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a, (x-1) . 0,5 = 7,5 : (x-1)
=> = ( x - 1 ) 0,5 = \(\frac{x-1}{2}\)
\(=\frac{7,5}{x-1}=\frac{15}{2\left(x-1\right)}\)
=> x = - 1 \(\sqrt{15}\)
x = \(\sqrt{15+1}\)
đề sao sao ý
Bài 1:
\(\text{a) }x.x^2.x^3.x^4.x^5.....x^{49}.x^{50}\)
\(=x^{1+2+3+4+5+...+49+50}\)
\(=x^{\frac{51.50}{2}}\)
\(=x^{1275}\)
\(\text{b) Ta có:}\)
\(4^{15}=\left(2^2\right)^{15}=2^{2.15}=2^{30}\)
\(8^{11}=\left(2^3\right)^{11}=2^{3.11}=2^{33}\)
\(\text{Vì }2^{30}< 2^{33}\text{ nên }4^{15}< 8^{11}\)
Bài 2: Tìm x
\(\left(x-1\right)^4:3^2=3^6\)
\(\Rightarrow\left(x-1\right)^4=3^6\times3^2\)
\(\Rightarrow\left(x-1\right)^4=3^8\)
\(\Rightarrow\left(x-1\right)^4=3^{2.4}\)
\(\Rightarrow\left(x-1\right)^4=\left(3^2\right)^4\)
\(\Rightarrow x-1=9\)
\(\Rightarrow x=10\)
Bài 3 và bài 4 mk làm sau
Bài 1 : a) \(x.x^2.x^3.x^4.....x^{49}.x^{50}=x^{1+2+3+...+49+50}\) (Dễ rồi tự tính)
b) \(\hept{\begin{cases}4^{15}=\left(2^2\right)^{15}=2^{30}\\8^{11}=\left(2^3\right)^{11}=2^{33}\end{cases}}\)Rồi tự so sánh đi
Bài 2 :
\(\left(x-1\right)^4\div3^2=3^6\Leftrightarrow\left(x-1\right)^4=3^8=\left(3^2\right)^4=9^4\Leftrightarrow x-1=9\Leftrightarrow x=10\)
Bài 3 :
\(\hept{\begin{cases}27^{15}=\left(3^3\right)^{15}=3^{45}\\81^{11}=\left(3^4\right)^{11}=3^{44}\end{cases}}\) nt
a, 3x - 2 ⋮ x + 3
=> 3x + 9 - 11 ⋮ x + 3
=> 3(x + 3) - 11 ⋮ x + 3
=> 11 ⋮ x + 3
b, x ⋮ 2x + 1
=> 2x ⋮ 2x + 1
=> 2x + 1 - 1 ⋮ 2x + 1
=> 1 ⋮ 2x + 1
c, 3x + 6 ⋮ x + 1
=> 3x + 3 + 3 ⋮ x + 1
=> 3(x + 1) + 3 ⋮ x + 1
=> 3 ⋮ x + 1
d, em không biết làm
câu a,b,c bn Cả Út lm r
mik làm câu d
\(x^2⋮x-2\)
\(\Rightarrow x\left(x-2\right)+2x⋮x-2\)
\(\Rightarrow2x⋮x-2\)
\(\Rightarrow2\left(x-2\right)+4⋮x-2\)
\(\Rightarrow4⋮x-2\)
\(\Rightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{3;1;4;0;6;-2\right\}\)
Vậy..............................
\(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Rightarrow-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}-\frac{1}{3}+1\)
\(\Rightarrow-7x=\frac{-1}{6}\)
\(\Rightarrow x=\frac{1}{42}\)
Vậy ...
\(\)
\(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Rightarrow3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\Rightarrow\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Rightarrow3x=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{18}\)
Vậy...
\(x+1⋮x-1\)
\(x-1+2⋮x-1\)
\(2⋮x-1\)hay \(x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x - 1 | 1 | -1 | 2 | -2 |
| x | 2 | 0 | 3 | -1 |
\(x-2⋮x+1\)
\(x+1-3⋮x+1\)
\(-3⋮x+1\)thự hiện tương tự nhé !
Bài 1:
a) \(x^{2015}=x\Leftrightarrow x^{2015}-x=0\Leftrightarrow x\left(x^{2014}-1\right)=0\)
\(\Leftrightarrow x=0\)hoặc \(x=1\)(do \(x\in N\))
b)\(3^{x+1}+3^x=108\Leftrightarrow3^x\left(3+1\right)=108\Leftrightarrow3^x=27\Leftrightarrow x=3\)
c) Do \(x\in N\)nên \(1+3x>0\). Ta có:
\(\left(1+3x\right)^4=256\Leftrightarrow\left(1+3x\right)^4=4^4\Leftrightarrow1+3x=4\Leftrightarrow x=1\)
Bài 2: \(A=1+2+2^2+2^3+...+2^{2015}\)\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{2013}+2^{2014}+2^{2015}\right)\)\(=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2013}\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\left(1+2^3+...+2^{2013}\right)=7.\left(1+2^3+...+2^{2013}\right)\)chia 7 dư 0
Bài 3: Mình chưa hiểu đề bài cho lắm. Nếu như tính GTBT thì C=1022; D=289; E=128.
p/s: Đây ms chỉ là gợi ý. Bạn nên tự làm lại để hiểu hơn.
\(2^{3x-1}=2^8\)
\(\Rightarrow3x-1=8\)
\(3x=9\)
\(x=3\)
23x-1 = 256
23x-1 = 28
=> 3x-1 = 8
3x = 8 + 1
3x = 9
x = 9 : 3
x = 3
23x-1=256
23x-1=28
3x-1=8
3x=8+1
3x=9
x=9:3=3.
NHỚ K NHA ÁNH!!!!!!!!!!!!!!!
Trả lời:
\(2^{3x-1}=256\)
\(2^{3x-1}=2^8\)
\(3x-1=8\)
\(3x=9\)
\(x=3\)
Vậy \(x=3\)
Hok tốt!
Vuong Dong Yet