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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

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

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......................

\(a,2x-138=2^3:\left(-3\right)^2\)

\(\Rightarrow2x-138=8:9\)

\(\Rightarrow2x=\frac{8}{9}+138\)

\(\Rightarrow2x=\frac{1250}{9}\)

\(\Rightarrow x=\frac{626}{9}\)

\(10+2x=\left(-4\right)^5:\left(-4\right)^3\)

\(10+2x=-1024:\left(-64\right)\)

\(10+2x=16\)

\(2x=16-10\)

\(2x=6\)

\(x=6:2=3\)

25 tháng 8 2018

\(a,\frac{27}{x+1}=\frac{x+1}{3}\)

\(\Leftrightarrow27.3=\left(x+1\right)\left(x+1\right)\)

\(\Leftrightarrow81=\left(x+1\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=9^2\\\left(x+1\right)^2=\left(-9\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}}\)

Vậy \(x\in\left\{8;-10\right\}\)

25 tháng 8 2018

a)\(\frac{27}{x+1}=\frac{x+1}{3}\)

Ta có : 27 . 3 = (x + 1)(x + 1)

=> 81 = (x + 1)2

=> 92 = (x + 1)2

=> \(\orbr{\begin{cases}x+1=9\\x+1=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-10\end{cases}}\)

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

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

10 tháng 7 2016

b, \(\left(x+1\right)^2=1\)

Vì \(1^2=1=>\left(x+1=1\right)\)

=> x=0

câu a không có máy tính nhác làm

a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(=>\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(=>x-\frac{1}{2}=\frac{1}{3}\)

\(=>x=\frac{5}{6}\)

b) \(\left(x+1\right)^2=1\)

\(=>\orbr{\begin{cases}x+1=-1\\x+1=1\end{cases}=>\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)

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