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 \(x=-\dfrac{1}{2}\)

a. \(\left(x-\dfrac{1}{2}\right)^2=0\)

<=> \(x-\dfrac{1}{2}=0\)

<=> \(x=\dfrac{1}{2}\)

b. (x - 2)² = 1

<=> (x - 2)² - 1² = 0

<=> (x - 2 - 1)(x - 2 + 1) = 0

<=> (x - 3)(x - 1) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c. (2x - 1)³ = -8

<=> (2x - 1)³ + 2³ = 0

<=> (2x - 1 + 2)\(\left[\left(2x-1\right)^2-2\left(2x-1\right)+2^2\right]=0\)

<=> (2x + 1)(4x² - 4x + 1 - 4x + 2 + 4) = 0

<=> (2x + 1)(4x² - 8x + 7) = 0

<=> \(\left[{}\begin{matrix}2x+1=0\\4x^2-8x+7=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\S=\varnothing\end{matrix}\right.\)

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

<=> \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{\left(\sqrt{6}\right)^2}=0\)

<=> \(\left(x+\dfrac{1}{2}-\dfrac{1}{\sqrt{6}}\right)\left(x+\dfrac{1}{2}+\dfrac{1}{\sqrt{6}}\right)=0\)

<=> \(\left(x+\dfrac{3-\sqrt{6}}{6}\right)\left(x+\dfrac{3+\sqrt{6}}{6}\right)=0\)

<=> \(\left[{}\begin{matrix}x+\dfrac{3-\sqrt{6}}{6}=0\\x+\dfrac{3+\sqrt{6}}{6}=0\end{matrix}\right.\)

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

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(x-\frac{1}{2}=0\)

\(x=0+\frac{1}{2}\)

\(x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(\left(x-2\right)^2=1^2\)

\(x-2=1\)

\(x=1+2\)

\(x=3\)

c) \(\left(2x-1\right)^3=\left(-8\right)\)

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

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

\(2x=\left(-2\right)+1\)

\(2x=-1\)

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

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)

\(x+\frac{1}{2}=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{1}{2}\)

\(x=-\frac{1}{4}\)

a) \(\left(x-\frac{1}{2}\right)^2=0\)

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

\(\Leftrightarrow x=\frac{1}{2}\)

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

c) \(\left(2x-1\right)^2=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)

\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

 \(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)

\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)

\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)

\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)

\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)

Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)

c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z

=> -2x ⋮ x + 1

=> -2x - 2 + 2 ⋮ x + 1

=> -2( x + 1 ) + 2 ⋮ x + 1

Vì -2( x + 1 ) ⋮ ( x + 1 )

=> 2 ⋮ x + 1

=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }

Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Vậy x ∈ { -3 ; -2 ; 0 ; 1 }

a) \(\left(x-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)

b) \(\left(x-2\right)^2=1\)

\(\Leftrightarrow\left(x-2\right)^2-1=0\)

\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

c) \(\left(2x-1\right)^3=-8\)

\(\Leftrightarrow2x-1=-2\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=\frac{-1}{2}\)

d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)

\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)

\(\Rightarrow x=-\frac{1}{4}\)

a>(x-1/2)^2=0

     (x-1/2)^2=0^2

=>x-1/2=0

   x=1/2

b>(x-2)^2=1

     (x-2)^2=1^2

    =>x-2=1

           x=3

(2x-1)^3=-8

=>(2x-1)^3=-2^3

  =>2x-1=-2

        2x=-1

         x=-0,5

d>(x+1/2)^2=1/16

   (x+1/2)^2=(1/4)^2

   =>x+1/2=1/4

             x=1/4-1/2

             x=-1/4

chuc ban may man trong cuoc song!!!!!

Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà

a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)

\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)

b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)

c Tương tự b

2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)

\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)

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