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a) (-5/9)^10 : x = (-5/9)^8
=> x = (-5/9)^10 : (-5/9)^8
=> x = (-5/9)^10-8 = (-5/9)^2
=> x = 25/81
b ) x : (-5/9)^8 = (-9/5)^8
=> x = (-9/5)^8 . (-5/9)^8
=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )
=> x = 1
C) x^3 = -8 =(-2)^3
=> x = -2
a) (-5/9)¹⁰ : x = (-5/9)⁸
x = (-5/9)¹⁰ : (-5/9)⁸
x = (-5/9)²
x = 25/81
b) x : (-5/9)⁸ = (-9/5)⁸
x = (-9/5)⁸ . (-5/9)⁸
x = [-9/5 . (-5/9)]⁸
x = 1⁸
x = 1
c) x³ = -8
x³ = (-2)³
x = -2
\(\frac{8}{3}:x=\frac{16}{9}:\frac{8}{3}\)
\(\frac{8}{3}:x=\frac{16}{9}.\frac{3}{8}\)
\(\frac{8}{3}:x=\frac{2}{3}\)
\(x=\frac{8}{3}:\frac{2}{3}\)
\(x=\frac{8}{3}.\frac{3}{2}\)
\(x=4\)
Vậy x = 4
Ta có: \(x+\frac{1}{9}=\frac{8}{x}-1\)
\(\Leftrightarrow x-\frac{8}{x}=-1-\frac{1}{9}\)
\(\Leftrightarrow\frac{x^2-8}{x}=-\frac{10}{9}\)
\(\Leftrightarrow9.\left(x^2-8\right)=-10x\)
\(\Leftrightarrow9x^2+10x-72=0\)
\(\Leftrightarrow\left(9x^2+10x+\frac{25}{9}\right)-\frac{673}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{5}{3}\right)^2=\frac{673}{9}\)
\(\Leftrightarrow3x+\frac{5}{3}=\pm\frac{\sqrt{673}}{3}\)
+ \(3x+\frac{5}{3}=\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{\sqrt{673}-5}{3}\)
\(\Leftrightarrow\)\(x=\frac{\sqrt{673}-5}{9}\)
+ \(3x+\frac{5}{3}=-\frac{\sqrt{673}}{3}\)\(\Leftrightarrow\)\(3x=\frac{-\sqrt{673}-5}{3}\)
\(\Leftrightarrow\)\(x=\frac{-\sqrt{673}-5}{9}\)
Vậy \(x\in\left\{\frac{\sqrt{673}-5}{9};\frac{-\sqrt{673}-5}{9}\right\}\)
\(\left(\dfrac{4}{9}\right)^{x+1}=\left(\dfrac{8}{27}\right)^6\)
\(\Leftrightarrow\left(\dfrac{2}{3}\right)^{2x+2}=\left(\dfrac{2}{3}\right)^{18}\)
\(\Leftrightarrow2x+2=18\Leftrightarrow2x=16\Leftrightarrow x=8\)
Từ đề bài=>(x+1)/10+(x+2)/9+(x+3)/8-(-3)=0
<=>......................................(như trên)+3=0
<=>(x+1)/10+1+(x+2)/9+1+(x+3)/8+1=0( vì 1+1+1=3)
<=>(x+1+10)/10+(x+2+9)/9+(x+3+8)/8=0
<=>(x+11)/10+(x+11)/9+(x+11)/8=0
<=>(x+11).(1/10+1/9+1/8)=0
Mà 1/10<1/9<1/8
=>1/10+1/9+1/8 khác 0
<=>x+11=0<=>x=-11
Vậy x=-11