K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
19 tháng 6 2019
\(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+16x^7-28x^6+12x^5\)
M
2
15 tháng 8 2015
a) \(-4x^5\cdot\left(x^3-4x^2+7x-3\right)=-4x^8+16x^7-28x^6+12x^5\)
b) \(4x^3y^2\cdot\left(-2x^2y+4x^4-3y^2\right)=-6x^5y^3+16x^7y^2-12x^3y^4\)
28 tháng 6 2021
a) (7x - 8)(7x + 8) - 10(2x + 3)2 + 5x(3x - 2)2 - 4x(x - 5)2
= 49x2 - 64 - 10(4x2 + 12x + 9) + 5x(9x2 - 12x + 4) - 4x(x2 - 10x + 25)
= 49x2 - 64 - 40x2 - 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x - 100x
= 41x3 - 51x2 - 160x - 154
b) (x2 - 3)(x2 + 3) - 5x2(x + 1)2 - (x2 - 3x)(x2 - 2x) + 4x(x + 2)2
= x4 - 9 - 5x2(x2 + 2x + 1) - x4 + 5x3 - 6x2 + 4x(x2 + 4x + 4)
= 5x3 - 6x2 - 5x4 - 10x3 - 5x2 + 4x3 + 16x2 + 16x - 9
= -5x4 - x3 + 5x2 + 16x - 9
QA
28 tháng 6 2021
Trả lời:
a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )2 + 5x ( 3x - 2 )2 - 4x ( x - 5 )2
= 49x2 - 64 - 10 ( 4x2 + 12x + 9 ) + 5x ( 9x2 - 12x + 4 ) - 4x ( x2 - 10x + 25 )
= 49x2 - 64 - 40x2 + 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x2 - 100x
= 41x3 - 11x2 + 40x - 154
b , ( x2 - 3 ) ( x2 + 3 ) - 5x2 ( x + 1 )2 - ( x2 - 3x ) ( x2 - 2x ) + 4x ( x + 2 )2
= x4 - 9 - 5x2 ( x2 + 2x + 1 ) - ( x4 - 2x3 - 3x3 + 6x2 ) + 4x ( x2 + 4x + 4 )
= x4 - 9 - 5x4 - 10x3 - 5x2 - x4 + 2x3 + 3x3 - 6x2 + 4x3 + 16x2 + 16x
= - 5x4 - x3 + 5x2 + 16x - 9
1 tháng 1 2018
a)\(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}:\dfrac{2\left(x-3\right)}{3\left(x+1\right)}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(x-3\right)\left(9+3x+x^2\right)3\left(x+1\right)}{5\left(x+1\right)2\left(x-3\right)}\)
\(=\dfrac{-\left(9+3x+x^2\right)3}{10}\)
b)\(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4\left(x^2-4\right):\dfrac{3\left(x+2\right)}{7x-2}\)
\(=4\left(x-2\right)\left(x+2\right)\cdot\dfrac{7x-2}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c)\(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x^3+1\right)}{x-1}:x^2-x+1\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{x-1}\cdot\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d)\(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{\left(1-x\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{2\left(2x+3y\right)}{x-1}\cdot\dfrac{-\left(x-1\right)\left(1+x+x^2\right)}{\left(2x+3y\right)^2}\)
\(=\dfrac{-2\left(1+x+x^2\right)}{2x+3y}\)
HN
1 tháng 1 2018
a) \(\dfrac{27-x^3}{5x+5}:\dfrac{2x-6}{3x+3}\)
\(=\dfrac{27-x^3}{5x+5}.\dfrac{3x+3}{2x-6}\)
\(=\dfrac{\left(3-x\right)\left(9+3x+x^2\right)}{5\left(x+1\right)}.\dfrac{3\left(x+1\right)}{2\left(x-3\right)}\)
\(=-\dfrac{3\left(x-3\right)\left(x^2+3x+9\right)\left(x+1\right)}{10\left(x+1\right)\left(x-3\right)}\)
\(=-\dfrac{3\left(x^2+3x+9\right)}{10}\)
b) \(4x^2-16:\dfrac{3x+6}{7x-2}\)
\(=4x^2-16.\dfrac{7x-2}{3x+6}\)
\(=\dfrac{4\left(x^2-4\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(x+2\right)\left(7x-2\right)}{3\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)\left(7x-2\right)}{3}\)
c) \(\dfrac{3x^3+3}{x-1}:x^2-x+1\)
\(=\dfrac{3x^3+3}{x-1}.\dfrac{1}{x^2-x+1}\)
\(=\dfrac{3\left(x^3+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3\left(x+1\right)}{x-1}\)
d) \(\dfrac{4x+6y}{x-1}:\dfrac{4x^2+12xy+9y^2}{1-x^3}\)
\(=\dfrac{4x+6y}{x-1}.\dfrac{1-x^3}{4x^2+12xy+9y^2}\)
\(=\dfrac{2\left(2x+3y\right)\left(1-x\right)\left(1+x+x^2\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(2x+3y\right)\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(2x+3y\right)^2}\)
\(=-\dfrac{2\left(x^2+x+1\right)}{2x+3y}\)
TN
23 tháng 7 2017
\(6x^2-7x-20\)
\(=6x^2-15x+8x-20\)
\(=6x\left(x-\dfrac{5}{2}\right)+8\left(x-\dfrac{5}{2}\right)\)
\(=\left(6x+8\right)\left(x-\dfrac{5}{2}\right)\)
\(=2\left(3x+4\right)\left(x-\dfrac{5}{2}\right)\)
\(b,x^3-4x^2-4x-5\)
\(=x^3-5x^2+x^2-5x+x-5\)
\(=x^2\left(x-5\right)+x\left(x-5\right)+\left(x-5\right)\)
\(=\left(x^2+x+1\right)\left(x-5\right)\)
23 tháng 7 2017
Bạn có thể dung máy tính CASIO fx-570ES PLUS để giải hệ phương trình và phương trình.
Bayh mình giải hộ bạn nha
a, 6x2 - 7x - 20 = 6x2 -15x + 8x - 20
= 3x(2x-5)+4(2x-5)
= (3x+4)(2x-5)
b, x3 - 4x2 - 4x - 5= x3 -5x2 +x2 -5x+x-5 = x2(x-5)+x(x-5)+x-5
= (x-5)(x2 +x+1) . Đến đây thì ko cần phân tích tiếp nữa vì x2 +x+1 phân tích sẽ ra VN và căn nha bạn.
c, x4 + 6x3 + 7x2 – 6x + 1 = x4 + 6x3 – 2x2 + 9x2 – 6x + 1
= x4 + 2x2(3x – 1) + (3x)2- 2.3x.1+12
= (x2)2 +2.x2 (3x-1) + (3x-1)2
= (x2 + 3x – 1)2
KT
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
NB
5
L
24 tháng 7 2016
1 )
=x3-2x2+6x2-12x+5x-10
=x2(x-2)+6x(x-2)+5(x-2)
=(x-2)(x2+6x+5)
=(x-2)(x2+x+5x+5)
=(x-2)[x(x+1)+5(x+1)]
=(x-2)(x+1)(x+5)
24 tháng 7 2016
toàn mũ lớn hơn 3 khó làm quá!!!! >.<
653645645645645676746784734746856876897684737547
-4x5(x3 - 4x2 + 7x - 3)
= -4x8 + 16x7 - 28x6 + 12x5
Ta có : \(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+4x^7-28x^6+12x^5\)
@Hoc tot@