Question

Bài4. Giải các phương trình

a/\(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

b/ \(x^3-\frac{x}{49}\) =0

c/ \(x^2-7x+12=0\)

d/ \(4x^2-3x-1=0\)

e/ \(x^3-2x-4=0\)

f/ \(x^3+8x^2+17x+10=0\)

Answer 1

a, (3x-1)² - (x+3)² = 0

<=> [(3x-1)-(x+3)][(3x-1)+(x+3)] = 0

<=> (3x-1-x-3)(3x-1+x+3) = 0

<=> (2x-4)(4x+2) = 0

=> 2x-4=0 hoặc 4x+2=0

=> 2x =4 hoặc 4x = -2

=> x = 2 hoặc x = \(\frac{-1}{2}\)

Answer 2

\(\begin{array}{l} a){\left( {3x - 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow \left( {3x - 1 + x + 3} \right)\left[ {3x - 1 - x - 3} \right] = 0\\ \Leftrightarrow \left( {4x + 2} \right)\left( {2x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 2 = 0\\ 2x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{2}\\ x = 2 \end{array} \right.\\ b){x^3} - \dfrac{x}{{49}} = 0\\ \Leftrightarrow 49{x^3} - x = 0\\ \Leftrightarrow x\left( {49{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 49{x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \pm \dfrac{1}{7} \end{array} \right.\\ c){x^2} - 7x + 12 = 0\\ \Leftrightarrow {x^2} - 3x - 4x + 12 = 0\\ \Leftrightarrow x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 3 = 0\\ x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 4 \end{array} \right.\\ d)4{x^2} - 3x - 1 = 0\\ \Leftrightarrow 4{x^2} + x - 4x - 1 = 0\\ \Leftrightarrow x\left( {4x + 1} \right) - \left( {4x + 1} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 1 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{4}\\ x = 1 \end{array} \right.\\ e){x^3} - 2x - 4 = 0\\ \Leftrightarrow {x^3} - 4x + 2x - 4 = 0\\ \Leftrightarrow x\left( {{x^2} - 4} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left[ {x\left( {x + 2} \right) + 2} \right] = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ {x^2} + 2x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ {x^2} + 2x + 2x = 0\left( {VN} \right) \end{array} \right.\\ f){x^3} + 8{x^2} + 17x + 10 = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 7x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 5x + 2x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left[ {x\left( {x + 5} \right) + 2\left( {x + 5} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ x + 5 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 5\\ x = - 2 \end{array} \right. \end{array}\)

Answer 3

a. \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)

\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+2=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=2\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{2};2\right\}\).

b. \(x^3-\frac{x}{49}=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{49}\right)=0\)

\(\Leftrightarrow x\left(x+\frac{1}{7}\right)\left(x-\frac{1}{7}\right)=0\)

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

Vậy phương trình có tập nghiệm \(S=\left\{0;\frac{-1}{7};\frac{1}{7}\right\}\).

Answer 4

c, x² - 7x + 12 = 0

=> x² - 4x - 3x +12= 0

=> (x²-4x)-(3x-12) = 0

=> x(x-4)-3(x-4) = 0

=> (x-3)(x-4) = 0

=> x-3=0 hoặc x-4=0

=> x =3 hoặc x =4

Answer 5

a) (3x - 1)² - (x + 3)² = 0

⇔ (3x - 1 - x - 3)(3x - 1+ x + 3) = 0

⇔ (2x - 4)(4x + 2) = 0

⇔\(\left[{}\begin{matrix}2x-4=0\\4x+2=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\x=\frac{-1}{2}\end{matrix}\right.\)

Answer 6

b) x³ - \(\frac{x}{49}\) = 0

⇔ 49x³ - x = 0

⇔ x(49x² - 1) = 0

⇔ x(7x - 1)(7x + 1) = 0

⇔\(\left[{}\begin{matrix}x=0\\7x-1=0\\7x+1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\\x=\frac{-1}{7}\end{matrix}\right.\)

Answer 7

c) x² - 7x + 12 = 0

⇔ x² - 3x - 4x + 12 = 0

⇔ x(x - 3) - 4(x - 3) = 0

⇔(x - 3)(x - 4) = 0

⇔\(\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Answer 8

d) 4x² - 3x - 1 = 0

⇔ 4x² - 4x + x - 1 = 0

⇔ 4x(x - 1) + (x - 1) = 0

⇔(x - 1)(4x + 1) = 0

⇔\(\left[{}\begin{matrix}x-1=0\\4x+1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=1\\x=\frac{-1}{4}\end{matrix}\right.\)

Answer 9

e) x³ - 2x - 4 = 0

⇔ x³ - 2x² + 2x² - 4x + 2x - 4 = 0

⇔ x²(x - 2) + 2x(x - 2) + 2(x - 2) = 0

⇔(x - 2)(x² + 2x + 2) = 0

⇔(x - 2)[(x + 1)² + 1] = 0

⇔\(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\\left(x+1\right)^2+1\ge1>0\end{matrix}\right.\)

Answer 10

f) x³ + 8x² + 17x + 10 = 0

⇔ x³ + 2x² + 6x² + 12x + 5x + 10 = 0

⇔x²(x + 2) + 6x(x + 2) + 5(x + 2) = 0

⇔(x + 2)(x² + 6x + 5) = 0

⇔(x + 2)(x + 1)(x + 5) = 0

⇔\(\left[{}\begin{matrix}x+2=0\\x+1=0\\x+5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=-2\\x=-1\\x=-5\end{matrix}\right.\)