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a) \(32< 2^x< 128\)
=> \(2^5< 2^x< 2^7\)
=> x = 6
b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)
=> 9.2x-1 = 9.25
=> 2x-1 = \(\frac{9\cdot2^5}{9}=2^5\)
=> x - 1 = 5 => x = 6
c) \(9\cdot27\le3^x\le243\)
=> \(243\le3^x\le243\)
=> x = 5
d) Giống câu b)
e) \(3^{x-1}+5\cdot3^{x-2}=216\)
=> 8.3x-2 = 216
=> 3x-2 = 27
=> 3x-2 = 33
=> x - 2 = 3 => x = 5
f) 27x-3 = 9x+3
=> 27x-3 = 9x+3
=> (33)x-3 = (32)x+3
=> 33x-9 = 32x + 6
=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên
g) x2019 = x => x2019 - x = 0 => x(x2018 - 1) = 0 => x = 0 hoặc x = 1
a)
\(2^5< 2^x< 2^7\)
\(5< x< 7\)
\(x=6\)
b)
\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
\(2^{x-1}+2^{2+x}=9\cdot2^5\)
\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\)
\(2^{x-1}\cdot9=9\cdot2^5\)
\(2^{x-1}=2^5\)
\(x-1=5\)
\(x=6\)
a) A = 3,4 . 10-8 = \(\frac{3,4}{10^{-8}}=\frac{34}{10^{-9}}\)
B = 34 . 10-9 = \(\frac{34}{10^{-9}}\)
A = B
b) \(\frac{A}{B}=\frac{\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}}{\frac{1}{10^9}}\)
\(=\left(\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}\right).10^9\)
\(=\frac{1+10+10^2}{10^4}.10^9\)
\(=111.10^5\)
=> A = 11100000B
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
a: TH1: x<2
Pt sẽ là 5-x+2-x=5x
=>5x=-2x+7
=>x=1(nhận)
TH2: 2<=x<5
Pt sẽ là 5x=x-2+5-x=3
=>x=3/5(loại)
TH3: x>=5
Pt sẽ là 5x=x-5+x-2=2x-7
=>3x=-7
=>x=-7/3(loại)
b: \(A=\dfrac{2^6\cdot5^2+2^{11}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)
\(=\dfrac{2^6\cdot5^2\left(1+2^5\cdot5^7\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{1+2^5\cdot5^7}{2^{10}\cdot5^5\cdot6}\)
nhờ 2 câu nha
tim x
(x2+1)2+3x (x2+1)2+2x2=0
x3+6x+12x +8x3 -21=0
\(1.\) sai đề rồi nha. dề đúng phải là \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
đặt \(\left(x^2+1\right)=k\) \(\Rightarrow\)biểu thức trên có dạng là \(k^2+3xk+2x^2=0\)
\(\Leftrightarrow\) \(\left(k+x\right)\left(k+2x\right)=0\)
suy ra \(k+x=0\) hoặc \(k+2x=0\)
\(x^2+1+x=0\) \(x^2+1+2x=0\)
bấm máy tình không ra \(\left(x+1\right)^2=0\)
nên ko có giá trị của x \(x+1=0\)
\(x=-1\)
Vậy \(x=-1\)
Bài 1:
a, \(A=3^{100}+3^{99}+...+3+1\)
\(\Rightarrow3A=3^{101}+3^{100}+...+3^2+3\)
\(\Rightarrow3A-A=\left(3^{101}+3^{100}+...+3^2+3\right)-\left(3^{100}+3^{99}+...+3+1\right)\)
\(\Rightarrow2A=3^{101}+1\Rightarrow A=\dfrac{3^{101}+1}{2}\)
b, \(B=\dfrac{15^9.2^{18}.9^8}{3^{15}.4^8.25^4}=\dfrac{3^9.5^9.2^{18}.3^{16}}{3^{15}.2^{16}.5^8}\)
\(=3^{10}.5.2^2=472392\)
c, \(C=\dfrac{2^{10}.10^{17}.7^9}{5^{15}.14^9.64^9}=\dfrac{2^{10}.2^{17}.5^{17}.7^9}{5^{15}.2^9.7^9.2^{54}}\)
\(=\dfrac{5^2}{2^{36}}\)
Chúc bạn học tốt!!!
1.
\(A=3^{100}+3^{99}+3^{98}+...+3^2+3+1\\ A=\dfrac{3-1}{2}\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)\\ =\dfrac{\left(3-1\right)\cdot\left(3^{100}+3^{99}+3^{98}+...+3^2+3+1\right)}{2}\\ =\dfrac{3^{101}-3^{100}+3^{100}-3^{99}+...+3^2-3+3-1}{2}\\ =\dfrac{3^{101}-1}{2}\)
\(B=\dfrac{15^9\cdot2^{18}\cdot9^8}{3^{15}\cdot4^8\cdot25^4}\\ =\dfrac{\left(3\cdot5\right)^9\cdot2^{18}\cdot\left(3^2\right)^8}{3^{15}\cdot\left(2^2\right)^8\cdot\left(5^2\right)^4}\\ =\dfrac{3^9\cdot5^9\cdot2^{18}\cdot3^{16}}{3^{15}\cdot2^{16}\cdot5^8}\\ =\dfrac{3^9\cdot5\cdot2^2\cdot3}{1\cdot1\cdot1}\\ =3^{10}\cdot5\cdot2^2\\ =59049\cdot5\cdot4\\ =59049\cdot\left(5\cdot4\right)\\ =59049\cdot20\\ =1180980\)
\(C=\dfrac{2^{10}\cdot10^{17}\cdot7^9}{5^{15}\cdot14^9\cdot64^9}\\ =\dfrac{2^{10}\cdot\left(2\cdot5\right)^{17}\cdot7^9}{5^{15}\cdot\left(2\cdot7\right)^9\cdot\left(2^6\right)^9}\\ =\dfrac{2^{10}\cdot2^{17}\cdot5^{17}\cdot7^9}{5^{15}\cdot2^9\cdot7^9\cdot2^{54}}\\ =\dfrac{2\cdot1\cdot5^2\cdot1}{1\cdot1\cdot1\cdot2^{37}}\\ =\dfrac{5^2}{2^{36}}\\ =\dfrac{25}{2^{36}}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)