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\(\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=\left(\pm2,3\right)^2\)
+) 0,4x - 1,3 = 2,3
0,4x = 3,6
x = 9
+) 0,4x - 1,3 = -2,3
0,4x = -1
x = -2,5
Vậy,........
(0,4x - 1,3)2 = 5,29
(0,4x - 1,3)2 = 2,32
=> \(\orbr{\begin{cases}0,4x-1,3=2,3\\0,4x-1,3=-2,3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0,4x=3,6\\0,4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-\frac{5}{2}\end{cases}}\)
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
TH1: \(0,4x-1,3=2,3\)
\(0,4x=3,6\)
\(x=9\)
TH2: \(0,4x-1,3=-2,3\)
\(0,4x=-1\)
\(x=-\dfrac{5}{2}\)
=.= hok tốt!!
Bài 10:
a) (1/3)n = 1/81
=> (1/3)n = (1/3)4
=> n = 4
b) -512/343 = (-8/7)n
=> (-8/7)3 = (-8/7)n
=> 3 = n (hay n = 3)
c) (-3/4)n = 81/256
=> (-3/4)n = (-3/4)4
=> n = 4
d) 64/(-2)n = (-2)3
=> 64/(-2)n = -8
=> (-2)n = -8
=> (-2)n = (-2)3
=> n = 3
Bài 11: (không có y để tìm nhé)
a) (0,4x - 1,3)2 = 5,29
=> (0,4x - 1,3)2 = (2,3)2
=> 0,4x - 1,3 = 2,3
=> 0,4x = 3,6
=> x = 9
b) (3/5 - 2/3x)3 = -64/125
=> (3/5 - 2/3x)3 = (-4/5)3
=> 3/5 - 2/3x = -4/5
=> 2/3x = 7/5
=> x = 21/10
(2,4x2 + 1,7y2 +2xy) - (0,4x2 -1,3y2 +xy)
= 2,4x2 + 1,7y2 +2xy - 0,4x2 +1,3y2 -xy
= (2,4x2 - 0,4x2 ) + (1,7y2 + 1,3y2 ) +(2xy - xy)
=2x2 + 3y2 +xy
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\\dfrac{2}{5}x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5:\dfrac{2}{5}=-\dfrac{25}{2}\end{matrix}\right.\)