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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-1/20)²=0

=> x-1/20=0

=> x=1/20

b. (x-2)²=1

=> (x-2)²=1²=(-1)²

+) x-2=1

=> x=3

+) x-2=-1

=> x=1

Vậy x \(\in\){1;3}

c. (2x-1)³=-8

=> (2x-1)³=(-2)³

=> 2x-1=-2

=> 2x=-1

=> x=-1/2

d. (x+1/2)²=1/16

=> (x+1/2)²=(1/4)²=(-1/4)²

+) x+1/2=1/4

=> x=-1/4

+) x+1/2=-1/4

=> x=-3/4

Vậy x \(\in\){-3/4; -1/4}

1) Các cách viết số 25 dưới dãng lũy thừa là: 25¹; 5²; (-5)²

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

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

b) (x - 2)² = 1

=> \(\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)

Vậy \(x\in\left\{3;1\right\}\)

c) (2x - 1)³ = -8

=> (2x - 1)³ = (-2)³

=> 2x - 1 = -2

=> 2x = -2 + 1

=> 2x = -1

=> \(x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}\)

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

=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{array}\right.\)

Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}\)

1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2

2) a)

=>

=>

Vậy

b) (x - 2)2 = 1

=> =>

Vậy

c) (2x - 1)3 = -8

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

=> 2x - 1 = -2

=> 2x = -2 + 1

=> 2x = -1

=>

Vậy

d)

=> =>

Vậy

\(a,5^x+5^{x+2}=650\Leftrightarrow5^x+5^x+5^2=650\Leftrightarrow5^x.26=650\Leftrightarrow5^x=5^2\Leftrightarrow x=2\) x=2

b,Với x=0 khi đó 3^0-1+5.3^0-1=2 (loại)

   Với x=1 khi đó 3^1+5.3^1=18 (loại)

   Với x=2 khi đó 5.3^x-1>16 (loại)

   Vậy không có x thỏa mãn

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.

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

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

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

b.

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

\(x-2=\pm1\)

TH1:

\(x-2=1\)

\(x=1+2\)

\(x=3\)

TH2:

\(x-2=-1\)

\(x=-1+2\)

\(x=1\)

Vậy x = 3 hoặc x = 1

c.

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

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

\(2x-1=-2\)

\(2x=-2+1\)

\(2x=-1\)

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

d.

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

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

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

TH1:

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

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

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

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

TH2:

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

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

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

\(x=-\frac{3}{4}\)

Vậy \(x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)

a) x + 16 = (x + 1) + 15 chia hết cho x + 1

Suy ra 15 chia hết cho x + 1 => x + 1 là Ư(15) = {1;3;5;15}

=> x thuộc {0; 2; 4; 14}

b) Tương tư câu a, tách x + 11 = (x + 2) + 9 

Để x + 11 chia hết cho (x+2) thi 9 chia hết cho (x+2) hay là x + 2 là Ư(9)

=> x + 2 thuộc {1; 3; 9} => x thuộc {1; 7}

Còn nếu x nguyên thì nhớ lấy cả ước âm nhé

a)x=1/2

b)x=3 hoặc x=1

c)x=-1/2

d)x=-1/4 hoặc x= -3/4