Question

Giải phương trình:

\(a,7\left(2x-0,5\right)-3\left(x+4\right)=4-5\left(x-0,7\right);\)

\(b,5x^3-2x^2-7x=0\).

Cakpan làm để mình kiểm tra cái nkaaa 

Answer 1

a. 7(2x - 0,5) - 3(x + 4) = 4 - 5(x - 0,7)

⇔ 14x - 4,5 - 3x - 12 = 4 - 5x + 3,5

⇔ 14x -3x + 5x = 4 + 4,5 + 3,5

⇔ 16x = 12

⇔ x = \(\dfrac{12}{16}=\dfrac{3}{4}\)

Answer 2

a. 7(2x - 0,5) - 3(x + 4) = 4 - 5(x - 0,7)

⇔ 14x - 3,5 - 3x - 12 = 4 - 5x + 3,5

⇔ 14x - 3x + 5x = 4 + 3,5 + 3,5

⇔ 16x = 11

⇔ x = \(\dfrac{11}{16}\)

 

Answer 3

a. \(7\left(2x-0,5\right)-3\left(x+4\right)=4-5\left(x-0,7\right)\)

\(\Rightarrow14x-3,5-3x-12=4-5x+3,5\)

\(\Rightarrow14x-3x+5x=4+3,5+3,5+12\)

\(\Rightarrow16x=23\)

\(\Rightarrow x=\dfrac{23}{16}\)

Vậy \(S=\left\{\dfrac{23}{16}\right\}\)

b. \(5x^3-2x^2-7x=0\)

\(\Rightarrow x\left(5x^2-2x-7\right)=0\)

\(\Rightarrow x\left(x-\dfrac{7}{5}\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{7}{5}=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)

Vậy \(S=\left\{0;\dfrac{7}{5};-1\right\}\)

Answer 4

a. 

<=> x= 23/16

 

 

Answer 5

b: Ta có: \(5x^3-2x^2-7x=0\)

\(\Leftrightarrow x\left(5x^2-2x-7\right)=0\)

\(\Leftrightarrow x\left(5x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\\x=-1\end{matrix}\right.\)