Question

Tìm số nguyên x biết :

1) 10+3.(x-1)=10+6x

2) 2.(x-2)+3.(3-x)=-4

3) (|x-2|+1).(x-3)=0

4) (x+5).(-3x-15)=0

Answer 1

a, Ta có : \(10+3\left(x-1\right)=10+6x\)

=> \(10+3x-3-10-6x=0\)

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

=> \(x=-1\)

b, Ta có : \(2\left(x-2\right)+3\left(3-x\right)=-4\)

=> \(2x-4+9-3x=-4\)

=> \(-x=-9\)

=> \(x=9\)

c, \(\left(\left|x-2\right|+1\right)\left(x-3\right)=0\)

TH₁ : \(x-2\ge0\left(x\ge2\right)\)

=> \(\left|x-2\right|=x-2\)

Nên ta có phương trình : \(\left(x-2+1\right)\left(x-3\right)=0\)

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

=> \(x=3\)

TH₂ : \(x-2< 0\left(x< 2\right)\)

=> \(\left|x-2\right|=2-x\)

Nên ta có phương trình : \(\left(2-x+1\right)\left(x-3\right)=0\)

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

d, Ta có : \(\left(x+5\right)\left(-3x-15\right)=0\)

=> \(\left[{}\begin{matrix}x+5=0\\-3x-15=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-5\\x=-5\end{matrix}\right.\)

Answer 2

1) 10 + 3 . (x - 1) = 10 + 6x

10 + 3x - 3 = 10 + 6x

3x - 6x = 10 - 10 + 3

-3x = 0 + 3

-3x = 3

x = 3 : (-3)

x = -1

Vậy x = -1

2) 2 . (x - 2) + 3 . (3 - x) = -4

2x - 4 + 9 - 3x = -4

2x - 3x = -4 + 4 - 9

-x = 0 - 9

-x = -9

=> x = 9

Vậy x = 9

3) (Ix - 2I + 1) . (x - 3) = 0

=> (Ix - 2I + 1) = 0 hoặc (x - 3) = 0

Ix - 2I = 0-1 x = 0 + 3

Ix - 2I = -1( loại ) x = 3( thỏa mãn )

Vậy x = 3

4) (x + 5) . (-3x - 15) = 0

=> (x + 5) = 0 hoặc (-3x - 15) = 0

x = 0 - 5 -3x = 0 + 15

x = -5 -3x = 15

x = 15 : (-3)

x = -5

Vậy x = -5

Tick cho mk nha