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a, ( 1+x )^3 = (2x)^3
b, ( x-1 )^2=16
c, (x+1)^2=25
d, 4x^3+15=47
e,(2x-1)^5=x^5
Mn giải nhanh giúp mk vs
a,\(\left(1+x\right)^3=\left(2x\right)^3\)
=>\(1+x=2x\)
=>\(x-2x=-1\)
=>\(-x=-1\)
=>\(x=1\)
vậy \(x=1\)
b,\(\left(x-1\right)^2=16\)
=>\(\left(x-1\right)^2=4^2\)
=>\(x-1=4\)
=>\(x=4+1\)
=>\(x=5\)
Vậy\(x=5\)
c,\(\left(x+1\right)^2=25\)
=>\(\left(x+1\right)^2=5^2\)
=>\(x+1=5\)
=>\(x=5-1\)
=>\(x=4\)
Vậy \(x=4\)
d,\(4x^3+15=47\)
=>\(4x^3=47-15\)
=>\(4x^3=32\)
=>\(x^3=32:4\)
=>\(x^3=8\)
=>\(x^3=2^3\)
=>\(x=2\)
Vậy\(x=2\)
e,\(\left(2x-1\right)^5=x^5\)
=>\(2x-1=x\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy\(x=1\)
ĐÚNG K MÌNH NHA
a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
1.
a) \(2^x=128\)
\(2^x=2^7\)
\(=>x=7\)
b) \(8^{x-1}=64\)
\(8^{x-1}=8^2\)
\(=>x-1=2\)
\(x=2+1\)
\(=>x=3\)
c) \(3+3^x=30\)
\(3^x=30-3\)
\(3^x=27=3^3\)
\(=>x=3\)
d) \(\left(x+2\right)=64\) -> đề có thiếu không vậy?
e) \(3^2.x=3^5\)
\(x=3^5:3^2\)
\(=>x=3^3=27\)
f) \(\left(2x-1\right)^3=343\)
\(\left(2x-1\right)^3=7^3\)
\(=>2x-1=7\)
\(2x=7+1\)
\(2x=8\)
\(x=8:2\)
\(=>x=4\)
\(#Wendy.Dang\)
a,\(2^x\)=128 b,\(8^{x-1}\)=64 c,3+\(3^x\)=30 d,x+2=64
\(2^7\)=128 \(8^{x-1}\)=\(8^2\) \(3^x\)=30-3 x=64-2
=>x=7 =>x-1=2 \(3^x\)=27 x=62
x=2+1=3 \(3^x\)=\(3^3\)
=>x=3
e,\(3^2\).x=\(3^5\) f,(2x-\(1^3\))=343
x=\(3^5\):\(3^2\) 2x=1+343
x=27 2x=344
x=344:2
x=172
a,(x+1)-(x+2)-(x+3)=24
=>x+1-x-2-x-3 =24
=>(x-x-x)+(1-2-3) =24
=> -x-4 =24
=> -x =24+4
=> -x =28
=> x =-28
Vậy x=-28
b,4x+2-3(x-1)=3x-5
=>4x+2-3x+3=3x-5
=>3x-4x+3x =2+3+5
=>2x =10
=>x =5
Vậy x=5
c,x-1-2(x-2)=x-11
=>x-1-2x+4=x-11
=>x-2x-x =-11+1-4
=>-2x =-14
=>x =7
Vậy x = 7
Ko cần đâu bn à mk mong bn đấy
a)\(\left(3x-1\right)\left(5-\frac{1}{2}x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)
b)\(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)
a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0 hay \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x = 1 I\(\Leftrightarrow\)\(\frac{-1}{2}\)x = -5
\(\Leftrightarrow\) x = \(\frac{1}{3}\) I\(\Leftrightarrow\) x = 10
b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\) I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\) hay \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{29}{24}\) I\(\Leftrightarrow\)\(\frac{1}{2}x\) = \(\frac{-13}{24}\)
\(\Leftrightarrow\) x = \(\frac{29}{12}\) I\(\Leftrightarrow\) x = \(\frac{-13}{12}\)
c) (2x +\(\frac{3}{5}\))2 - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))2 = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\) = \(\frac{3}{5}\) hay 2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x = 0 I \(\Leftrightarrow\)2x = \(\frac{-6}{5}\)
\(\Leftrightarrow\) x = 0 I \(\Leftrightarrow\) x = \(\frac{-3}{5}\)
d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
Bài 1:
a) =25,97+(6,54+103,46)
=25,97+110
=135,97
b)136x75+75x64
=75x(136+64)
=75x200
=15 000
c) (21/8+1/2):5/16
=(21/8+4/8)x16/5
=25/8x16/5
=10
d)3/17-4/5+14/17
=(3/17+14/17)-4/5
=1-4/5
=1/5
Bài 2:
a)720:\([41-(2x-5)]\)=120
41 - (2x-5) =720:120
41 - (2x-5) =6
2x-5 =41-6
2x-5 =35
2x =35+5
2x =40
x =40:2
x =20
b)2/3 x X +3/4=3
2/3 x X =3-3/4
2/3 x X =12/4-3/4
2/3 x X =9/4
x =9/4:2/3
x =9/4x3/2
x =27/8
c) x+0,34=1,19x1,02
x+0,34=1,2138
x =1,2138-0,34
x =0,8738
a) \(2^x=16\) b) \(3^{2x-1}=243\)
\(\Rightarrow2^4=16\) \(2x-1=243:3\)
Vì : 2 . 2 . 2 . 2 = 16 \(2x-1=81\)
Vậy x = 4 \(2x=81+1\)
\(2x=82\)
=> x = \(\varnothing\)
b. \(2x=\dfrac{-1}{5}\)
\(x=-0,25\)
d. \(\dfrac{2}{x+1}=\dfrac{x+1}{16}\)
\(\Leftrightarrow\left(x+1\right)^2=32\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\sqrt{2}\\x+1=-4\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\sqrt{2}-1\\x=-4\sqrt{2}-1\end{matrix}\right.\)