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a) 3x = 81 b) 2x . 16 = 128 c) 3x : 9 = 27 d) x4 = x
3x = 34 2x = 128 : 16 3x = 27 : 9 => x = 1
=> x = 4 2x = 8 3x = 3 Vậy x= 1
Vậy x = 4 x = 4 => x = 1
Vậy x = 4 Vậy x = 1
e) ( 2x + 1 )3 = 27
( 2x + 1 )3 = 33
=> 2x + 1 = 3
2x = 3 - 1
2x = 2
x = 1
Vậy x = 1
a, 3x=81 b, 2x*16=128 c, 3x:9=27 d, x4=x
=> 3x=34 => 2x=128:16 => 3x=27.9 => x=0 hoặc x=1
=> x=4 => 2x=8 => 3x=243
=> x=4 => 3x=35
=> x=5
e, (2x+1)3=27 f, (x-2)2=(x-2)4
=> (2x+1)3=33 +, TH1: x-2=1 => (x-2)2=(x-2)4=1 => x-2=x-2=1 => x=3
=> 2x+1=3 +, TH2: x-2=0 => (x-2)2=(x-2)4=0 => x-2=x-2=0 => x=2
=> 2x=2
=> x=2:2
=> x=1
g, 25 <= 5x < 625
=> 52 <= 5x < 54
=> x={2;3}
a, \(5^x=625\Rightarrow5^x=5^4\Rightarrow x=4\)
b, \(4^{2x-6}=1\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\Rightarrow x=3\)
c, \(\left(3x-1\right)^3=8\Rightarrow3x-1=2\)
\(\Rightarrow x=1\)
d, \(49.7^n=2401\Rightarrow7^{n+2}=7^4\)
\(\Rightarrow n+2=4\Rightarrow n=2\)
e, \(x^4.x-27.x=0\)
\(\Rightarrow x\left(x^4-27\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^4-27=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{27}\end{cases}}\)
f, \(\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow\left(x-6\right)^3-\left(x-6\right)^2=0\)
\(\Rightarrow\left(x-6\right)^2\left(x-6-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-6=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)
Chúc bạn học tốt!!!
\(a,5^x=625\) \(b,4^{2x-6}=1\)
\(\Rightarrow5^x=5^4\) \(\Rightarrow2x-6=0\) (Vì mọi số mũ không bằng 1)
\(\Rightarrow x=4\) \(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
\(c,\left(3x-1\right)^3=8\) \(d,49.7^n=2401\)
\(\Rightarrow\left(3x-1\right)^3=2^3\) \(\Rightarrow7^n=2401:49\)
\(\Rightarrow3x-1=2\) \(\Rightarrow7^n=49\)
\(\Rightarrow3x=3\) \(\Rightarrow7^n=7^2\)
\(\Rightarrow x=1\) \(\Rightarrow n=2\)
\(e,x^4.x-27.x=0\)
\(\Rightarrow x.\left(x^4-27\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^4-27=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^4=27\end{cases}}}\)
\(f,\left(x-6\right)^3=\left(x-6\right)^2\)
\(\Rightarrow x.\left(x^4-27\right)=0\) \(\Rightarrow\hept{\begin{cases}\left(x-6\right)^3=\left(x-6\right)^2\\\left(x-6\right)^3=\left(x-6\right)^2\end{cases}\Rightarrow\hept{\begin{cases}\left(x-6\right)^3:\left(x-6\right)^2=1\\\left(x-6\right)^3-\left(x-6\right)^2=0\end{cases}\Rightarrow}\hept{\begin{cases}x-6=1\\x-6=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=6\end{cases}}}\)
a) 3x.812x+1 = 81
3x.(34)2x+1 = 81
3x.38x+4 = 81
3x+8x+4 = 81 = 34
=> x + 8x + 4 = 4
9x + 4 = 4
9x = 0
x = 0
b) 2.3x+3x = 27
3x.(2+1) = 27
3x.3 = 27
3x = 9 = 32
=> x = 2
c) 4x -25 = 89
4x = 114
=>...
( câu c bn thử tìm xem có 4 mũ bao nhiêu = 114 ko nha! câu này mk ko tìm giúp bn đk)
3x.812x + 1 = 81
3x.38x + 4 = 81
39x + 4 = 81
9x + 4 = 4
9x = 0
x = 0
a) 2x . 4 = 128
2x = 128 : 4
2x = 32
2x = 25
=> x = 5
b) 2x . 24 = 26
=> x + 4 = 6
x = 6 - 4
x = 2
c) 5x + x = 39 - 311 : 39
6x = 39 - 32
6x = 39 - 9
6x = 30
x = 30 : 6
x = 5
d) 9x - 1 = 81
9x - 1 = 92
=> x - 1 = 2
x = 2 + 1
x = 3
e) 6x = 521 : 519 + 3 . 22 - 70
6x = 52 + 3 . 4 - 1
6x = 25 + 12 - 1
6x = 37 - 1
6x = 36
x = 36 : 6
x = 6
1) Tìm x
\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(\Rightarrow x\left(\frac{11}{2}+\frac{1}{3}\right)=1\)
\(\Rightarrow x\left(\frac{33}{6}+\frac{2}{6}\right)=1\)
\(\Rightarrow x.\frac{35}{6}=1\)
\(\Rightarrow x=\frac{6}{35}\)
2) So sánh
\(\frac{59}{40}< \frac{50}{31}\)( cái này bạn quy đồng là ra, mik chỉ ghi kq, bạn tự tính )
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}\right)-\frac{1}{2}\)
\(=-\frac{1}{3}+\frac{3}{14}-\frac{1}{2}\)
\(=-\frac{13}{21}\)
1)\(\frac{11}{2}.x+\frac{1}{3}.x=1\)
\(x.\left(\frac{11}{2}+\frac{1}{3}=1\right)\)
\(x.\frac{35}{6}=1\)
\(x=1:\frac{35}{6}\)
\(x=\frac{6}{35}\)
2) Ta có:
\(\frac{59}{40}=\frac{1829}{1240}\)
\(\frac{50}{31}=\frac{2000}{1240}\)
Vì \(2000>1829\Rightarrow\frac{2000}{1240}>\frac{1829}{1240}\Rightarrow\frac{50}{31}>\frac{59}{40}\)
3)\(\frac{1}{3}+\frac{4}{7}-\frac{5}{14}-\frac{1}{2}-\frac{2}{3}\)
\(=\left(\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{4}{7}-\frac{5}{14}-\frac{1}{2}\right)\)
\(=-\frac{1}{3}+\left(\frac{8}{14}-\frac{5}{14}-\frac{7}{14}\right)\)
\(=\frac{-1}{3}+\frac{-4}{14}\)
\(=\frac{-1}{3}+\frac{-2}{7}\)
\(=\frac{-7}{21}+\frac{-6}{21}\)
\(=\frac{-13}{21}\)
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
\(a,\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x-1=24\)
\(\Rightarrow x=25\)
\(b,-\frac{x}{4}=-\frac{9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x^2+x=72\)
\(\Leftrightarrow x\left(x+1\right)=72..\)
ấn nhầm: lm tiếp nhé!
\(x\left(x+1\right)=72\)
\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)
\(\Leftrightarrow x=8\)
a) \(\left(19x+2\times5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Rightarrow\left(19x+50\right):14=5^2-4^2=25-16=9\)
\(\Rightarrow19x+50=126\)
\(\Rightarrow19x=76\Rightarrow x=4\)
Vậy x = 4
b) \(2\times3^2=10\times3^{12}+8\times27^4\)
\(\Rightarrow2\times3^2=10\times\left(3^3\right)^4+8\times27^4\)
\(\Rightarrow2\times3^2=27^4\times\left(10+8\right)\)
\(\Rightarrow18=27^4\times18\)
\(\Rightarrow27^4\times18-18=0\Rightarrow18\times\left(27^4-1\right)=0\)
=> Không thấy biến x đâu cả
c) Ta thấy 33 = 27
\(\Rightarrow3^{2x-5}=3^3\Rightarrow2x-5=3\Rightarrow2x=8\Rightarrow x=4\)
Vậy x = 4
d) \(3^{x+1}-x=80\Rightarrow3^{x+1}=81\)
Ta thấy 34 = 81
\(\Rightarrow3^{x+1}=3^4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a) \(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
b) \(2^x.16=128\)
\(2^x=128:16\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
c) \(3^x:9=27\)
\(3^x=27.9\)
\(3^x=243\)
\(3^x=3^5\)
\(\Rightarrow x=5\)
d) \(x^4=x\)
\(\Rightarrow x=0\)hoac \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
e) \(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
f) \(\left(x-2\right)^2=\left(x-2\right)^4\)
\(\left(x-2\right)^2-\left(x-2\right)^4=0\)
\(\left(x-2\right)^2-\left(x-2\right)^2.\left(x-2\right)^2=0\)
\(\left(x-2\right)^2\left[1-\left(x-2\right)^2\right]=0\)
\(\left(x-2\right)^2\left(1-x+2\right)\left(1+x-2\right)=0\)
\(\Rightarrow\left(x-2\right)^2=0\)hoac \(\orbr{\begin{cases}3-x=0\\x-1=0\end{cases}}\)
\(\Rightarrow x-2=0\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
a) \(3^x=81\Leftrightarrow3^x=3^4\Rightarrow x=4\)
b)\(2^x\times16=128\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
c) \(3^x\div9=27\Leftrightarrow3^x\div3^2=3^3\Rightarrow x=5\)
d) \(x^4=x\Leftrightarrow x=1\)
e) \(\left(2x+1\right)^3=27\Leftrightarrow\left(2x+1\right)^3=3^3\Rightarrow2x+1=3 \)
\(\Rightarrow2x=3+1\Leftrightarrow2x=4\Rightarrow x=2\)
F)