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\(a,12⋮x-1\)
\(x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
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
`#3107`
b)
`2.3^x = 162`
`\Rightarrow 3^x = 162 \div 2`
`\Rightarrow 3^x = 81`
`\Rightarrow 3^x = 3^4`
`\Rightarrow x = 4`
Vậy, `x = 4`
c)
`(2x - 15)^5 = (2 - 15)^3`
\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`
\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)
Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)
`d)`
\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!
`e)`
\(7\cdot4^{x-1}+4^{x-1}=23?\)
\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)
Bạn xem lại đề!
`f)`
\(2\cdot2^{2x}+4^3\cdot4^x=1056\)
\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy, `x = 2`
_____
\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)
\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)
\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)
\(\Rightarrow\left(x\div3+17\right)\div10=2\)
\(\Rightarrow x\div3+17=20\)
\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)
Vậy, `x = 9.`
\(a,12⋮x-1\)
\(x-1\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm12\right\}\)
Ta lập bảng xét giá trị
x - 1 1 -1 2 -2 3 -3 4 -4 12 -12
x 2 0 3 -1 4 -2 5 -3 13 -11
\(c,x+15⋮x+3\)
\(x+3+12⋮x+3\)
\(12⋮x+3\)
Tự lập bảng , lười ~~~
\(d,\left(x+1\right)\left(y-1\right)=3\)
Ta lập bảng
| x+1 | 1 | -1 | 3 | -3 |
| y-1 | 3 | -3 | 1 | -1 |
| x | 2 | 0 | 2 | -4 |
| y | 4 | -2 | 2 | 0 |
i, Theo bài ra ta có : ( olm thiếu dấu và == nên trình bày kiủ nài )
\(x⋮10,x⋮12,x⋮15\)và \(100< x< 150\)
Gợi ý : Phân tích thừa số nguyên tố r xét ''BC'' ( chắc là BC )
:>> Hc tốt
a) \(x=10\)
b) \(x=11\)
Nhớ k cho mình với nhá, chúc bạn may mắn!
Bài 2:a, x+17=109
x =109-17
x =92
b,x+34=56+42+43
x+34=141
x =141-34
x =107
c,2x+45=x+94
2*x =x+94-45
x+x =x+49
suy ra x=49
d,x+109=1+2+3+........+99
x+109=495
x =495-109
x =386
Bài 1 mik ko có làm được.
\(x\left(x+1\right)=156\)
\(\Rightarrow x^2+x=156\)
\(\Rightarrow x^2+x-156=0\)
\(\Rightarrow x^2+13x-12x-156=0\)
\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)
\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)
___________________
\(x\left(x+1\right)=342\)
\(\Rightarrow x^2+x=342\)
\(\Rightarrow x^2+x-342=0\)
\(\Rightarrow x^2+19x-18x-342=0\)
\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)
\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)
__________________
\(x\left(x+1\right)=650\)
\(\Rightarrow x^2+x=650\)
\(\Rightarrow x^2-x+650=0\)
\(\Rightarrow x^2+26x-25x-650=0\)
\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)
\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)
______________________
\(x\left(x+1\right)=380\)
\(\Rightarrow x^2+x=380\)
\(\Rightarrow x^2+x-380=0\)
\(\Rightarrow x^2+20x-19x-380=0\)
\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)
\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)
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