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\(1+3+5+...+x=3200\)
\(\Leftrightarrow\left(x+1\right)+\left(3+x-2\right)+...+...=3200\)
\(\Leftrightarrow\left(x+1\right)+\left(x+1\right)+...+...=3200\)
Có các số hạng là:
\(\frac{x-1}{2}+1=\frac{x-1}{2}+1=\frac{x+1}{2}\)
Có số cặp \(\left(1+x\right)\)là:
\(\frac{x+1}{2}.\frac{1}{2}=\frac{x+1}{4}\)
\(\Leftrightarrow\left(1+x\right)\frac{x+1}{4}=\frac{\left(x+1\right)^2}{4}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=4.3200=\left(2.40\sqrt{2}\right)\)
\(\Leftrightarrow x+1=2.40\sqrt{2}\)
\(\Leftrightarrow x=80\sqrt{2}-1\)
(x-y)(2y+1)= 11
=> x-y \(\in\)Ư(11) ={ 1;11; -1; -11}
Nếu x-y = 1 thì 2y+1= 11 => 2y= 10 => y=5 => x= 6
Nếu x-y= 11 thì 2y+1 = 1 => 2y=0 => y=0 => x= 11
Nếu x-y = -1 thì 2y+1= -11 => 2y= -12 => y= -6 => x= -7
.............................
Vậy....
(x-y)(2y+1)=11
x,y nguyên => x-y; 2y+1 = nguyên
=> x-y; 2y+1 thuộc Ư (11)={-11;-1;1;11}
Ta có bảng
2y+1 | -11 | -1 | 1 | 11 |
y | -6 | -1 | 0 | 5 |
x-y | -1 | -11 | 11 | 1 |
x | -7 | -12 | 11 | 6 |
a) x : 123 = 12
x = 12 . 123
x = 124
b) 81 . x = 94
81 . x = 6561
x = 6561 : 81
x = 81
c) 2x - 26 = 6
2x = 6 + 26
2x = 32
2x = 25
x = 5
d) 3x = 81
3x = 34
x = 4
Ta có : \(\frac{x-1}{12}=\frac{3}{x-1}\)
\(\Rightarrow\left(x-1\right).\left(x-1\right)=12.3\)
\(\Rightarrow\left(x-1\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=6^2\\\left(x-1\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}\)
Vậy \(x=7;x=-5\)
\(\frac{x-1}{12}=\frac{3}{x-1}ĐKXĐ\left(x\ne1\right)\)
\(\left(x-1\right)^2=36\)
\(\left(x-1\right)^2=6^2\)
\(\Rightarrow\orbr{\begin{cases}x-1=6\\x-1=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x=-5\end{cases}}}\)tm ))
Bài 1:
a) \(x^{10}=1^x\Rightarrow\orbr{\begin{cases}x=1\\x=10\end{cases}}\)
b) \(x^{10}=x\Rightarrow x=1\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\left(2x-15\right)^5.\left(2x-15\right)^3=\left(2x-15\right)^3\)
\(\left(2x-15\right)^2=1\Rightarrow x=8\)
Bài 2:
\(a;2^{16}=2^{13}\cdot2^3=2^{13}\cdot8>7\cdot2^{13}\)
\(b;49^8\cdot27^5=7^{16}\cdot3^{15}=21^{15}\cdot7>21^5\)
C;Ta có:\(199^{20}< 200^{20}=2^{20}\cdot10^{40}=2^{15}\cdot10^{40}\cdot2^5\)
\(2003^{15}>2000^{15}=2^{15}\cdot10^{45}=2^{15}\cdot10^{40}\cdot10^5\)
Vì 25<105 nên 19920<200315
\(d;3^{39}< 3^{40}=9^{20}< 11^{20}< 11^{21}\)
x – (11 – x) = -48 + (-12 + x)
=> x−11+x=−48+−12+x
=> −11+x=−48−12
=> -11+x=-60
=> x=-49
Vậy x=-49
\(x-\left(11-x\right)=-48+\left(-12+x\right)\)
\(\Leftrightarrow x-11+x=-48-12+x\)
\(\Leftrightarrow x=-49\)
Vậy x = -49