Question

Giúp tớ bài này nhaaaaaa!!~~~~~~~~~~~~

a)(-5) . (x-2)²+360 = (-150) . 3 +43.5

b)(x+5)-(3x+9) = -16

c) 3. (x+2)-(15-x) . 6 = 160+(-1)¹⁰⁰⁵

d)x.(x-1). (x² - 4) = 0

Pls~~~~~~~~!!!!

Answer 1

USE máy tính

Answer 2

a) \(\left(-5\right)\left(x-2\right)^2+360=\left(-150\right)\cdot3+43\cdot5\)

\(-5\cdot\left(x-2\right)^2+360=-235\)

\(-5\cdot\left(x-2\right)^2=-595\)

\(\left(x-2\right)^2=119\)

\(\left(x-2\right)^2=\left(\pm\sqrt{199}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{119}\\x-2=-\sqrt{119}\end{cases}\Rightarrow\orbr{\begin{cases}x=\sqrt{119}+2\\x=2-\sqrt{119}\end{cases}}}\)

b) \(\left(x+5\right)-\left(3x+9\right)=-16\)

\(x+5-3x-9=-16\)

\(-2x-4=-16\)

\(-2x=-12\)

\(x=6\)

c) \(3\left(x+2\right)-\left(15-x\right)\cdot6=160+\left(-1\right)^{1005}\)

\(3x+6-90+6x=160-1\)

\(9x-84=159\)

\(9x=243\)

\(x=27\)

d) \(x\left(x-1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x\left(x-1\right)=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\left\{0;1\right\}\\x=\left\{2;-2\right\}\end{cases}}}\)