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a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vũ Hồng Linh bạn check lại bài đầu dùm =_="
\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
Bài 2:
x=13 nên x+1=14
\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)
\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)
=14-x=1
x=13 nên x+1=14
f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14
=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14
=14-x=1
\(\frac{1}{\left(x-1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+7\right)}+....+\frac{1}{\left(x+76\right).\left(x+80\right)}=-\frac{81}{320}\)
\(\Rightarrow\frac{4}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x+3\right)\left(x+7\right)}+...+\frac{4}{\left(x+76\right)\left(x+80\right)}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+7}+...+\frac{1}{x+76}-\frac{1}{x+80}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+90}=\frac{-81}{80}\)
Vì \(\frac{-81}{80}< 0\Rightarrow\frac{1}{x-1}< \frac{1}{x+90}\)
\(\Leftrightarrow x-1>x+90\)( luôn sai \(\forall x\in R\))
Vậy không tìm được x
-1/33 . x = 1/81
-1/27 . x = 1/81
x = 1/81 : (-1/27)
x = 1/81 . (-27)
x = -1/3
Vậy x = -1/3
(x+1/2)mũ 3 : 3 = -1/81
(x+1/2)mũ 3 = -1/81 . 3
(x+1/2)mũ 3 = -1/27
(x+1/2)mũ 3 = (-1/3)mũ3
x+1/2 = -1/3
x = -1/3 - 1/2
x = -5/6
`(x+1/2)^3 = -1/81 xx 3`
`(x+1/2)^3 = -1/27`
`x + 1/2 = -1/3`
`x = -1/3 - 1/2 = -5/6`
(1/30^x=(1/3)^3
x=3