\(\left(x+1\right)^3+\left(x-2\right)^3-2x^2\left(x-1,5\right)=3\)

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3\)

\(\Leftrightarrow15x=10\Leftrightarrow x=\dfrac{2}{3}\)

Mình làm hơi tắt phần sau chút bạn nhé !

a) \(2^x=8\)

⇔ \(2^x=2^3\)

⇒ \(x=3\)

b) \(3^x=27\)

⇔ \(3^x=3^3\)

⇒ \(x=3\)

c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)

⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)

d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)

⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)

d) \(\left(x+1\right)^3=-125\)

⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)

⇔ \(x+1=-5\)

⇔ \(x=-5-1=-6\)

2:

a: (x-1,2)^2=4

=>x-1,2=2 hoặc x-1,2=-2

=>x=3,2(loại) hoặc x=-0,8(loại)

b: (x-1,5)^2=9

=>x-1,5=3 hoặc x-1,5=-3

=>x=-1,5(loại) hoặc x=4,5(loại)

c: (x-2)^3=64

=>(x-2)^3=4^3

=>x-2=4

=>x=6(nhận)

332323232323232= 3
⇔15x−12=0
⇔15x=10

⇔x= 2/3

Trả lời:

( x + 1 )³ + ( x - 2 )³ - 2x² ( x - 1,5 ) = 3

<=> x³ + 3x² + 3x + 1 + x³ - 6x² + 12x - 8 - 2x³ + 3x² = 3

<=> 15x - 7 = 3

<=> 15x = 10

<=> x = 2/3

Vậy x = 2/3 là nghiệm của pt.

mình làm hơi lỗi :<<

\(\Rightarrow x+\dfrac{2}{3}x=1,5+3,5\Rightarrow\dfrac{5}{3}x=5\Rightarrow x=5:\dfrac{5}{3}=3\)

⇒x+23x=1,5+3,5⇒53x=5⇒x=5:53=3

yggucbsgfuyvfbsudy

????????

1) \(|5x-3|=|7-x|\)

\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)

Vậy...

2) \(2.|3x-1|-3x=7\)

\(\Leftrightarrow2.|3x-1|=7+3x\)

\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)

Vậy...

5) \(\text{1,6 - |x + 1,5| = 0}\)

\(|x + 1,5| = 1,6-0\)

\(\text{|x + 1,5| = 1,6}\)

\( |x | = 1,6-1,5\)

\(|x|=0,1\)

a)\(\left|x-3,5\right|=0,25\)

\(\Leftrightarrow\orbr{\begin{cases}x-3,5=0,25\\x-3,5=-0,25\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3,75\\x=3,25\end{cases}}}\)

b)\(\left|1,5-x\right|=1,75\)

\(\Leftrightarrow\orbr{\begin{cases}1,5-x=1,75\\1,5-x=-1,75\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-0,25\\x=3,25\end{cases}}}\)

c)\(\left|x-1,5\right|=1,5\)

\(\Leftrightarrow\orbr{\begin{cases}x-1,5=1,5\\x-1,5=-1,5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=0\end{cases}}}\)

d)\(\left|x-2,5\right|+\left|x+1,5\right|=3\)

\(\Leftrightarrow\left|x-2,5\right|+x+1,5=3\)

\(\Leftrightarrow\left|x-2,5\right|=1,5-x\)

\(\Leftrightarrow\orbr{\begin{cases}x-2,5=1,5-x\\x-2,5=x-1,5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\varnothing\end{cases}}}\)