a)\(\left(x-1\right)^3=125\)
b)\(\left(2.x+1\right)^3=343\)
c)\(\left(7.x-11\right)^3=2^5.5^2+200\)
1. Tìm x:
a) \(2^x-15=17\)
b) \(\left(7x-11\right)^3=2^5.5^2+200\)
c) \(x^{10}=1^x\)
d) \(x^{10}=x\)
e)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
2. Tính:
\(A=\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
a) 2x - 15 = 17
2x = 25
b) ( 7x - 11 )3 = 25. 52 + 200
(7x - 11)3 = 32 . 25 + 200
(7x -11)3 = 1000
(7x-11)3 = 103
7x - 11 = 10
7x = 10+11
7x = 21
x = 21 : 7
x = 3
like nha
Bài 1:
a. \(\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)
b. \(\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)
c. \(\frac{3^6.7^2-3^7.7}{3^7.21}\)
Bài 2:
a. \(\left(x-1,2\right)^2=4
\)
b.\(\left(x+1\right)^3=-125\)
c. \(\left(x-5\right)^3=2^6\)
d. \(\left(2x+1\right)^{x+1}=5^{x+1}\)
e. \(\left(5-x\right)^2+\left(5-x\right)^5=0\)
f. \(\left(1-2x\right)^4-\left(1-2x\right)^6=0\)
g. \(3^{4-x}+1=26\)
h. \(\left(7x+2\right)^{-1}=3^{-2}\)
\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)
\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)
\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)
\(a,\left(x-1,2\right)^2=4\)
\(\Rightarrow x-1,2=2\)
\(\Rightarrow x=3,2\)
\(b,\left(x+1\right)^3=-125\)
\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Rightarrow x+1=-5\Rightarrow x=-6\)
\(c,\left(x-5\right)^3=2^6\)
\(\Rightarrow\left(x-5\right)^3=4^3\)
\(\Rightarrow x-5=4\Rightarrow x=9\)
\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
\(e,\left(5-x\right)^2+\left(5-x\right)^5=0\)
\(\Rightarrow\left(5-x\right)^2\left(1+\left(5-x\right)^3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left(5-x\right)^2=0\\1+\left(5-x\right)^3=0\end{cases}\Rightarrow}\orbr{\begin{cases}5-x=0\\5-x=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}}\)
\(f,\left(1-2x\right)^4-\left(1-2x\right)^6=0\)
\(\Rightarrow\left(1-2x\right)^4\left(1-\left(1-2x\right)^2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-2x=0\\1-2x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\0\end{cases}}\)
\(g,3^{4-x}-1=26\)
\(\Rightarrow3^{4-x}=27\Rightarrow3^{4-x}=3^3\Rightarrow4-x=3\Rightarrow x=1\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
1. Tìm x,biết:
a) \(2^x\)- 15 = 17
b) \(\left(7x-11\right)^3\)= \(2^5.5^2+200\)
c) \(^{x^{10}=1^x}\)
d) \(x^{10}=x\)
e) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
2. Tính:
\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
a) 2x-15= 17
2x = 17-15
2x = 2
2x = 21
x = 1
tim \(x\in\)N biet
a . \(x^{10}=x\)
b . \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
c . \(\left(7\times n-11\right)=2^5.5^2+200\)
a) \(x^{10}=x\)
\(\Rightarrow x^{10}-x=0\)
\(\Rightarrow x\left(x^9-1\right)=0\)
=> x=0 HOẶC \(x^9-1=0\)
=>x=0 HOẶC x=1
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
Tìm x, biết:
a)\(x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2};\) b)\(x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9};\)
c)\({\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9};\) d)\(x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\)
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
Tìm \(x\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=1000\)
\(\Leftrightarrow7x-11=10\)
\(\Leftrightarrow7x=21\)
\(\Leftrightarrow x=3\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=800+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=800+200\)
\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=3\)
1,tìm x
a.\(3^x\)=81
b.\(5^{x+2}\)=125
c.\(2^3\)*\(2^{x-1}\)=64
d. 7*\(7^{n+1}\)=343
e.\(\left(x+3\right)^5\)=243
f.\(\left(2x-3\right)^6\)=64
g.\(\left(x-6\right)^3\)=\(\left(x-6\right)^2\)
h.\(\left\{\left[\left(2x+14\right):2^2-3\right]:3\right\}\)-1=0
i.\(\left\{\left[\left(x:3+17\right):10+3\cdot2^4\right]:10\right\}=5\)
2,tính
a.\(7^{3^1}\) b.\(7^{1^3}\) c.\(6^{1^{2^{3^4}}}\) d.\(2017^{2^{0^{1^0}}}\) e.\(6^3:6^2-6^2:6\) f.\(\frac{11\cdot2^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4