x3 - x = 57

10x + 3 - x = 57

10x - x = 57 - 3

9x = 54

x = 54 : 9

x = 6

mình cảm ơn nam hỏi giúp mình nha

\(ĐK:x\ge0\\ PT\Leftrightarrow\left(x-\dfrac{3}{4}\right)\left(x^2+\dfrac{3}{4}x+\dfrac{9}{16}\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\left(n\right)\\\sqrt{x}=3\left(n\right)\\x^2+2\cdot\dfrac{3}{8}x+\dfrac{9}{64}+\dfrac{27}{64}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\\\left(x+\dfrac{3}{8}\right)^2+\dfrac{27}{64}=0\left(\text{vô nghiệm}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=9\end{matrix}\right.\)

Tính: A=x^3+y^3

A=x^3+y^3

A=(x+y)(x^2-xy+y^2)

A=3 . [(x^2+y^2)-xy]

A=3 . (5-xy)

A=15-3xy

2.Cho x-y=5 và x^2+y^2=15

Tính B= x^3-y^3

B=x^3-y^3

B=(x-y)(x^2+xy+y^2)

B=5 . [(x^2+y^2)+xy]

B=5 . (15+xy)

B=75+5xy

Ta có: \(x+y=3\)

\(\Rightarrow\left(x+y\right)^2=9\)

\(\Rightarrow x^2+y^2+2xy=9\)

\(\Rightarrow2xy+5=9\Rightarrow2xy=4\Rightarrow xy=2\)

a) \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(5-2\right)=9\)

\(x^2+y^2=5\\ \Leftrightarrow\left(x+y\right)^2-2xy=5\\ \Leftrightarrow9-2xy=5\\ \Leftrightarrow xy=2\\ a,x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ =3^3-3\cdot2\cdot3=27-18=9\)

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)

\(\Leftrightarrow x^3+9x+2=x^3+8\)

\(\Leftrightarrow x^3+9x=x^3+8-2\)

\(\Leftrightarrow x^3+9x=x^3+6\)

\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)

\(\Leftrightarrow\frac{2}{3}\)

b) \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^4-4=8x-16+16\)

\(\Leftrightarrow x^2+12=8x\)

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

\(\Leftrightarrow x^2-8x+12=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

x/2+x+x/3+x+x+x/4=23/4

⇒ 6x/12+12x/12+4x/12+12x/12+12x/12+3x/12=23/4

⇒ (6x+12x+4x+12x+12x+3x)/12=23/4

⇒ 49x/12=23/4

⇒ 49x=23/4.12

⇒ 49x=69

⇒ x=69/49

Giải:

\(\dfrac{x}{2}+x+\dfrac{x}{3}+x+x+\dfrac{x}{4}=\dfrac{23}{4}\)

\(x.\left(\dfrac{1}{2}+1+\dfrac{1}{3}+1+1+\dfrac{1}{4}\right)=\dfrac{23}{4}\) 

                               \(x.\dfrac{49}{12}=\dfrac{23}{4}\)

                                      \(x=\dfrac{23}{4}:\dfrac{49}{12}\) 

                                      \(x=\dfrac{69}{49}\)