(0,4x - 1,3)2 = 5,29
Bạn ơi, giúp mình bài này với, cảm ơn!
(0,4x-1,3)2=5,29
(0,4x - 1,3)^2 = 5,29
giúp me với ai nhanh thì mk sẽ tk cho 3 tk đó
\(\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=\left(\pm2,3\right)^2\)
+) 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
=> \(\orbr{\begin{cases}0,4x-1,3=2,3\\0,4x-1,3=-2,3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}0,4x=3,6\\0,4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-\frac{5}{2}\end{cases}}\)
\(\left(0,4x-1,3\right)^2=5,29\)
Ta xét 2th:
Th1: \(0,4x-1,3=\sqrt{5,29}\)
\(0,4x-1,3=\sqrt{5,29}+1,3\)
=> x = 9
Th2: \(0,4x-1,3=-\sqrt{5,29}\)
\(0,4x-1,3=-\sqrt{5,29}+1,3\)
=> x = -2,5
\(\Rightarrow\orbr{\begin{cases}x=9\\x=-2,5\end{cases}}\)
a) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\\ \)
b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)
c) \(\left(0,4x-1,3\right)^2=5,29\)
Tìm x,y biết :
a)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
b)\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
c)\(\left(0,4x-1,3\right)^2=5,29\)
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
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:Tìm số nguyên n
a) (1/3)n=1/81 b) -512/343=(-8/7)n c)(-3/4)n=81/256 d)64/(-2)n=(-2)3
Bài 11:Tìm x,y biết
a) (0,4x-1,3)2=5,29
b) (3/5 -2/3x)3=-64/125
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
Tìm x , y:
a , \(\dfrac{2}{3}.3^{x+1}-7.3^x=405\)
b , \(\left(0,4x-1,3\right)^2=5,29\)
c , \(5.2^{x+1}.2^{-2}-2^x=384\)
d , \(4^x+4^{x+3}=4160\)
e , \(2^{x-1}+5.2^{x-2}=\dfrac{7}{32}\)
a: \(\Leftrightarrow3^x\cdot\left(\dfrac{2}{3}\cdot3-7\right)=405\)
\(\Leftrightarrow3^x=-81\)(vô lý)
b: \(\left(0,4x-1,3\right)^2=5,29\)
=>0,4x-1,3=2,3 hoặc 0,4x-1,3=-2,3
=>0,4x=3,6 hoặc 0,4x=-1
=>x=9 hoặc x=-2,5
c: \(5\cdot2^{x+1}\cdot2^{-2}-2^x=284\)
\(\Leftrightarrow2^x\cdot5\cdot2\cdot2^{-2}-2^x=284\)
\(\Leftrightarrow2^x\cdot\left(\dfrac{5}{2}-1\right)=284\)
\(\Leftrightarrow2^x=\dfrac{568}{3}\)(vô lý)
d: \(\Leftrightarrow4^x\left(1+4^3\right)=4160\)
\(\Leftrightarrow4^x=64\)
hay x=3
Tìm x , y :
a , \(\frac{2}{3}.3^{x+1}-7.3^x=-405\)
b , \(\left(0,4x-1,3\right)^2=5,29\)
c , \(5.2^{x+1}.2^{-2}-2^x=384\)
d , \(3^{x+2}.5^y=45^x\)
e , \(4^x+4^{x+3}=4160\)
f , \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
a/ \(\frac{2}{3}.3^{x+1}-7.3^x=405\)
<=> 2.3x-7.3x=-405
<=> 5.3x=405
<=> 3x=81 = 34
=> x=4
b/ (0,4x-1,3)2=5,29=(2,3)2
=> \(\hept{\begin{cases}0,4x-1,3=2,3\\0,4x-1,3=-2,3\end{cases}}\)=> \(\hept{\begin{cases}x=9\\x=-\frac{5}{2}\end{cases}}\)
c/ 5.2x+1.2-2-2x=384
<=> 5.2x-1-2.2x-1=384
<=> 3.2x-1=384
<=> 2x-1=128=27
=> x-1=7 => x=8
d/ 3x+2.5y=45x
<=> 3x+2.5y=32x.5x
=> \(\hept{\begin{cases}x+2=2x\\x=y\end{cases}}\)=> x=y=2
(0,4-1,3)2=5,29
(x-3)10=(x-3)30
(x+1,5)8+(2,7-y)12=0
a, đề sai
b) \(\left(x-3\right)^{10}=\left(x-3\right)^{30}\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=-1\\x-3=0\\x-3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)
Vậy .....
c) \(\left(x+1,5\right)^8+\left(2,7-y\right)^{12}=0\)
Vì: \(\left\{{}\begin{matrix}\left(x+1,5\right)^8\ge0\forall x\\\left(2,7-y\right)^{12}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\) để bt = 0
=>\(\left\{{}\begin{matrix}\left(x+1,5\right)^8=0\\\left(2,7-y\right)^{12}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+1,5=0\\2,7-y=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1,5\\y=2,7\end{matrix}\right.\)
Vậy.............
Tìm x : 60% + 0,4x = 2
\(x:60\%+0,4x=2\)
\(\Rightarrow\dfrac{5}{3}x+0,4x=2\)
\(\Rightarrow\left(\dfrac{5}{3}+0,4\right)x=2\)
\(\Rightarrow\dfrac{31}{15}x=2\)
\(\Rightarrow x=2:\dfrac{31}{15}\)
\(\Rightarrow x=\dfrac{30}{31}\)
#DarkPegasus
60% + 0,4x=2
0,4x=2-60%
0,4x=2-0,6
0,4x=1,4
x=1,4:0,4
x=3,5
Vậy:...
nhớ ticks nhe
\(\dfrac{60}{100}+0,4x=2\)
\(\dfrac{3}{5}+0,4x=2\)
\(0,4x=2-\dfrac{3}{5}\)
\(0,4x=\dfrac{7}{5}\)
\(x=\dfrac{7}{5}:0,4\)
\(x=3,5\)