...

\(\text{x=9341/10000}\)

x=1

x=2639/2500

tìm 2 số nguyên a và b biết :a+b=-1 và a.b=-12.Giup mình nha

\(3x-5^3\div5^2=106\)

\(3x-5=106\)

\(3x=106+5\)

\(3x=111\)

\(x=111\div3\)

\(x=37\)

\(12-2x=2^5\div2^3\)

\(12-2x=2^2\)

\(12-2x=4\)

\(2x=12-4\)

\(2x=8\)

\(x=8\div2=4\)

\(15x-55=5^3.5^2\)

\(15x-55=5^5\)

\(15x-55=3125\)

\(15x=3125+55\)

\(15x=3180\)

\(x=3180\div15\)

\(x=212\)

Dạng 3 :

a) 3x - 10 = 2x + 13

=> 3x - 2x = 13 - 10

=> x = 3

b) x + 12 = -5 - x

=> x + x = -5 - 12

=> 2x = -17

=> x = -8,5

c) x + 5 = 10 - x 

=> x + x = 10 - 5

=> 2x = 5

=> x = 2,5

d) 6x + 23 = 2x - 12

=> 2x - 6x = 23 + 12

=> -4x = 35

=> x = -8,75

e) 12 - x = x + 1

=> x + x = 12 - 1

=> 2x = 11

=> x = 5,5

f) 14 + 4x = 3x + 20

=> 4x - 3x = 20 - 14

=> x = 6

b) Tìm $x$

$10(8x+9x)+8(2x-1)-2(5-6x)=20$

$\Leftrightarrow 170x+16x-8-10+12x=20$

$\Leftrightarrow 198x-18=20$

$\Leftrightarrow 198x=38$

$\Leftrightarrow x=\dfrac{19}{99}$.

Vậy: $x=\dfrac{19}{99}$.

c) Tìm $x$

$(5x^2)(194+20x)=50$

$\Leftrightarrow x^2(194+20x)=10$

$\Leftrightarrow 10x^3+97x^2-5=0$

Thử $x=\dfrac15$:

$10\left(\dfrac15\right)^3+97\left(\dfrac15\right)^2-5=0$.

Suy ra: $(5x-1)(2x^2+39x+5)=0$.

Do đó: $x=\dfrac15$ hoặc $x=\dfrac{-39\pm\sqrt{39^2-4\cdot2\cdot5}}{4}$

$=\dfrac{-39\pm\sqrt{1481}}{4}$.

Vậy: $x=\dfrac15\ \text{hoặc}\ x=\dfrac{-39+\sqrt{1481}}4\ \text{hoặc}\ x=\dfrac{-39-\sqrt{1481}}4.$

\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)

\(720:\left[41-\left(2x-5\right)\right]=8.5\)

\(720:\left[41-\left(2x-5\right)\right]=40\)

\(41-\left(2x-5\right)=720:40\)

\(41- \left(2x-5\right)=18\)

\(2x-5=41-18\)

\(2x-5=23\)

\(2x=23+5\)

\(2x=28\)

\(x=28:2\)

\(\Rightarrow x=14\)

kết quả =18 và 14