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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}
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, \(5^x=625\Rightarrow5^x=5^4\Rightarrow x=4\)
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,ĐK : x \(\ne\)3/7
\(\frac{24}{7x-3}=-\frac{4}{25}\Leftrightarrow600=-28x+12\Leftrightarrow-28x=588\Leftrightarrow x=-21\)
b, ĐK : x;y \(\ne\)6
Xét : \(\frac{4}{x-6}=-\frac{12}{18}\Leftrightarrow72=-12x+72\Leftrightarrow x=0\)
Xét : \(\frac{y}{24}=-\frac{12}{18}\Leftrightarrow18y=-288\Leftrightarrow y=-16\)
\(\frac{24}{7.x-3}=-\frac{4}{25}\)
24.25=7.x-3.-4
600=7.x-3.-4
7.x-3.-4=600
7.x-3=600:-4
7.x-3=-150
7.x=-150+3
7.x=-147
x=-147:7
x=-21
vậy x=-21
Xin lỗi bạn mình k làm đầy đủ đc ạ :
2) a) Vì (x-3)(2y+1) = 7
=> x-3 và 2y + 1 \(\in\)Ư(7) = { 1;7}
Ta có bảng :
x-3 | 1 | 7 |
x | 4 | 10 |
2y+1 | 7 | 1 |
y | 3 | 0 |
Vậy...
b) (2x+1)(3y-2) = -55
=> 2x +1 và 3y - 2 \(\in\)Ư(-55) = { 1; 5 ; 11 ; 55}
Ta có bảng :
2x+1 | 1 | 55 | 5 | 11 | |||
x | 0 | 27 | 2 | 5 | |||
3y-2 | 55 | 1 | 11 | 5 | |||
y | 19 | 1 | ktm | ktm |
Sr kẻ bảng thừa cột :))
Vậy...
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)
lấy máy tính tính 2 vế
xong thay x vào để thỏa mãn điều kiện
hok tốt
Ta có :
\(\frac{1}{5}+\frac{2}{30}+\frac{121}{165}\le x\le\frac{1}{2}+\frac{156}{72}+\frac{1}{3}\)
\(\Leftrightarrow\)\(\frac{3}{15}+\frac{1}{15}+\frac{11}{15}\le x\le\frac{3}{6}+\frac{13}{6}+\frac{2}{6}\)
\(\Leftrightarrow\)\(\frac{15}{15}\le x\le\frac{18}{6}\)
\(\Leftrightarrow\)\(1\le x\le3\)
\(\Rightarrow\)\(x\in\left\{1;2;3\right\}\)
Vậy \(x\in\left\{1;2;3\right\}\)
Chúc bạn học tốt ~
Ta có :
\(25\le5^x< 625\)
\(\Rightarrow5^2\le5^x< 5^4\)
\(\Rightarrow5^x\in\left(5^2;5^3\right)\)
\(\Rightarrow x\in\left(2;3\right)\)