Tìm số tự nhiên x, biết:
A, x * ( x + 1 ) = 156
B, x * ( x + 1 ) = 342
C, x * ( x + 1 ) = 650
D, x* ( x + 1 ) = 380
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a)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
a) \(5^x+5^{x+2}=650\)
\(5^x+5^x.5^2=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=25\)
\(5^x=5^2\)
\(\Rightarrow x=2\)
Vậy x = 2
b) \(3^{x-1}+5.3^{x-1}=162\)
\(3^{x-1}.\left(1+5\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(x=3+1\)
\(x=4\)
Vậy x = 4
a) 5x+5x + 2 = 650
=> 5x+ 5x.52 = 650
=> 5x(1+ 52) = 650
=> 5x.26 = 650
=> 5x = 650:26
=> 5x = 25
=> 5x = 52
=> x = 2
Vậy x = 2
b) 3x-1 + 5.3x-1 = 162
=> 3x-1(1+5) = 162
=> 3x-1. 6 = 162
=> 3x-1 = 162
=> x ko có giá trị
Vậy x ko tìm đc giá trị thỏa mãn đề bài.
Ta có: a, (x - 35) - 120 = 0
(x - 35) = 0 + 120
x - 35 = 120
x = 120 + 35
x = 155
b, 124 + ( 118 - x) = 217
(118 - x) = 217 - 124
118 - x = 93
x = 118 - 93
x = 25
c, 156 - ( x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
a,=>x-35=120
x=120+35
x=165
b,=>118-x=217-124
118-x=93
x=118-93
x=25
c,=>x+61=156-82
x+61=74
x=74-61
x=13
\(a,12⋮x-1\)
\(x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Tự lập bảng nha
\(b,28⋮2x+1\)
\(2x+1\inƯ\left(28\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
Ta có bảng
2x+1 | 1 | -1 | 2 | -2 | 7 | -7 | 14 | -14 |
2x | 0 | -2 | 1 | -3 | 6 | -8 | 13 | -15 |
x | 0 | -1 | 1/2 | -3/2 | 3 | -4 | 13/2 | -15/2 |
\(c,x+15⋮x+3\)
\(x+3+12⋮x+3\)
\(12⋮x+3\)
\(\Rightarrow x+3\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
Tự lập bảng
\(d,\left(x+1\right)\left(y-1\right)=3\)
\(\Rightarrow x+1;y-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta lập bảng
x+1 | 1 | -1 | 3 | -3 |
y-1 | 3 | -3 | 1 | -1 |
x | 0 | -2 | 2 | -4 |
y | 4 | -2 | 2 | 0 |
a)\(x^{2016}=x^{2017}\)
\(\Leftrightarrow x^{2017}-x^{2016}=0\)
\(\Leftrightarrow x^{2016}.\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^{2016}=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vay ...
b) \(2y.\left(x+1\right)-x-7=0\)
\(\Leftrightarrow2y.\left(x+1\right)-\left(x+1\right)=6\)
\(\Leftrightarrow\left(x+1\right).\left(2y+1\right)=6\)
Đến chỗ này bạn tự tìm các cặp x,y nha
\(x\left(x+1\right)=156\)
\(\Rightarrow x^2+x=156\)
\(\Rightarrow x^2+x-156=0\)
\(\Rightarrow x^2+13x-12x-156=0\)
\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)
\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)
___________________
\(x\left(x+1\right)=342\)
\(\Rightarrow x^2+x=342\)
\(\Rightarrow x^2+x-342=0\)
\(\Rightarrow x^2+19x-18x-342=0\)
\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)
\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)
__________________
\(x\left(x+1\right)=650\)
\(\Rightarrow x^2+x=650\)
\(\Rightarrow x^2-x+650=0\)
\(\Rightarrow x^2+26x-25x-650=0\)
\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)
\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)
______________________
\(x\left(x+1\right)=380\)
\(\Rightarrow x^2+x=380\)
\(\Rightarrow x^2+x-380=0\)
\(\Rightarrow x^2+20x-19x-380=0\)
\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)
\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)
a, \(x\).(\(x\) + 1) = 156
156 = 22.3.13 = 12.13
Vậy \(x\).(\(x\) + 1) = 12.13
Vậy \(x\) = 12
b, \(x.\)(\(x\) + 1) = 342
342 = 2.32.19 = 18.19
\(x\).(\(x+1\)) = 18.19
\(x\) = 18
c, \(x\).(\(x\) + 1) = 650
650 = 2.52.13 = 25.26
\(x\).(\(x\) +1) = 25.26
\(x\) = 25
d, \(x\).(\(x\) +1) = 380
380 = 22.5.19 = 19.20
\(x\).(\(x\) + 1) = 19.20
\(x\) = 19