Tính (2x^3-2bx-24):(x^2+4x+3)
Những câu hỏi liên quan
Cho các đa thức: M(x)=ax^3+2x^3-4x+c-5
N(x)=-2x^3-2bx+2
(trong đó a,b,c là hằng số)
Xác định các hệ số a, b, c để: M(x) = N(x)
Thực hiện phép tính:
a. (19x^2-14x^3+9-20x+2x^4) : (1+x^2-4x)
b. (3x^4-2x^3-2x^2+4x+8) : x^2-2
c. (2x^3-26x-24) : (x^2+4x+3)
Giúp mình với mn ơi :(((
Tính:
a) (2x3-26x-24) : (x2+4x+3)
b) (x3-7x+6) : (x+3)
Tính:
a) \((8{x^3} + 2{x^2} - 6x):(4x)\);
b) \((5{x^3} - 4x):( - 2x)\);
c) \(( - 15{x^6} - 24{x^3}):( - 3{x^2})\).
a) \(\begin{array}{l}(8{x^3} + 2{x^2} - 6x):(4x) = 8{x^3}:(4x) + 2{x^2}:(4x) - (6x):(4x)\\ = (8:4).({x^3}:x) + (2:4).({x^2}:x) - (6:4).(x:x)\\ = 2{x^2} + \dfrac{1}{2}x - \dfrac{3}{2}\end{array}\)
b) \(\begin{array}{l}(5{x^3} - 4x):( - 2x) = 5{x^3}:( - 2x) - 4x:( - 2x) = (5: - 2).({x^3}:x) - (4: - 2).(x:x)\\ = - \dfrac{5}{2}{x^{3 - 1}} - ( - 2) = - \dfrac{5}{2}{x^2} + 2\end{array}\)
c) \(\begin{array}{l}( - 15{x^6} - 24{x^3}):( - 3{x^2}) = ( - 15{x^6}):( - 3{x^2}) + ( - 24{x^3}):( - 3{x^2})\\ = ( - 15: - 3).({x^6}:{x^2}) + ( - 24: - 3).({x^3}:{x^2})\\ = 5.{x^{6 - 2}} + 8.{x^{3 - 2}} = 5{x^4} + 8x\end{array}\)
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1. Viết thành tích:
a) x2 - 7xy + 10y2
b) xm+4 - xm+3 - x +1
c) 2x2a +2bx2 +2ca2 -4axy + 2y2a +2by2 + 2cy2 - 4bxy - 4cxy
2. Giải: x4 - 4x3 - 2x2 +4x +1 = 0
Giải phương trình:
1. (x+2)(x-3)(x2+2x-24) =16x2
2. (4x+7)(4x+5)(x+1)(2x+1) =9
3. (4x-5)2(2x-3)(x-1)=1,5
1.
PT \(\Leftrightarrow (x+2)(x-3)(x-4)(x+6)=16x^2\)
\(\Leftrightarrow [(x+2)(x+6)][(x-3)(x-4)]=16x^2\)
\(\Leftrightarrow (x^2+8x+12)(x^2-7x+12)=16x^2\)
\(\Leftrightarrow (a+8x)(a-7x)=16x^2\) (đặt \(x^2+12=a\) )
\(\Leftrightarrow a^2+ax-72x^2=0\)
\(\Leftrightarrow (a-8x)(a+9x)=0\Rightarrow \left[\begin{matrix} a-8x=0\\ a+9x=0\end{matrix}\right.\)
Nếu \(a-8x=0\Leftrightarrow x^2+12-8x=0\Leftrightarrow (x-2)(x-6)=0\Rightarrow \left[\begin{matrix} x=2\\ x=6\end{matrix}\right.\)
Nếu \(a+9x=0\Leftrightarrow x^2+12+9x=0\Leftrightarrow x=\frac{-9\pm \sqrt{33}}{2}\)
Vậy...........
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2.
PT \(\Leftrightarrow [(4x+7)(2x+1)][(4x+5)(x+1)]=9\)
\(\Leftrightarrow (8x^2+18x+7)(4x^2+9x+5)=9\)
\(\Leftrightarrow (2a+7)(a+5)=9\) (đặt \(a=4x^2+9x\) )
\(\Leftrightarrow 2a^2+17a+26=0\)
\(\Leftrightarrow (a+2)(2a+13)=0 \)\(\Rightarrow \left[\begin{matrix} a+2=0\\ 2a+13=0\end{matrix}\right.\)
Nếu \(a+2=0\Leftrightarrow 4x^2+9x+2=0\Leftrightarrow (4x+1)(x+2)=0\)
\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{4}\\ x=-2\end{matrix}\right.\)
Nếu \(2a+13=0\Leftrightarrow 8x^2+18x+13=0\) (pt này dễ thấy vô nghiệm)
Vậy.........
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3.
PT \(\Leftrightarrow (16x^2-40x+25)(2x^2-5x+3)=1,5\)
\(\Leftrightarrow [8(2x^2-5x)+25](2x^2-5x+3)=1,5\)
\(\Leftrightarrow (8a+25)(a+3)=1,5\) (đặt \(a=2x^2-5x\))
\(\Leftrightarrow 8a^2+49a+73,5=0\)
\(\Leftrightarrow 16a^2+98a+147=0\)
\(\Leftrightarrow (8a+21)(2a+7)=0\) \(\Rightarrow \left[\begin{matrix} 8a+21=0\\ 2a+7=0\end{matrix}\right.\)
Nếu \(8a+21=0\Leftrightarrow 16x^2-40x+21=0\)
\(\Leftrightarrow (4x-7)(4x-3)=0\Rightarrow \left[\begin{matrix} x=\frac{7}{4}\\ x=\frac{3}{4}\end{matrix}\right.\)
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Bài 1: Tìm a, b, c biết
3x^2(a^2-2bx-3c)=3x^4-12x^3+27x2
Bài 2:Tìm x biết
a) 4(x+2)-7(2x-1)+3(3x-4)=30
b) 2(5x-8)-3(4x-5)=4(3x-4)+11
c) 5x(1-2x)-3x(x+18)=0
d) (x+2)(x+3)-(x-2)(x+5)=0
Bài 3. Tìm x:
a) (3 – x)^2 – x(x – 4) = 2x – 5
b) x^2 – 2x + 1 = 25x^2
c) 4x^2 – 4x = 24
b) x2 - 2x + 1 = 25x2
<=> (x - 1)2 - 25x2 = 0
<=> (x - 1 - 5x)(x - 1 + 5x) = 0
<=> (-4x - 1)(6x - 1) = 0
<=> \(\orbr{\begin{cases}-4x-1=0\\6x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{6}\end{cases}}\)
c) 4x2 - 4x = 24
<=> x2 - x - 6 = 0
<=> x2 - 3x + 2x - 6 = 0
<=> x(x - 3) + 2(x - 3) = 0
<=> (x + 2)(x - 3) = 0
<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
a) (3 - x)2 - x(x - 4) = 2x - 5
<=> x2 - 6x + 9- x2 + 4x = 2x - 5
<=> -2x + 9 = 2x - 5
<=> 2x + 2x = 9 + 5
<=> 4x = 14
<=> x = 7/2
Thực hiện phép tính chia
a. (6x3 + 3x2 + 4x + 2) : (3x2 + 2)
b. (2x3 - 26x - 24) : (x2 + 4x + 3)
c. (x3 - 7x + 6) : (x + 3)
a) 6x3 + 3x2 + 4x + 2
= ( 6x3 + 3x2 ) + ( 4x + 2 )
= 3x2( 2x + 1 ) + 2( 2x + 1 )
= ( 2x + 1 )( 3x2 + 2 )
=> ( 6x3 + 3x2 + 4x + 2 ) : ( 3x2 + 2 ) = 2x + 1
b) 2x3 - 26x - 24
= 2( x3 - 13x - 12 )
= 2( x3 + 4x2 - 4x2 + 3x - 16x - 12 )
= 2[ ( x3 + 4x2 + 3x ) - ( 4x2 + 16x + 12 ) ]
= 2[ x( x2 + 4x + 3 ) - 4( x2 + 4x + 3 ) ]
= 2( x2 + 4x + 3 )( x - 4 )
=> ( 2x3 - 26x - 24 ) : ( x2 + 4x + 3 ) = 2( x - 4 ) = 2x - 8
c) x3 - 7x + 6
= x3 - 3x2 + 3x2 + 2x - 9x - 6
= ( x3 - 3x2 + 2x ) + ( 3x2 - 9x + 6 )
= x( x2 - 3x + 2 ) + 3( x2 - 3x + 2 )
= ( x2 - 3x + 2 )( x + 3 )
=> ( x3 - 7x + 6 ) : ( x + 3 ) = x2 - 3x + 2
a,\(\left(6x^3+3x^2+4x+2\right)\div\left(3x^2+2\right)\)
\(=\left[3x^2\left(2x+1\right)+2\left(2x+1\right)\right]\div\left(3x^2+2\right)\)
\(=\left[\left(3x^2+2\right)\left(2x+1\right)\right]\div\left(3x^2+2\right)\)
\(=2x+1\)