Chứng minh:
(x-2)(x2-5x+1) -x(x2+11)= -(7x2+2)
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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2024
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10
20 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`1.`
\(\left(-4xy\right)\cdot\left(2xy^2-3x^2y\right)\)
`=`\(\left(-4xy\right)\left(2xy^2\right)+\left(-4xy\right)\left(-3x^2y\right)\)
`=`\(-8\left(x\cdot x\right)\left(y\cdot y^2\right)+12\left(x\cdot x^2\right)\left(y\cdot y\right)\)
`=`\(-8x^2y^3+12x^3y^2\)
`2.`
\(\left(-5x\right)\left(3x^3+7x^2-x\right)\)
`=`\(\left(-5x\right)\left(3x^3\right)+\left(-5x\right)\left(7x^2\right)+\left(-5x\right)\left(-x\right)\)
`=`\(-15x^4-35x^3+5x^2\)
`3.`
\(\left(3x-2\right)\left(4x+5\right)-6x\left(2x-1\right)\)
`=`\(3x\left(4x+5\right)-2\left(4x+5\right)-12x^2+6x\)
`=`\(12x^2+15x-8x-10-12x^2+6x\)
`=`\(\left(12x^2-12x^2\right)+\left(15x-8x+6x\right)-10\)
`=`\(13x-10\)
`4.`
\(2x^2\left(x^2-7x+9\right)\)
`=`\(2x^2\cdot x^2+2x^2\cdot\left(-7x\right)+2x^2\cdot9\)
`=`\(2x^4-14x^3+18x^2\)
`5.`
\(\left(3x-5\right)\left(x^2-5x+7\right)\)
`=`\(3x\left(x^2-5x+7\right)-5\left(x^2-5x+7\right)\)
`=`\(3x^3-15x^2+21x-5x^2+25x-35\)
`=`\(3x^3-20x^2+46x-35\)
11 tháng 11 2021
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
H9
HT.Phong (9A5)
CTVHS
27 tháng 6 2023
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
16 tháng 11 2021
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
16 tháng 11 2021
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
H9
HT.Phong (9A5)
CTVHS
16 tháng 11 2023
1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)
b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)
\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)
16 tháng 11 2023
2: \(x^2-8x+7=0\)
=>\(x^2-x-7x+7=0\)
=>\(x\left(x-1\right)-7\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
1:
a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)
b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)
\(=x^2+1\)
2 tháng 8 2020
2. -x2 + x - 33 = -x2 + x - 1/4 - 131/4 = -( x2 - x + 1/4 ) - 131/4 = -( x - 1/2 )2 - 131/4
-( x - 1/2 )2 ≤ 0 ∀ x => -( x - 1/2 )2 - 131/4 ≤ -131/4 < 0 ∀ x ( đpcm )
3. x2 + 4x + 33 = x2 + 4x + 4 + 29 = ( x + 2 )2 + 29
( x + 2 )2 ≥ 0 ∀ x => ( x + 2 )2 + 29 ≥ 29 > 0 ∀ x ( đpcm )
4. x2 + 8x = x2 + 8x + 16 - 16 = ( x + 4 )2 - 16
( x + 4 )2 ≥ 0 ∀ x => ( x + 4 )2 - 16 ≥ -16 ∀ x
Đẳng thức xảy ra <=> x + 4 = 0 => x = -4
Vậy GTNN của biểu thức = -16, đạt được khi x = -4
9 tháng 11 2021
\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left(2x+1\right)\left(3x^2-5x+2\right):\left(2x+1\right)=3x^2-5x+2\\ b,=\left(x^4-2x^3+3x^2+x^3-2x^2+3x\right):\left(x^2-2x+3\right)\\ =\left(x^2-2x+3\right)\left(x^2+x\right):\left(x^2-2x+3\right)=x^2+x\)
Ta có:
\(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2\)
\(=\left(-7x^2+2\right)\)