1.Tìm x biết:\(\dfrac{\left(-3\right)^x}{81}\)=-27
Hình học;
1.Qua A ở ngoài đường thẳng a, vẽ 101 đường thẳng phân biệt.CMR: có ít nhất 100 đường thẳng cắt a.
các bn có thể giải bài này xog trog ngày mai đc không
\(\left(x+\dfrac{1}{3}\right)+\left(x+\dfrac{1}{9}\right)+\left(x+\dfrac{1}{27}\right)+\left(x+\dfrac{1}{81}\right)=\dfrac{56}{81}\)
Tham khảo link: https://olm.vn/hoi-dap/detail/55111422944.html
`(x+1/3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81`
`x+x+x+x+1/3+1/9+1/27=56/81-1/81`
`4x+13/27=55/81`
`4x=55/81-13/27`
`4x=55/81-52/81`
`4x=16/81`
`x=4/108`
Vậy `x=4/108`
Tìm x ∈ N biết :
a) \(8< 2^x\le2^9.2^{-5}\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-3}{5}\right)^2\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{1}{5}\)
1.Có bao nhiêu giá trị của \(\left|2x-1\right|=5\) thỏa mãn
A. 0 B. 1 C. 3 D.2
2. Tìm x, biết \(\dfrac{\left(-3\right)^x}{81}=-27\)
A. x=7 B. x=2187 C. x=3 D. x = -7
\(1,\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\1-2x=5\end{matrix}\right.\Leftrightarrow D\\ 2,\Leftrightarrow\left(-3\right)^x=-27\cdot81=-2187=\left(-3\right)^7\\ \Leftrightarrow x=7\left(A\right)\)
1. \(\left(y+\dfrac{1}{3}\right)\)+\(\left(y+\dfrac{1}{9}\right)\)+\(\left(y+\dfrac{1}{27}\right)\)+\(\left(y+\dfrac{1}{81}\right)\)=\(\dfrac{56}{81}\)
2. 18:\(\dfrac{Xx0,4+0,32}{X}\)+5=14
3. \(\dfrac{3xX}{2}\)=\(\dfrac{2}{5}+\)X\(+\dfrac{1}{3}\)
4. X-\(\dfrac{11}{15}\)=\(\dfrac{3+X}{5}\)
Bài 1:
$(y+\frac{1}{3})+(y+\frac{1}{9})+(y+\frac{1}{27})+(y+\frac{1}{81})=\frac{56}{81}$
$(y+y+y+y)+(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81})=\frac{56}{81}$
$4\times y+\frac{40}{81}=\frac{56}{81}$
$4\times y=\frac{56}{81}-\frac{40}{81}=\frac{16}{81}$
$y=\frac{16}{81}:4=\frac{4}{81}$
Bài 2:
$18: \frac{x\times 0,4+0,32}{x}+5=14$
$18: \frac{x\times 0,4+0,32}{x}=14-5=9$
$\frac{x\times 0,4+0,32}{x}=18:9=2$
$x\times 0,4+0,32=2\times x$
$2\times x-x\times 0,4=0,32$
$x\times (2-0,4)=0,32$
$x\times 1,6=0,32$
$x=0,32:1,6=0,2$
Bài 3:
$\frac{3\times x}{2}=\frac{2}{5}+x+\frac{1}{3}$
$1,5\times x=x+\frac{11}{15}$
$1,5\times x-x=\frac{11}{15}$
$x\times (1,5-1)=\frac{11}{15}$
$x\times 0,5=\frac{11}{15}$
$x=\frac{11}{15}: 0,5=\frac{22}{15}$
Tìm x biết:
a) \(\dfrac{-32}{\left(-2\right)^x}=4\) f) \(\left(3x-1\right)^3=\dfrac{-8}{27}\)
b) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\) g) \(\left(2x+3\right)^2=\dfrac{9}{121}\)
c) \(\dfrac{1}{9}.27^x=3^x\) h) \(5^x+5^{x+2}=650\)
d) \(9^x:3^3=\dfrac{1}{243}\) i) \(\left(x-7\right)^{x+1}-\left(x-7\right)=0\)
e) \(\dfrac{x7}{81}=27\) m) \(\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
tìm x ∈ Q biết : \(\dfrac{3\left(x+1\right)}{2}\)=\(\dfrac{8}{27\left(x-1\right)}\)
Tìm x
1) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\) 2) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\) 3) \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
4) \(\left(\dfrac{4}{9}\right)^x=\left(\dfrac{8}{27}\right)^{10}\) 5) \(2^x=4^5.4^3\) help me !!!!
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
Tìm x biết
a) \(2.3^x-405=3^{x-1}\)
b) \(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
c) \(\left|3x-5\right|=\left|\dfrac{1}{2}x+3\right|\)
d) \(\left(\dfrac{1}{81}\right)^x.27^{2x}=\left(-9\right)^4\)
a, Theo đề ta có:
\(2.3^x-405=3^{x-1}\)
=> \(2.3^x-405=3^x:3\)
=> \(405=(2.3^x)-(3^x:3)\)
=>\(405=(2.3^x)-(3^x.\dfrac{1}{3})\)
=> \(405=3^x(2-\dfrac{1}{3})\)
=>\(405=3^x(\dfrac{6}{3}-\dfrac{1}{3})\)
=> \(405=3^x.\dfrac{5}{3}\)
=> \(3^x=405:\dfrac{5}{3}\)
=>\(3^x=405.\dfrac{3}{5}\)
=> \(3^x=81.3\)
=> \(3^x=243\)
=> \(3^x=3^5\)
=> x=5
Vậy:..............................