\((x+3)^3=\frac{81}{27}\)
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=> 9.81 = 27.3x+2
32.34 = 33.3x.32
36 = 33+x.32
36:32 = 33+x
34 = 33+x
<=> 3+x = 4
x = 4-3
x = 1
\(\frac{9}{27^x}=\frac{3^{x+2}}{81}\Rightarrow\frac{9}{27^x}=\frac{3^x.3^2}{9.9}\Rightarrow\frac{9}{27^x}=\frac{3^x.9}{9.9}\Rightarrow\frac{9}{27^x}=\frac{3^x}{9}\Rightarrow9.9=3^x.27^x\Rightarrow3^2.3^2=3^x.\left(3^3\right)^x\Rightarrow3^{2+2}=\left(3^{1+3}\right)^x\)
\(\Rightarrow3^4=\left(3^4\right)^x\Rightarrow x=1\)
Sửa đề: \(\dfrac{3}{x^2+6x+9}-\dfrac{3}{x^2-6x+9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
\(=\dfrac{3x^2-18x+27-3x^2-18x-27+x^2+30x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{x^2-6x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}=\dfrac{\left(x-9\right)\left(x+3\right)}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{\left(x-9\right)}{\left(x^2-9\right)\left(x-3\right)}\)
\(\Rightarrow\frac{27^x}{3^x}=\frac{\left(3^3\right)^x}{3^3}=\frac{3^{3x}}{3^3}=3^4\)
33x=34:33
33x=3
<=> 3x=1
x=1:3
x=1/3
\(\frac{27^x}{3^x}\)=81
Ta có:27x=27x27x27x...x27 (x thừa số 27)
3x=3x3x3x3x...x3(x thừa số 3)
=>\(\frac{27^x}{3^x}\)=27x27x27x27x...27(x thừa số 27):3x3x3x3x...x3(x thừa số 3)
=9x9x9x9x...x9(x thừa số 9)
mà 81=92
=> x=2
Vậy x=2
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)
\(x\times4+\frac{40}{81}=\frac{51}{81}\)
\(x\times4=\frac{51}{81}-\frac{40}{81}\)
\(x\times4=\frac{11}{81}\)
\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{11}{324}\)
[ 61 + ( 53 - x ) ] . 17 = 1785
61 + ( 53 - x ) = 1785 : 17
61 + ( 53 - x ) = 105
( 53 - x ) = 105 - 61
53 - x = 44
=> x = 53 - 44
=> x = 9
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
\(\left(x+3\right)^3=\frac{81}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=3\)
\(\Leftrightarrow x+3=\sqrt[3]{3}\)
\(\Leftrightarrow x=-3+\sqrt[3]{3}\)