## *Phân tích thành nhân tử a)A=x ³-3x ²-3x+1 b)B=x ³+x ²-36 *Tìm x biết a)x ³-9x ²+14x=0 b)x ³-5x ²+8x-4=0 c)x $x^{4}$ -2x ³+x ²=10 d)2x ³+x ²-4x-2=0 Gi

Question

*Phân tích thành nhân tử
a)A=x ³-3x ²-3x+1
b)B=x ³+x ²-36
*Tìm x biết
a)x ³-9x ²+14x=0
b)x ³-5x ²+8x-4=0
c)x $x^{4}$ -2x ³+x ²=10
d)2x ³+x ²-4x-2=0
Giúp mình với

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3 tháng 2021-09-09T21:45:00+00:00 2 Answers 3 views 0

## Answers ( )

1. Đáp án:

2) d. $$\left[ \begin{array}{l} x = – \dfrac{1}{2}\\ x = \sqrt 2 \\ x = – \sqrt 2 \end{array} \right.$$

Giải thích các bước giải:

$$\begin{array}{l} 1)a.A = {x^3} + {x^2} – 4{x^2} – 4x + x + 1\\ = {x^2}\left( {x + 1} \right) – 4x\left( {x + 1} \right) + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} – 4x + 1} \right)\\ b.B = {x^3} + {x^2} – 36\\ = {x^3} – 3{x^2} + 4{x^2} – 12x + 12x – 36\\ = {x^2}\left( {x – 3} \right) + 4x\left( {x – 3} \right) + 12\left( {x – 3} \right)\\ = \left( {x – 3} \right)\left( {{x^2} + 4x + 12} \right)\\ 2)a.{x^3} – 9{x^2} + 14x = 0\\ \to x\left( {{x^2} – 9x + 14} \right) = 0\\ \to \left[ \begin{array}{l} x = 0\\ {x^2} – 7x – 2x + 14 = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 0\\ x\left( {x – 7} \right) – 2\left( {x – 7} \right) = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 0\\ x – 7 = 0\\ x – 2 = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 0\\ x = 7\\ x = 2 \end{array} \right.\\ b.{x^3} – 5{x^2} + 8x – 4 = 0\\ \to {x^3} – {x^2} – 4{x^2} + 4x + 4x – 4 = 0\\ \to {x^2}\left( {x – 1} \right) – 4x\left( {x – 1} \right) + 4\left( {x – 1} \right) = 0\\ \to \left[ \begin{array}{l} x – 1 = 0\\ {x^2} – 4x + 4 = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 1\\ {\left( {x – 2} \right)^2} = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = 1\\ x = 2 \end{array} \right.\\ c.{x^4} – 2{x^3} + {x^2} = 10\\ \to {x^2}\left( {{x^2} – 2x + 1} \right) = 10\\ \to {x^2}{\left( {x – 1} \right)^2} = 10\\ \to \left[ \begin{array}{l} x\left( {x – 1} \right) = \sqrt {10} \\ x\left( {x – 1} \right) = – \sqrt {10} \end{array} \right.\\ \to \left[ \begin{array}{l} {x^2} – x – \sqrt {10} = 0\\ {x^2} – x + \sqrt {10} = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} {x^2} – 2x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{1}{4} – \sqrt {10} = 0\\ {x^2} – 2x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{1}{4} + \sqrt {10} = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} {\left( {x – \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + \sqrt {10} \\ {\left( {x – \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} – \sqrt {10} \left( l \right)\left( {Do:{{\left( {x – \dfrac{1}{2}} \right)}^2} \ge \forall x} \right) \end{array} \right.\\ \to {\left( {x – \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + \sqrt {10} \\ \to \left[ \begin{array}{l} x – \dfrac{1}{2} = \sqrt {\dfrac{1}{4} + \sqrt {10} } \\ x – \dfrac{1}{2} = – \sqrt {\dfrac{1}{4} + \sqrt {10} } \end{array} \right.\\ \to \left[ \begin{array}{l} x = \dfrac{1}{2} + \sqrt {\dfrac{1}{4} + \sqrt {10} } \\ x = \dfrac{1}{2} – \sqrt {\dfrac{1}{4} + \sqrt {10} } \end{array} \right.\\ d.2{x^3} + {x^2} – 4x – 2 = 0\\ \to {x^2}\left( {2x + 1} \right) – 2\left( {2x + 1} \right) = 0\\ \to \left[ \begin{array}{l} 2x + 1 = 0\\ {x^2} – 2 = 0 \end{array} \right.\\ \to \left[ \begin{array}{l} x = – \dfrac{1}{2}\\ x = \sqrt 2 \\ x = – \sqrt 2 \end{array} \right. \end{array}$$

2. a, A = x^3 – 3x^2 – 3x +1

⇔ A = (x^3 + 1) – (3x^2 + 3x)

⇔ A = (x+1)(x^2 – x + 1)  – 3x(x+1)

⇔A = (x+1)(x^2-x+1-3x)

⇔ A =(x+1)(x^2 – 4x +1)

b, B = x^3 + x^2-36

⇔ B = x^3 – 3x^2 + 4x^2 – 12x + 12x – 36

⇔ B = x^2(x-3) + 4x(x-3) + 12(x-3)

⇔ B = (x-3)(x^2 + 4x + 12)

* Tìm x

a, x^3 – 9x^2 + 14x = 0

⇔ x^3 – 7x^2 – 2x^2 + 14 = 0

⇔ x(x-2)(x-7) = 0

⇔ x = 0 hoặc x = 2 hoặc x =7

b, x^3 – 5x^2+8x-4=0

⇔ x^3 – 2x^2 – 3x^2 + 6x +2x – 4 = 0

⇔ (x-2)^2(x-1) = 0

⇔$$\left[ \begin{array}{l}(x-2)^2=0\\x-1=0\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}x=2\\x=1\end{array} \right.$$

c, 2x^3 + x^2 – 4x -2 = 0

⇔ (2x +1)(x^2-2) = 0

⇔ $$\left[ \begin{array}{l}2x+1=0\\x^2-2=0\end{array} \right.$$

⇔ $$\left[ \begin{array}{l}x=-1/2\\x=x=√2\end{array} \right.$$