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10 tháng 8 2020
a) M = -x2 - 4x + 2 = -x2 - 4x - 4 + 6 = -( x2 + 4x + 4 ) + 6 = -( x + 2 )2 + 6
\(-\left(x+2\right)^2\le0\forall x\Rightarrow-\left(x+2\right)^2+6\le6\)
Dấu " = " xảy ra <=> x + 2 = 0 => x = -2
Vậy MMax = 6 , đạt được khi x = -2
b) N = -2y2 - 3y + 5 = -2( y2 + 3/2y + 9/16 ) + 49/8 = -2( y + 3/4 )2 + 49/8
\(-2\left(y+\frac{3}{4}\right)^2\le0\forall y\Rightarrow-2\left(y+\frac{3}{4}\right)^2+\frac{49}{8}\le\frac{49}{8}\)
Dấu " = " xảy ra <=> y + 3/4 = 0 => y = -3/4
Vậy NMax = 49/8 , đạt được khi y = -3/4
c) P = ( 2 -x )( x + 4 ) = -x2 - 2x + 8 = -x2 - 2x - 1 + 9 = -( x2 + 2x + 1 ) + 9 = -( x + 1 )2 + 9
\(-\left(x+1\right)^2\le0\forall x\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy PMax = 9 , đạt được khi x = -1
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
A
22 tháng 10 2017
4.a)n2(n+1)+2n(n+1)=(n+1)(n2+2n)=n(n+1)(n+2)
n,(n+1),(n+2) là ba số nguyên liên tiếp nên chia hết cho 2 và 3
\(\Rightarrow\)n(n+1)(n+2) chia hết cho 6
22 tháng 10 2017
4 Chứng minh rằng:
a)\(n^2+\left(n+1\right)+2n\left(n+1\right)\) chia hết cho 6
Ta có:
\(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=n^3+3n^2+2n\)
\(=n\left(n^2+3n+2\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
Ta thấy n , n+1 và n+2 là ba số tự nhiên liên tiếp
=> n(n+1) (n+2)\(⋮\)6
=> đpcm
b)\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8
Ta có:
\(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=\left(2n-1\right).2\left(n-1\right).2n\)
\(=4n\left(2n-1\right)\left(n-1\right)\)
=>\(4n\left(2n-1\right)\left(n-1\right)⋮4\left(1\right)\)
Mà(2n-1)(n-1)=(n+n-1)(n-1)
=>\(\left(2n-1\right)\left(n-1\right)⋮2\left(2\right)\)
Từ (1) và (2)=> Đpcm
c)\(\left(n+7\right)^2-\left(n-5\right)^2\) chia hết cho 24
Câu hỏi của Ngoc An Pham - Toán lớp 8 | Học trực tuyến
Chúc bạn học tốt!^^
TG
3 tháng 8 2020
Bài 1:
a) \(4x\left(3x-1\right)-2\left(3x+1\right)-\left(x+3\right)\)
\(=12x^2-4x-6x-2-x-3\)
\(=12x^2-11x-5\)
b) \(=\left(-2x^2-1xy+2y^2\right)\left(-1x^2y\right)\)
\(=\left[\left(-1x^2y\right)\left(-2x^2\right)\right]-\left[\left(-1x^2y\right).1xy\right]+\left[\left(-1x^2y\right).2y^2\right]\)
\(=\left(2x^4y\right)-\left(-1x^3y^2\right)+\left(-2x^2y^3\right)\)
\(=2x^4y+1x^3y^2-2x^2y^3\)
c) \(4x\left(3x^2-x\right)-\left(2x+3\right)^2\left(6x^2-3x+1\right)\)
\(=\left(4x.3x^2\right)-\left(4x.x\right)-\left[\left(2x\right)^2+2.2x.3+3^2\right]\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left(4x^2+12x+9\right)\left(6x^2-3x+1\right)\)
\(=12x^3-4x^2-\left[4x^2\left(6x^2-3x+1\right)+12x\left(6x^2-3x+1\right)+9\left(6x^2-3x+1\right)\right]\)
\(=12x^3-4x^2-\left[\left(24x^4-12x^3+4x^2\right)+\left(72x^3-36x^2+12x\right)+\left(36x^2-27x+9\right)\right]\)
\(=12x^3-4x^2-24x^4+12x^3-4x^2-72x^3+36x^2-12x-36x^2+27x-9\)
\(=-48x^3-8x^2-24x^4+15x-9\)
9 tháng 10 2017
1/ a/ 3x(4x2 - 3xy + 5y)
= (3x.4x2)-[3x.3xy)]+(3x.5y)
= 12x3 - 9x2y + 15xy
b/ (2x - 1).(x2 + 5x + 2)
= 2x3 + 10x2 + 4x - x2 - 5x - 2
= 2x3 + 9x2 - x - 2.
2/ a/ 4x - 8y = 4.(x-2y)
b/ x3 - x2y - x + y
= x2.(x - y) - (x - y)
= (x - y).(x2 - 1)
= (x - y).(x - 1).(x + 1)
3/ a/ x3 - 4x = 0
x..(x2 - 4) = 0
x.(x+2).(x-2) = 0
=> x = 0; x + 2 = 0; x - 2 = 0
hay x = 0; x = -2; x = 2.
b/ (2x+3)2 - x(4x+3) = 18
4x2 + 6x + 9 - 4x2 - 3x - 18 = 0
(4x2 - 4x2) + (6x - 3x) + (9-18) = 0
3x - 9 = 0
=> 3x = 9 => x = 3.
LT
1
13 tháng 8 2017
Tách tách tách :v
$(15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52$
$=>(60x+15-8x^2-2x)-(26x-39-8x^2+12x)-(x^2+3x+2)+x^2=52$
$=>60x+15-8x^2-2x-26x+39+8x^2-12x-x^2-3x-2+x^2=52$
$=>(8x^2-8x^2+x^2-x^2)+(60x-2x-26x-12x-3x)+(15+39-2)=52$
$=>17x+52=52$
$=>x=0$
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: \(\left(x+1\right)^2-\left(x-1\right)^2+3x^2-3x\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1+3x^2-3x\left(x^2-1\right)\)
\(=3x^2+4x-3x^3+3x\)
\(=-3x^3+3x^2+7x\)
3 tháng 11 2017
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
18 tháng 8 2020
a) Ta có: \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
b) Ta có: \(16x-5x^2-3\)
\(=-5x^2+16x-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
c) Ta có: \(2x^2+7x+5\)
\(=2x^2+2x+5x+5\)
\(=2x\left(x+1\right)+5\left(x+1\right)\)
\(=\left(x+1\right)\left(2x+5\right)\)
d) Ta có: \(2x^2+3x-5\)
\(=2x^2+5x-2x-5\)
\(=x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(x-1\right)\)
e) Ta có: \(x^3-3x^2+1-3x\)
\(=\left(x+1\right)\cdot\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
f) Ta có: \(x^2-4x-5\)
\(=x^2-4x+4-9\)
\(=\left(x-2\right)^2-3^2\)
\(=\left(x-2-3\right)\left(x-2+3\right)\)
\(=\left(x-5\right)\left(x+1\right)\)
g) Ta có: \(\left(a^2+1\right)^2-4a^2\)
\(=\left(a^2+1\right)^2-\left(2a\right)^2\)
\(=\left(a^2+1-2a\right)\left(a^2+1+2a\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)^2\)
h) Ta có: \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
k) Ta có: \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
l) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
m) Ta có: \(x^4+4x^2-5\)
\(=x^4-x^2+5x^2-5\)
\(=x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) Ta có:
\(M=4x^2-2x+1\)
\(=\left(2x\right)^2-2x.2.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(2x\right)^2-2x.2.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta lại có: \(\left(2x-\dfrac{1}{2}\right)^2\ge0\)
\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow M\ge\dfrac{3}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow\left(2x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy \(Min_M=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(M=4x^2-2x+1=\left(4x^2-2x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của M là \(\dfrac{3}{4}\) khi x = \(\dfrac{1}{4}\)
\(N=-x^2+x-2=-\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{7}{4}=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{7}{4}\le-\dfrac{7}{4}\)
Vậy GTLN của N là \(-\dfrac{7}{4}\) khi x = \(\dfrac{1}{2}\)