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1+3+5+...+x=1600
=(x+1).[(x-1):2+1] /2 =1600
=(x+1).(x+1) /2 =1600
=(x+1)^2:2=40^2
=(x+1):2=40
=x+1=80
=x=79
Bg
c) 9 < 3x : 3 < 81
=> 32 < 3x - 1 < 34
=> x - 1 = {2; 3; 4}
=> x = {3; 4; 5}
d) 5x . 5x + 1 . 5 x + 2 < 218 . 518 : 218
=> 5x + x + 1 + x + 2 < 218 : 218 . 518
=> 53x + 3 < 1.518
=> 53.(x + 1) < 518
=> 3.(x + 1) < 18
=> x + 1 < 18 : 3
=> x + 1 < 6
=> x < 6 - 1
=> x < 5
c. \(9\le3^x:3\le81\)
\(\Rightarrow3^2\le3^{x-1}\le3^4\)
\(\Rightarrow3^{x-1}\in\left\{3^2;3^3;3^4\right\}\)
\(\Rightarrow x-1\in\left\{2;3;4\right\}\)
\(\Rightarrow x\in\left\{3;4;5\right\}\)
d. Thêm đk : x thuộc N
\(5^x.5^{x+1}.5^{x+2}\le2^{18}.5^{18}:2^{18}\)
\(\Rightarrow5^{x+x+1+x+2}\le5^{18}\)
\(\Rightarrow x+x+x+1+2\le18\)
\(\Rightarrow3x+3\le18\)
\(\Rightarrow3\left(x+1\right)\le18\)
\(\Rightarrow x+1\le6\)
\(\Rightarrow x\le5\)
\(\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
A.(x+2)x-1=150
=>A.(x+2)x-1=1
=> x + 2 = 1 hoặc x + 2 = -1 hoặc x - 1 = 0
=> x = -1 hoặc x = -3 hoặc x = 1.
B. (5-x)x=1(x<5)
=> 5 - x = 1 hoặc 5 - x = -1 hoặc x = 0
=> x = 4 hoặc x = 6 hoặc x = 0.
C.15x-2=225
=> 15x-2=152
=> x - 2 = 2 => x = 4.
D.(x+2)2.(x+1)=64
=>(x+2).(x+2).(x+1)=64 = 1.2.32 = 2.2.16 = ...
Mà x + 2 và x + 2 và x + 1 chỉ hơn kém nhau 1 đơn vị nên không có x nào thỏa mãn.
E.(x-5)3.(x-5)=16
=>(x-5)4=16=24
=>x-5=2=>x=7.
2|x - 5| = 8
\(\Rightarrow\) |x - 5| = 4
\(\Rightarrow\) |x - 5| = \(\left\{{}\begin{matrix}4\\-4\end{matrix}\right.\)
\(\Rightarrow\) x = \(\left\{{}\begin{matrix}9\\1\end{matrix}\right.\)
a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)