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H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
p) \(\left(9-x\right)\left(x^2+2x-3\right)\)
\(=9\left(x^2+2x-3\right)-x\left(x^2+2x-3\right)\)
\(=9x^2+18x-27-x^3-2x^2+3x\)
\(=-x^3+7x^2+21x-27\)
n) \(\left(-x+3\right)\left(x^2+x+1\right)\)
\(=-x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)
\(=-x^3-x^2-x+3x^2+3x+3\)
\(=-x^2+2x^2+2x+3\)
o) \(\left(-6x+\dfrac{1}{2}\right)\left(x^2-4x+2\right)\)
\(=-6x\left(x^2-4x+2\right)+\dfrac{1}{2}\left(x^2-4x+2\right)\)
\(=-6x^3+24x^2-12x+\dfrac{1}{2}x^2-2x+1\)
\(=-6x^3+\dfrac{49}{2}x^2-14x+1\)
q) \(\left(6x+1\right)\left(x^2-2x-3\right)\)
\(=6x\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=6x^3-12x^2-18x+x^2-2x-3\)
\(=6x^3-11x^2-20x-3\)
r) \(\left(2x+1\right)\left(-x^2-3x+1\right)\)
\(=2x\left(-x^2-3x+1\right)+\left(-x^2-3x+1\right)\)
\(=-2x^3-6x^2+2x-x^2-3x+1\)
\(=-2x^3-7x^2-x+1\)
u) \(\left(2x-3\right)\left(-x^2+x+6\right)\)
\(=2x\left(-x^2+x+6\right)-3\left(-x^2+x+6\right)\)
\(=-2x^3+2x^2+12x+3x^2-3x-18\)
\(=-2x^3+5x^2+9x-18\)
s) \(\left(-4x+5\right)\left(x^2+3x-2\right)\)
\(=-4x\left(x^2+3x-2\right)+5\left(x^2+3x-2\right)\)
\(=-4x^3-12x^2+8x+5x^2+15x-10\)
\(=-4x^3-7x^2+23x-10\)
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4x^3\right)\)
\(=-\dfrac{1}{2}x\left(2x+6-4x^3\right)+3\left(2x+6-4x^3\right)\)
\(=-x^2-3+2x^4+6x+18-12x^3\)
\(=2x^4-12x^3-x^2+6x+15\)
30 tháng 7 2023
p: (-x+9)(x^2+2x-3)
=-x^3-2x^2+3x+9x^2+18x-27
=-x^3+7x^2+21x-27
n: (-x+3)(x^2+x+1)
=-x^3-x^2-x+3x^2+3x+3
=-x^3+2x^2+2x+3
o: (-6x+1/2)(x^2-4x+2)
=-6x^3+24x^2-12x+1/2x^2-2x+1
=-64x^3+49/2x^2-14x+1
q: (6x+1)(x^2-2x-3)
=6x^3-12x^2-18x+x^2-2x-3
=6x^3-11x^2-20x-3
r: (2x+1)(-x^2-3x+1)
=-2x^3-6x^2+2x-x^2-3x+1
=-2x^3-7x^2-x+1
u: =-2x^3+2x^2+12x+3x^2-3x-18
=-2x^3+5x^2+9x-18
s: =-4x^3-12x^2+8x+5x^2+15x-10
=-4x^3-7x^2+23x-10
19 tháng 11 2017
\(C=\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+7}{\left(x+2\right)\left(\left(4x+7\right)\right)}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+8}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4}{4x+7}\)
\(D=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\\ D=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\\ D=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{\left(3x-2\right)2x}{\left(2x-1\right)2x}-\dfrac{3x-2}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(3x-2\right)2x-\left(3x-2\right)}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left[\left(3x-2\right)2x-\left(3x-2\right)\right]}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(3x-2\right)\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=\dfrac{\left[\left(1-3x\right)+\left(3x-2\right)\right]\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=-\dfrac{1}{2x}\)
19 tháng 7 2016
\(M=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\) Do \(\left(x-2\right)^2\ge0=>\left(x-2\right)^2-3\ge-3\)
Vậy min M=-3 khi x=2
\(N=x^2+10x+50=x^2+10x+25+25=\left(x+5\right)^2+25.\) Do \(\left(x+5\right)^2\ge0\Rightarrow\left(x+5\right)^2+25\ge25\Rightarrow N_{min}=25\) khi x=-5
\(P=x^2+12x-1=x^2+12x+36-37=\left(x+6\right)^2-37\) Do \(\left(x+6\right)^2\ge0\Rightarrow\left(x+6\right)^2-37\ge-37\Rightarrow P_{min}=-37\) khi x=-6
\(Q=x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\) Do \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\Rightarrow Q_{min}=\frac{3}{4}\) khi \(x=\frac{1}{2}\)
\(R=x^2-3x+2=\left(x^2-3x+2,25\right)-0,5=\left(x-1,5\right)^2-0,5\) Do \(\left(x-1,5\right)^2\ge0\Rightarrow\left(x-1,5\right)^2-0,5\ge-0,5\Rightarrow R_{min}=-0,5\) khi x=1,5
\(S=2x^2-8x+1=2\left(x^2-4x+2\right)-3=2\left(x-2\right)^2-3\) Do \(2\left(x-2\right)^2\ge0\Rightarrow2\left(x-2\right)^2-3\ge-3\Rightarrow S_{min}=-3\) khi x=2
\(T=2x^2+6x+1=2\left(x^2+3x+2,25\right)-3,5=2\left(x+1,5\right)^2-3,5\) Do \(2\left(x+1,5\right)^2\ge0\Rightarrow2\left(x+1,5\right)^2-3,5\ge-3,5\Rightarrow T_{min}=-3,5\) khi x=-1,5
\(V=3x^2+x+2=3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)+\frac{23}{24}=3\left(x+\frac{1}{6}\right)^2+\frac{23}{24}\) Do\(3\left(x+\frac{1}{6}\right)^2\ge0\Rightarrow3\left(x+\frac{1}{6}\right)^2+\frac{23}{24}\ge\frac{23}{24}\Rightarrow V_{min}=\frac{23}{24}\) khi \(x=\frac{1}{6}\)
16 tháng 12 2020
Bài 1.
a)\(\frac{4x-4}{x^2-4x+4}\div\frac{x^2-1}{\left(2-x\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\div\frac{\left(x-1\right)\left(x+1\right)}{\left(x-2\right)^2}=\frac{4\left(x-1\right)}{\left(x-2\right)^2}\times\frac{\left(x-2\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4}{x+1}\)
b) \(\frac{2x+1}{2x^2-x}+\frac{32x^2}{1-4x^2}+\frac{1-2x}{2x^2+x}=\frac{2x+1}{x\left(2x-1\right)}+\frac{-32x^2}{4x^2-1}+\frac{1-2x}{x\left(2x+1\right)}\)
\(=\frac{\left(2x+1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{\left(1-2x\right)\left(2x-1\right)}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-32x^3}{x\left(2x-1\right)\left(2x+1\right)}+\frac{-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{4x^2+4x+1-32x^3-4x^2+4x-1}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-32x^3+8x}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\frac{-8x\left(4x^2-1\right)}{x\left(2x-1\right)\left(2x+1\right)}=\frac{-8x\left(2x-1\right)\left(2x+1\right)}{x\left(2x-1\right)\left(2x+1\right)}=-8\)
c) \(\left(\frac{1}{x+1}+\frac{1}{x-1}-\frac{2x}{1-x^2}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{1}{x+1}+\frac{1}{x-1}+\frac{2x}{x^2-1}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\left(\frac{x-1+x+1+2x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-1}{4x}\)
\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}\times\frac{x-1}{4x}=\frac{1}{x+1}\)
Bài 3.
N = ( 4x + 3 )2 - 2x( x + 6 ) - 5( x - 2 )( x + 2 )
= 16x2 + 24x + 9 - 2x2 - 12x - 5( x2 - 4 )
= 14x2 + 12x + 9 - 5x2 + 20
= 9x2 + 12x + 29
= 9( x2 + 4/3x + 4/9 ) + 25
= 9( x + 2/3 )2 + 25 ≥ 25 > 0 ∀ x
=> đpcm
29 tháng 5 2018
a) Ta có: P(x) = 3y + 6 có nghiệm khi
3y + 6 = 0
3y = -6
y = -2
Vậy đa thức P(y) có nghiệm là y = -2.
b) Q(y) = y4 + 2
Ta có: y4 có giá trị lớn hơn hoặc bằng 0 với mọi y
Nên y4 + 2 có giá trị lớn hơn 0 với mọi y
Tức là Q(y) ≠ 0 với mọi y
Vậy Q(y) không có nghiệm.
1 tháng 10 2021
a: Ta có: \(\left(x^2-2x+2\right)\left(x^2-2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left(x^4-4\right)\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)\)
\(=\left(x^4-4\right)\cdot\left(x^4+4\right)\)
\(=x^8-16\)
b: Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)
14 tháng 11 2017
a, N = 2 + 6/x^2-8x+22
Có : x^2-8x+22 = (x-4)^2 + 6 >= 6 => 6/x^2-8x+22 <= 6/6 = 1 => N <= 2+1=3
Dấu "=" xảy ra <=> x-4 = 0 <=> x=4
Vậy Max N =3 <=> x=4
k mk nha
14 tháng 11 2017
Cảm ơn bạn đã giúp mink nhưng bạn làm kiểu thế mink ko hiểu. Mong bạn sửa lại !
20 tháng 5 2022
a: Ta có \(x^3-4x^2+x-n⋮x-4\)
\(\Leftrightarrow x^2\left(x-4\right)+x-4+n+4⋮x-4\)
=>n+4=0
hay n=-4
b: ta có: \(4x^3-2x^2+2x+n⋮2x+1\)
\(\Leftrightarrow4x^3+2x^2-4x^2-2x+4x+2+n-2⋮2x+1\)
=>n-2=0
hay n=2
c: \(\Leftrightarrow x^4-3x^3+3x^3-9x^2+6x^2-18x+21x-63-n+63⋮x-3\)
=>63-n=0
hay n=63
DT
1
HN
29 tháng 3 2017
a/ \(M=\dfrac{x^2-x+1}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3x^2-6x+3}{x^2+2x+1}=\dfrac{1}{4}+\dfrac{3\left(x-1\right)^2}{x^2+2x+1}\ge\dfrac{1}{4}\)
b/ \(N=\dfrac{3x^2+4x}{x^2+1}=4-\dfrac{x^2-4x+4}{x^2+1}=4-\dfrac{\left(x-2\right)^2}{x^2+1}\le4\)