a) \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)

b) \(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

c) \(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Leftrightarrow2x-1=-3\Leftrightarrow x=-1\)

d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{6}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=-\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)

b: Ta có: \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c: Ta có: \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

hay \(x=-\dfrac{1}{2}\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)

\(1,\\ a,\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow\left[{}\begin{matrix}x-4=4\\x-4=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=0\end{matrix}\right.\\ c,\Leftrightarrow2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\\ 2,\\ a,=1\\ b,=\left(\dfrac{13}{4}\right)^2=\dfrac{169}{16}\\ c,=\left(-\dfrac{7}{4}\right)^2=\dfrac{49}{16}\\ d,=\left(\dfrac{3}{7}\right)^{20}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^8=...\\ e,=\left(3\cdot5\cdot\dfrac{2}{3}\right)^2=10^2=100\)

\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)

a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)

=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)

=>\(6x-\dfrac{39}{4}=1\)

=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)

=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)

b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)

=>\(2x-6=3x+6-x+1\)

=>2x-6=2x+7

=>-6=7(vô lý)

c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)

=>\(x^2+3x+x^2-2x=2x^2-2x\)

=>3x-2x=-2x

=>3x=0

=>x=0

d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)

=>\(3x^2-3x-2x-4-2x=x^2-x\)

=>\(3x^2-7x-4-x^2+x=0\)

=>\(2x^2-6x-4=0\)

=>\(x^2-3x-2=0\)

=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

Em hc bảng xét dáu chx ??

Lớp 7 chưa học đâu em

aanh giúp em câu hóa này vs

a) \(2x+\frac{3}{15}=\frac{7}{5}\) 

=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)

=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)

b) \(x-\frac{2}{9}=\frac{8}{3}\)

=> \(x=\frac{8}{3}+\frac{2}{9}\)

=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)

c) \(\frac{-8}{x}=\frac{-x}{18}\)

=> x(-x) = (-8).18

=> -x² = -144

=> x² = 144(bỏ dấu âm)

=> x = \(\pm\)12

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)

=> 5(2x + 3) = 6(x - 2)

=> 10x + 15 = 6x - 12

=> 10x + 15 - 6x + 12 = 0

=> 4x + 27 = 0

=> 4x = -27

=> x = -27/4

e) \(\frac{x+1}{22}=\frac{6}{x}\)

=> x(x + 1) = 132

=> x(x + 1) = 11.12

=> x = 11

f) \(\frac{2x-1}{2}=\frac{5}{x}\)

=> x(2x - 1) = 10

=> 2x² - x = 10

=> 2x² - x - 10 = 0

tới đây tự làm đi nhé

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)

=> (2x - 1)(2x + 1) = 63

=> 4x² - 1 = 63

=> 4x² = 64

=> x² = 16

=> x = \(\pm\)4

h) Tương tự

a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)

b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)

c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)

e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)

g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)