Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
a: \(575-\left(2x+70\right)=445\)
=>\(2x+70=575-445=130\)
=>\(2x=130-70=60\)
=>x=60/2=30
b: \(575-2\left(x+70\right)=445\)
=>\(2\left(x+70\right)=575-445=130\)
=>x+70=130/2=65
=>x=65-70=-5
c: \(x^5=32\)
=>\(x^5=2^5\)
=>x=2
d: \(\left(3x-1\right)^3=8\)
=>\(\left(3x-1\right)^3=2^3\)
=>3x-1=2
=>3x=3
=>\(x=\dfrac{3}{3}=1\)
e: \(\left(x-2\right)^3=27\)
=>\(\left(x-2\right)^3=3^3\)
=>x-2=3
=>x=5
f: \(\left(2x-3\right)^2=9\)
=>\(\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
g: \(2x+5=3^4:3^2\)
=>\(2x+5=3^2\)
=>2x+5=9
=>2x=9-5=4
=>x=4/2=2
h: \(\left(4x-5^2\right)\cdot7^3=7^4\)
=>\(4x-25=\dfrac{7^4}{7^3}=7\)
=>4x=25+7=32
=>\(x=\dfrac{32}{4}=8\)
ta có:
\(\left(7\frac{1}{2}.8\frac{3}{70}+8\frac{3}{70}.\frac{9}{4}+\frac{19}{4}.8\frac{3}{70}+5\frac{1}{2}.8\frac{3}{70}\right):x=1126\)
\(8\frac{3}{70}\left(7\frac{1}{2}+\frac{9}{4}+\frac{19}{4}+5\frac{1}{2}\right):x=1126\)
\(\frac{563}{70}.\left(\frac{15}{2}+7+\frac{11}{2}\right):x=1126\)
\(\frac{563}{70}.20:x=1126\)
\(\frac{1126}{70}:x=1126\)
\(=>x=\frac{1126}{7}:1126\)
\(=>x=\frac{1}{7}\)
cho mình nha các bạn.
a) TH1: 2x - 1 > 0 => x > 1 nên ta có :
2x - 1 = x - 5
=> 2x - x = 1 - 5
=>x = -4 ( không thỏa mãn vì x phải > 1 )
TH2 : 2x - 1 < 0 = - ( 2x - 1 ) => x < 1 nên ta có :
- 2x + 1 = x - 5
=> - 2x - x = -1 - 5
=> -3 x = - 6
=> x = 2 ( không thỏa mãn )
Vậy phương trình trên vô nghiệm
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
3.(5x - 1) - 2 = 70
<=> 3.(5x - 1) = 70 + 2
<=> 3.(5x - 1) = 72
<=> 5x - 1 = 72 : 3
<=> 5x - 1 = 24
<=> 5x = 24 + 1
<=> 5x = 25
<=> 5x = 52
<=> x = 2
\(a,12\left(x-1\right)=0\\ x-1=0\\ x=1\\ b,45+5\left(x-3\right)=70\\ 5\left(x-3\right)=25\\ x-3=5\\ x=8\\ c,3.x-18:2=12\\ 3.x-9=12\\ 3.x=21\\ x=7\)
\(5^{x-1}\cdot3-2=70\)
\(5^{x-1}\cdot3=72\)
\(5^x\cdot5=72:3=24\)
đề hơi sai sai
5x-1x3-2=70
= 5x-1x3=70+2=72
= 5x-1=72/3=24
=> không có giá trị tương ứng của x
( 5^x - 1 ) . 3 - 2 = 70
( 5^x - 1 ) . 3 = 70 + 2
( 5^x - 1 ) . 3 = 72
( 5^x - 1 ) = 72 : 3
( 5^x - 1 ) = 24
5^x = 24 + 1
5^x = 25
5^x = 5^2
x = 2
Vậy x = 2
TK mình nha !!! Cảm ơn
(5x - 1 ).3 -2 =70
(5x -1 ).3 =70+2
(5x -1).3 =72
5x -1 =72:3
5x -1 = 24
5x = 24 +1
5x = 25
5x = 52
=) x=2
Vậy x=2
chúc các bạn học thật giỏi các môn nhé !