Bài 4: Phương trình tích

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Dân Nguyễn Chí

Giải phương trình:

1. (x - 4)2 - 25 = 0

2. (x - 3)2 - (x - 1)2 = 0

3. (x2 - 4)(2x +3) = (x2 - 4)(x - 1)

4. (x2 - 1) - (x + 1)(2 - 3x) = 0

5. x3 + x2 + x + 1 = 0

6. x3 + x2 - x - 1 = 0

7. 2x3 + 3x2 + 6x + 5 = 0

8. x4 - 4x3 - 19x2 + 106x - 120 = 0

9. (x2 - 3x + 2)(x2 + 15x + 56) + 8 = 0

Võ Đông Anh Tuấn
10 tháng 1 2018 lúc 17:33

1 ) \(\left(x-4\right)^2-25=0\)

\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow x=2.\)

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

\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)

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

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

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

\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)

5 ) \(x^3+x^2+x+1=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)

6 ) \(x^3+x^2-x-1=0\)

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

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

7 ) \(2x^3+3x^2+6x+5=0\)

\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)

\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=-1.\)

8 ) \(x^4-4x^3-19x^2+106x-120=0\)

\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)

\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)

\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)

Đặt \(x^2+6x-7=t\)

\(\Leftrightarrow t\left(t-9\right)+8=0\)

\(\Leftrightarrow t^2-9t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)

Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)

Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)

Vậy ........


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