Question

[Vũ Trà My]

Bài 1 giải các phương trình

a) 2(x + 1) = 3 + 2x

b) 4x (1 - x) - 8 = 1 - (4x² + 3)

c) x³ + 1 = x(x + 1)

d) 3x - 2/6 - 5 = 3 - 2(x + 7)/4

e) x² -13x + 12 = 0

f) (2x - 1)² - (x + 3)² = 0

g) 2x/x - 1 + 4/x² + 2x - 3 = 2x - 5/x + 3

(P/S: Mn cố gắng làm hết những câu này giúp mk nha!!! Mk cảm ơn trc!!! Mn giúp mk ko bị chết đuối nha!!!)

Answer 1

[Nguyễn Thành Trương]

\(a)2\left(x+1\right)=3+2x\\ \Leftrightarrow2x+2=3+2x\\ \Leftrightarrow2x-2x=3-1\\ \Leftrightarrow0x=2\left(VN\right)\)

Vậy phương trình vô nghiệm

\(b)4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-8=-2\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\)

Vậy \(S=\left\{\dfrac{3}{2}\right\}\)

Answer 2

[Nguyễn Thành Trương]

\(c)x^3+1=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=x\left(x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-x+1-x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2-2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^2-2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy \(S=\left\{-1;1\right\}\)

\(d)\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow 12\left(\dfrac{3x-2}{6}-5\right)=12.\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow 6x-4-60=9-6\left(x+7\right)\)

\(\Leftrightarrow 6x-64=9-6x-42\)

\(\Leftrightarrow 6x-64=-6x-33\)

\(\Leftrightarrow 6x+6x=-33+64\\\Leftrightarrow 12x=31\\\Leftrightarrow x=\dfrac{31}{12}\)

Vậy \(S=\left\{\dfrac{31}{12}\right\}\)