32^x x 16^x=517
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a ) − 13 30 ≤ x < 1 = > x = 0. b ) 5 9 < x ≤ 3 = > x ∈ {1;2;3}
a) − 2 5 + 1 6 + − 1 5 ≤ x < − 3 4 + 9 7 + − 1 4 + 5 7 ⇔ − 13 30 ≤ x ≤ 1 ⇔ x ∈ 0 ; 1
b) 5 17 + − 4 9 + 12 17 < x ≤ − 3 7 + 7 15 + 4 − 7 + 8 15 + 9 3 ⇔ 5 9 < x ≤ 3 ⇔ x ∈ 1 ; 2 ; 3
Bài 2:
a: =>x-45=-20
hay x=25
b: =>35+30-5x=-12+112
=>65-5x=100
=>5x=-35
hay x=-7
c: =>x-124=1000
hay x=1124
d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)
\(\Leftrightarrow\left(3x-2\right)^3=64\)
=>3x-2=4
hay x=2
Bài 2:
a: =>x-45=-20
hay x=25
b: =>35+30-5x=-12+112=100
=>65-5x=100
=>5x=-35
hay x=-7
c: =>x-124=1000
hay x=1124
d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)
\(\Leftrightarrow\left(3x-2\right)^3=64\)
=>3x-2=4
hay x=2
\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)
\(X+16=-5\)
\(X=-21\)
\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\dfrac{32}{1-x^{32}}\)
1)517......
X= 151481 : 517
X= 293
2)195.......
x=195906 : 634
x=309
\(32^x.16^x=512\) (sửa lại 517 thành 512 mới đúng, bạn xem lại đề)
\(\Rightarrow2^{5x}.2^{4x}=512=2^9\)
\(\Rightarrow2^{9x}=2^9\Rightarrow9x=9\Rightarrow x=1\)
thank you cậu