)
\(\Leftrightarrow5B=5+5^2+5^3+5^4+...+5^{2018}\)
\(\Leftrightarrow5B-B=5^{2018}-1\)
\(\Leftrightarrow4B=5^{2018}-1\)
\(\Leftrightarrow B=\dfrac{5^{2018}-1}{4}\)
Vậy \(B=\dfrac{5^{2018}-1}{4}\).
Chúc bạn học tốt!
a)A= 1 + 22+23 + 24 +....+22018
2A = 22 + 23 + 24 +......+22018 + 22019
_
A= 1 + 22+23 + 24 +....+22018
A= 22019 - 1
A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90
2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100
2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - ( 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 )
A = 2^100 - 2^3
B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50
5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51
5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )
4B = 5^51 - 1
B = 5^51 - 1 / 4
a: \(=\left\{145-\left[130-10\right]:2\right\}\cdot5\)
\(=\left\{145-60\right\}\cdot5=85\cdot5=425\)
b: \(=100:\left\{250:\left[450-4\cdot125+4\cdot25\right]\right\}\)
\(=\dfrac{100}{250:\left[450-500+100\right]}=\dfrac{100}{250:50}=\dfrac{100}{5}=20\)
c: \(=355-5\cdot\left[64-\left(27-25\right)\right]=355-5\cdot\left[64-2\right]\)
\(=355-310=45\)
a: =5-78*32
=5-2496
=-2491
b: \(=6\left(9-6\right)=6\cdot3=18\)
c: \(=46\cdot\dfrac{\left(123-42\right)}{81}=46\)
d: \(=181+3-84+8\cdot25\)
=100+200
=300
e: \(=64\cdot35+140\cdot84-1=2240-1+11760\)
=14000-1
=13999
f: \(=3^3+25\cdot8-1=26+200=226\)
g: \(=3+2^4+1=16+4=20\)
h: \(=36:4\cdot3+2\cdot25-1=27+50-1=27+49=76\)
1; 73.52.54.76:(55.78)
= (73.76).(52.54) : (55.78)
= 79.56: (55.78)
= (79:78).(56:55)
= 7.5
= 35
2; 33.a7.3.a2:(34.a6)
= (33.3).(a7.a2): (34.a6)
= 34.a9: (34.a6)
= (34:34).(a9:a6)
= a3
\(5^6:5^4+2^3+2^2-1^{2018}\)
\(=5^2+8+4-1\)
\(=25+11\)
\(=36\)
56 : 54 + 23 + 22 - 12018
= 52 + 23 + 22 - 12018
= 25 + 8 + 4 - 1
= 36
Học tốt nhé bạn !