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câu1
(3x-1).(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
=>x=1/3 =>x=0
câu2
1/4+1/3 :(2x-1)=5
=> 1/3:(2x-1)=19/4
=>2x-1 =57/4
=>2x=61/4
=>x=61/8
còn hai câu sau bn ghi đề mik ko hỉu
1.
a)(3x-1)(1/2x5)=0
=>3x-1=0 hoặc 1/2x5=0
3x=0+1 x=0:1/2:5
x=1/3 x=0
Vậy x=1/3 hoặc x=0
b)1/4+1/3:(2x-1)=5
1/3:(2x-1)=5-1/4=20/4-1/4=19/4
2x-1=1/3:19/4=1/3*4/19=4/57
2x=4/57+1=4/57+57/57=61/57
x=61/57:2=61/57*1/2=61/114
Vậy x=61/114
c)(2x+2/5)2-9/25=0=02-9/25
=>2x+2/5=0
2x=0-2/5
x=-2/5:2=-2/5*1/2
x=-1/5
Vậy x=-1/5
d)(3x-1/2)3+1/9=0=03+1/9
=>3x-1/2=0
3x=0+1/2
x=1/2:3=1/2*1/3
x=1/6
Vậy x=1/6

a) Vì x+1 chia hết cho x+1 => 2x+2 chia hết cho x+1
Ta có 2x+5 chai hết cho x+1
=> 2x+2+3 chia hết cho x+1
=> 3 chia hết cho x+1 ( do 2x+2 chia hết cho x+1)
=> x+1 thuộc Ư(3)={1;3}
=> x thuộc { 0:2}
Thử lại : x=0. Có 5 chia hết cho 1 (đúng )
x=2 . Có 9 chia hết cho 3 ( đúng )
b)

Con " Nguyễn Huyền Trang " đéo biết thì trả lời làm cái l*n gì

a)
Để \(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)=> 3x-1=0 hoặc \(-\frac{1}{2}x+5=0\)
=> x= \(\frac{1}{3}\) hoăc \(x=10\)
b)
\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=5\) => \(\frac{1}{3}:\left(2x-1\right)=5-\frac{1}{4}=\frac{19}{4}=>2x-1=\frac{1}{3}:\frac{19}{4}=\frac{4}{57}=>x=\frac{61}{114}\)
c) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0=>\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)\(=>2x+\frac{3}{5}\in\left\{\pm\frac{3}{5}\right\}=>2x\in\left\{0;\frac{-6}{5}\right\}=>x\in\left\{0;\frac{-3}{5}\right\}\)
d) Xem lại đề
a) để (3x-1).(\(-\dfrac{1}{2}x+5\))=0
=> 3x-1 hoặc \(-\dfrac{1}{2}x+5\) =0
TH1 : 3x-1=0
3x = 0+1=1
x = 1:3 = \(\dfrac{1}{3}\)
TH2 : \(-\dfrac{1}{2}x+5\)= 0
\(-\dfrac{1}{2}x\)= 0 -5 = -5
x= -5 : \(-\dfrac{1}{2}\)
x= 10

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.

1,
a,\(2020-\left(249+2020\right)+\left(249-573\right)\)
\(=2020-249-2020+249-573\)
\(=-573\)
b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)
\(=257+1-18-257\)
\(=1-18=-17\)
\(c,25.\left(85-47\right)-85.\left(47+25\right)\)
\(=25.85-25.47-47.85+85.25\)
\(=85.\left(25-47+25\right)-25.47\)
\(=85.3-25.47\)
\(=-920\)
2,
\(a,15-5.\left(x+2\right)=-30\)
\(=>5.\left(x+2\right)=15+30=45\)
\(=>x+2=\frac{45}{5}=9\)
\(=>x=7\)
\(b,\left(x+2\right)^2+5=105\)
\(=>\left(x+2\right)^2=100\)
\(=>\left(x+2\right)^2=10^2\)
\(=>x+2=10\)
\(=>x=8\)
\(c,\left|2x-5\right|-\left(-6\right)=11\)
\(=>\left|2x-5\right|=11-6=5\)
\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)
\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)

( 3x - 1 )^2 = 25 = (+-5)^2
+) 3x - 1 = 5
3x = 6
x = 2
+) 3x - 1 = -5
3x = -4
x = -4/3
Vậy,.........
Các câu còn lại tương tự
2x + 3x = \(5^6\): 25
5x = \(5^4\)
5x = 625
x = 625 : 5
x = 125
\(2x+3x=5^6:25\)
\(2x+3x=5^6:5^2\)
\(2x+3x=5^4\)
\(\Leftrightarrow x.5=5^4\)
\(\Leftrightarrow x=5^3\)