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) \(\left(x-\frac{1}{2}\right)^2=0\)

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

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!!!!!

a, [x+1]² + [y+5]² = 16

Theo đề, ta có: 0 \(\le\)[x+1]² \(\le\)16; 0\(\le\)[y+5]² \(\le\)16

Dễ dàng nhận thấy [x+1]² và [y+5]² là hai số chính phương, mà từ 0 - 16 chỉ có hai số chính phương 0 và 16 là có tổng là 16

=> Có hai trường hợp:

* \(\hept{\begin{cases}\left[x+1\right]^2=0\\\left[y+5\right]^2=16\end{cases}\Rightarrow}\hept{\begin{cases}x+1=0\\\hept{\begin{cases}y+5=4\\y+5=-4\end{cases}}\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases};}\hept{\begin{cases}x=-1\\y=-9\sqrt[]{}\sqrt[]{}\end{cases}}}\)

ghi câu hỏi rõ bạn ơi

Bài 1 : Tính nhanh

a) 16.(38−2)−38(16−1)16.(38−2)−38(16−1)

b) (−41).(59+2)+59(41−2)(−41).(59+2)+59(41−2)

Bài 2 :

Tìm các số x ; y ; x biết rằng :

x + y = 2 ;  y + z = 3 ;  z + x = -5

Bài 3 : Tìm x ; y ∈∈ Z biết rằng :

( y + 1 ) . xy - 1 ) = 3

a) 16.(38−2)−38(16−1)

= 16.38 - 16.2 - 38.16 - 38.1

=-16.2 - 38.1

=-32-38

=-70

a,\((x+4)^2-(x+1)(x-1)=16\)

 \(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)

b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)

\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)

\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)

\(\Rightarrow -x^2-10x-244=0\)

\(\Rightarrow -(x^2-10x+25)-219=0\)

\(\Rightarrow -(x-5)^2-219=0\)

\(\Rightarrow (x-5)^2+219=0\)

Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm

\(< =>\left(\left(x+2\right)^3-\left(x-2\right)^3\right)-16-108=0\)

\(< =>\left(x+2-x+2\right)\left(\left(x+2\right)^2+\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\right)-16-108=0\)

\(< =>4\left(x^2+4x+4+x^2-4+x^2-4x+4\right)-16-108=0\)

\(< =>4\left(3x^2+4\right)-16-108=0\)

\(< =>12x^2+16-16-108=0\)

\(< =>12x^2-108=0\)

\(< =>12\left(x^2-9\right)=0\)

\(< =>x^2-9=0\)

\( < =>\left(x-3\right)\left(x+3\right)=0\)

\(< =>\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}< =>\orbr{\begin{cases}x=3\\x=-3\end{cases}}}\)

(x+2)³-(x-2)³-16=108

x³+3x².2+3.x.2²+2³-x³+3x².2-3.x.2²+2³=124

(x³-x³)+(6x²+6x²)+(12x-12x)+(8+8)=124

12x²+16=124

12x²=108

x²=9

x=-3 hoặc x=3

x³ +6x² + 12x + 8 - x³ +6x² - 12x +8 -16 = 108

12x²  = 108

x²  = 9

x = +_- 3

a) \(25x^2-9=0\)

\(\Leftrightarrow\left(5x\right)^2-3^2=0\)

\(\Leftrightarrow\left(5x+3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)

Vậy \(S=\left\{\frac{3}{5};\frac{-3}{5}\right\}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x+17=16\)

\(\Leftrightarrow8x=-1\)

\(\Leftrightarrow x=-\frac{1}{8}\)

Vậy.........

c)\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

\(\Leftrightarrow x=-127,5\)

Vậy.............

có j sai xót mong m.n bỏ qua☺

a) \(25x^2-9=0\)                      

<=> \(\left(5x\right)^2=9\)

<=> \(\left(5x\right)^2=3^2\)

<=> \(5x=3\)

<=> \(x=\frac{3}{5}\)

b) \(\left(x+4\right)^2-\left(x-1\right)\left(x+1\right)=16\)

<=> \(x^2+2.x.4+4^2-\left(x^2-1^2\right)=16\)

<=> \(x^2+8x+16-x^2+1=16\)

<=> \(\left(x^2-x^2\right)+8x+\left(16+1\right)=16\)

<=> \(8x+17=16\)

<=> \(8x=-1\)

<=> \(x=\frac{-1}{8}\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(\left(2x\right)^2-2.2x.1+1^2+x^2+2.x.3+3^2-5\left(x^2-7^2\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+5.7^2=0\)

<=> \(\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+5.7^2\right)=0\)

<=> \(2x+245=0\)

<=> \(2x=-245\)

<=> \(x=\frac{-245}{2}\)

a) \(25x^2-9=0\)

\(\Rightarrow25x^2-3^2=0\)

\(\Rightarrow\left(25x+3\right).\left(25x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}25x+3=0\\25x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}25x=-3\\25x=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{25}\\x=\frac{3}{25}\end{cases}}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow\left(x+4\right)^2-\left[\left(x+1\right)^2-\left(x-1\right)^2\right]=16\)

\(\Rightarrow\left(x+4\right)^2=16\)

\(\Rightarrow\left(x+4\right)^2=4^2\)

\(\Rightarrow x+4=4\)

\(\Rightarrow x=0\)

c) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Rightarrow\left(2x-1\right)+2\left(2x-1\right)\left(x+3\right)+\left(x+3\right)^2-5.\left(x+7\right)^2-\left(x-7\right)^2=0\)

\(\Rightarrow2\left(x+3\right)^3=0\)

\(\Rightarrow\left(x+3\right)^2=0\)

\(\Rightarrow x=3\)