tính giá trị biểu thức R = 799 x2 - 3x 2 + 403 x + 1198 x3 - 1023 với x = 400
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Nếu \(x=400\Rightarrow\left\{{}\begin{matrix}799=2x-1\\403=x+3\\1198=3x-2\\1203=3\left(x+1\right)\end{matrix}\right.\)( * )
Thay ( * ) vào R , ta được :
\(R=\left(2x-1\right)x^2-3x^4+\left(x+3\right)x+\left(3x-2\right)x^3-3\left(x+1\right)\)
\(=2x^3-x^2-3x^4+x^2+3x+3x^4-2x^3-3x-3\)
\(=\left(2x^3-2x^3\right)+\left(3x^4-3x^4\right)+\left(x^2-x^2\right)+\left(3x-3x\right)-3\)
\(=-3\)
Vậy \(R=-3\) tại \(x=400\)
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a) P(x) = 7x2 . (x2 – 5x + 2 ) – 5x .(x3 – 7x2 + 3x)
= 7x2 . x2 + 7x2 . (-5x) + 7x2 . 2 – [5x. x3 + 5x . (-7x2) + 5x . 3x]
= 7. (x2 . x2) + [7.(-5)] . (x2 . x) + (7.2).x2 – {5. (x.x3) + [5.(-7)]. (x.x2) + (5.3).(x.x)}
= 7x4 + (-35). x3 + 14x2 – [ 5x4 + (-35)x3 + 15x2 ]
= 7x4 + (-35). x3 + 14x2 - 5x4 + 35x3 - 15x2
= (7x4 – 5x4) + [(-35). x3 + 35x3 ] + (14x2 - 15x2 )
= 2x4 + 0 - x2
= 2x4 – x2
b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:
P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))4 – (\( - \dfrac{1}{2}\))2 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
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Ta có
M = x ( x 3 + x 2 – 3 x – 2 ) - ( x 2 – 2 ) ( x 2 + x – 1 ) = x . x 3 + x . x 2 – 3 x . x – 2 . x – ( x 2 . x 2 + x 2 . x – x 2 – 2 x 2 – 2 x + 2 ) = x 4 + x 3 – 3 x 2 – 2 x – ( x 4 + x 3 – 3 x 2 – 2 x + 2 ) = x 4 + x 3 – 3 x 2 – 2 x – x 4 – x 3 + 3 x 2 + 2 x – 2
= - 2
Vậy M = -2
Đáp án cần chọn là: D
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Đa thức \(P\left(x\right)=x^3-3x+1\)có ba nghiệm phân biệt \(x_1,x_2,x_3\) có:
\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_3x_1=-3\\x_1x_2x_3=-1\end{cases}}\)
\(E=Q\left(x_1\right)Q\left(x_2\right)Q\left(x_3\right)=\left(x_1^2-1\right)\left(x_2^2-1\right)\left(x_3^2-1\right)\)
\(=\left(x_1x_2x_3\right)^2-\left(x_1^2x_2^2+x_2^2x_3^2+x_3^2x_1^2\right)+\left(x_1^2+x_2^2+x_3^2\right)-1\)
\(=\left(x_1x_2x_3\right)^2-\left[\left(x_1x_2+x_2x_3+x_3x_1\right)^2-2x_1x_2x_3\left(x_1+x_2+x_3\right)\right]+\left[\left(x_1+x_2+x_3\right)^2-2\left(x_1x_2+x_2x_3+x_3x_1\right)\right]-1\)
\(=\left(-1\right)^2-3^2+2.3-1=-3\)
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x 3 - 3 x 2 + 3x - 1 tại x = 101.
= x 3 - 3. x 2 .1 + 3.x. 1 2 - 1 3 = x - 1 3
= 101 - 1 3 = 100 3 = 1000000
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Ta có A = x 3 + 3 x ( x + 1 ) + 1 3 − x 2 − 4 x + 4 = x + 1 − x − 2 .
A= 3 với x ≥ 2. Do đó A= 3 khi x = 2017 3 .
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x^2-2\right)\left(x^2+x+1\right)-x\left(x^3+x^2-3x-2\right)=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x=2\left(đpcm\right)\)
R = 799 x ^2 - 3x ^2 + 403 x + 1198 x ^3 - 1023
=1198x3 796 x 2+403x -1023
thay vào ruj tính