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20 tháng 8 2020
a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
PN
20 tháng 8 2020
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d
TT
18 tháng 8 2020
a) tính bình thường thôi
b)\(\left(3x-4\right)\times\left(x-1\right)^3=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
c) \(2^{2x-1}:4=8^3\)
\(\Leftrightarrow2^{2x-1}=2048\Leftrightarrow2^{2x-1}=2^{11}\Leftrightarrow2x-1=11\Leftrightarrow x=6\)
d) \(x^{17}=x\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
e) \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
vậy........
XO
18 tháng 8 2020
a) (x : 23 + 45) . 37 - 22 = 24.105
=> (x : 23 + 45).37 - 22 = 1680
=> (x : 23 + 45).37 = 1702
=> x : 23 + 45 = 46
=> x : 23 = 1
=> x = 23
b) (3x - 4).(x - 1)3 = 0
=> \(\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
Vậy \(x\in\left\{\frac{4}{3};1\right\}\)
c) 22x - 1 : 4 = 83
=> 22x - 1 : 22 = (23)3
=> 22x - 1 : 22 = 29
=> 22x - 1 = 211
=> 2x - 1 = 11
=> 2x = 12
=> x = 6
d) x17 = x
=> x17 - x = 0
=> x(x16 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{16}-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^{16}=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{0;1;-1\right\}\)
e) (x - 5)4 = (x - 5)6
=> (x - 5)6 - (x - 5)4 = 0
=> (x - 5)4[(x - 5)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1^2\end{cases}}\Rightarrow\orbr{\begin{cases}x-5=0\\x-5=\pm1\end{cases}}\Rightarrow x-5\in\left\{0;1;-1\right\}\)
=> \(x\in\left\{5;6;4\right\}\)
PN
13 tháng 7 2017
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
KP
2
8 tháng 10 2020
Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)
\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)
\(\Leftrightarrow3^{3x+1}\cdot4=324\)
\(\Leftrightarrow3^{3x+1}=81=3^4\)
\(\Rightarrow3x+1=4\)
\(\Leftrightarrow x=1\)
8 tháng 10 2020
\(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(3^{x+3+2x-1}+3^{2x+x+1}=324\)
\(3^{3x+2}+3^{3x+1}=324\)
\(3^{3x+1}\cdot\left(3+1\right)=324\)
\(3^{3x+1}\cdot4=324\)
\(3^{3x+1}=324:4\)
\(3^{3x+1}=81\)
\(3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(3x=4-1\)
\(3x=3\)
\(x=3:3\)
\(x=1\)