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7 tháng 7 2019
\(a,2x-138=2^3:\left(-3\right)^2\)
\(\Rightarrow2x-138=8:9\)
\(\Rightarrow2x=\frac{8}{9}+138\)
\(\Rightarrow2x=\frac{1250}{9}\)
\(\Rightarrow x=\frac{626}{9}\)
7 tháng 7 2019
\(10+2x=\left(-4\right)^5:\left(-4\right)^3\)
\(10+2x=-1024:\left(-64\right)\)
\(10+2x=16\)
\(2x=16-10\)
\(2x=6\)
\(x=6:2=3\)
MS
6 tháng 6 2016
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
6 tháng 6 2016
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
21 tháng 7 2022
a: =>15-(x-2)=-13-27=-40
=>x-2=15+40=55
hay x=57
b: =>5-x=-114+12=-102
=>x=107
c: \(\Leftrightarrow\left|x\right|=-1-5=-6\)(vô lý)
d: \(\Leftrightarrow\left|x-3\right|=3\)
=>x-3=3 hoặc x-3=-3
=>x=6 hoặc x=0
3 tháng 9 2018
a) 25 : x = x
25 = x^2
x^2 = ( +-5 )^2
b) 3358 : 23 = 2x - 6
146 = 2x - 6
2x = 152
x = 76
c) ( 2x + 1 )^3 = 27 = 3^3
=> 2x + 1 = 3
=> 2x = 2
=> x = 1
d) ( x - 2 )^3 = ( x - 2 )^2
( x - 2 )^2 . ( x - 2 ) - ( x -2 )^2 = 0
( x - 2 )^2 . [ ( x - 2 ) - 1 ] = 0
+) x - 2 = 0
=> x = 2
+) x - 2 - 1 = 0
x - 3 = 0
x = 3
3 tháng 9 2018
\(25\div x=x\Rightarrow x.x=25\Rightarrow x^2=25\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
\(3358\div23=2x-6\)
\(\Rightarrow2x-6=146\)
\(\Rightarrow2x=152\)
\(\Rightarrow x=\frac{152}{2}=76\)
\(\left(2x+1\right)^3=27\)
Mà \(3^3=27\)
Nên \(\left(2x+1\right)^3=3^3\)
\(\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
\(\left(x-2\right)^3=\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)^3-\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right)^2.\left(x-2-1\right)=0\)
\(\Rightarrow\left(x-2\right)^2.\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
Vậy......................
KL
8 tháng 8 2017
1. a, 3x + 2 \(⋮2x-1\)
Có 3(2x - 1) \(⋮2x-1\)
Và 2(3x - 2) \(⋮2x-1\)
=> 6x - 4 - 6x + 3 \(⋮2x-1\)
<=> -1 \(⋮2x-1\)
=> 2x - 1 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> 2x = 2; 0
=> x = 1; 0 (thỏa mãn)
@Lớp 6B Đoàn Kết
KL
8 tháng 8 2017
1. b, x2 - 2x + 3 \(⋮x-1\)
<=> x(x - 2) + 3 \(⋮x-1\)
<=> x(x - 1) - x + 3 \(⋮x-1\)
<=> x(x - 1) - (x - 1) - 2 \(⋮x-1\)
<=> (x - 1)2 - 2 \(⋮x-1\)
<=> -2 \(⋮x-1\)
=> x - 1 \(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x = 2; 0; 3; -1 (thỏa mãn)
@Lớp 6B Đoàn Kết
20 tháng 7 2018
a/ \(\dfrac{1}{3}+\dfrac{1}{2}:x=4\)
\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{2x}=4\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{11}{3}\)
\(\Leftrightarrow22x=3\)
\(\Leftrightarrow x=\dfrac{3}{22}\)
Vậy ...
b/ \(\left(x-1\right)^3=27\)
\(\Leftrightarrow\left(x-1\right)^3=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
Vậy ...
c/ \(\left(2x+1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^2=4^2\\\left(2x+1\right)^2=\left(-4\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=4\\2x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy..
d/ \(\dfrac{x+3}{2}=\dfrac{3}{5}\)
\(\Leftrightarrow5\left(x+3\right)=6\)
\(\Leftrightarrow5x+15=6\)
\(\Leftrightarrow x=-\dfrac{9}{5}\)
Vậy..