(4x4 + 4x3 - 16x2 - 16x) : ( 4x + 4)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 10 2021
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
5 tháng 10 2021
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)
5 tháng 10 2021
\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
5 tháng 10 2021
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
\(\Leftrightarrow x^5-1=4x^4+4x^3+4x^2+4x+4\)
\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x^4+x^3+x^2+x+1\right)=0\)
\(\Leftrightarrow x=5\)
TE
1
22 tháng 8 2019
a, \(x^4-4x^3-6x^2-4x+1=0\)(*)
<=> \(x^4+4x^2+1-4x^3-4x+2x^2-12x^2=0\)
<=> \(\left(x^2-2x+1\right)^2=12x^2\)
<=>\(\left(x-1\right)^4=12x^2\) <=> \(\left[{}\begin{matrix}\left(x-1\right)^2=\sqrt{12}x\\\left(x-1\right)^2=-\sqrt{12}x\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x^2-2x+1-\sqrt{12}x=0\left(1\right)\\x^2-2x+1+\sqrt{12}x=0\left(2\right)\end{matrix}\right.\)
Giải (1) có: \(x^2-2x+1-\sqrt{12}x=0\)
<=> \(x^2-2x\left(1+\sqrt{3}\right)+\left(1+\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2+1=0\)
<=> \(\left(x-1-\sqrt{3}\right)^2-3-2\sqrt{3}=0\)
<=> \(\left(x-1-\sqrt{3}\right)^2=3+2\sqrt{3}\) <=> \(\left[{}\begin{matrix}x-1-\sqrt{3}=\sqrt{3+2\sqrt{3}}\\x-1-\sqrt{3}=-\sqrt{3+2\sqrt{3}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(ktm\right)\\x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(tm\right)\end{matrix}\right.\)
=> \(x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)
Giải (2) có: \(x^2-2x+1+\sqrt{12}x=0\)
<=> \(x^2-2x\left(1-\sqrt{3}\right)+\left(1-\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2+1=0\)
<=> \(\left(x+\sqrt{3}-1\right)^2=3-2\sqrt{3}\) .Có VP<0 => PT (2) vô nghiệm
Vậy pt (*) có nghiệm x=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)
H
8 tháng 8 2023
`4-x=2(x-4)^2`
`<=>4-x=2(x^2-8x+16)`
`<=> 4-x=2x^2 - 16x+32`
`<=> 4-x-2x^2+16x-32=0`
`<=> -2x^2 +15x-28=0`
`<=> -(2x^2-15x+28)=0`
`<=>-(2x^2-7x-8x+28)=0`
`<=> - [x(2x-7) - 4(2x-7)]=0`
`<=> -(2x-7)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-7\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
__
`(x^2 +1) (x-2)+2x=4`
`<=> x^3 -2x^2 +x-2+2x-4=0`
`<=> x^3 -2x^2 +3x-6=0`
`<=> (x^3+3x)-(2x^2+6)=0`
`<=> x(x^2 +3) -2(x^2+3)=0`
`<=>(x^2+3)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)
__
`x^4 -16x^2=0`
`<=> x^2 (x^2 -16)=0`
`<=>x^2(x-4)(x+4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
\(4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)
\(\Leftrightarrow4-x=2x^2-16x+32\)
\(\Leftrightarrow2x^2-15x+28=0\)
\(\Leftrightarrow2x^2-7x-8x+28=0\)
\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
___________
\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Leftrightarrow x^3-2x^2+x-2+2x=4\)
\(\Leftrightarrow x^3-2x^2+3x-2-4=0\)
\(\Leftrightarrow x^3-2x^2+3x-6=0\)
\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
________________
\(x^4-16x^2=0\)
\(\Leftrightarrow\left(x^2\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)x\left(x+4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
16 tháng 8 2021
a: Ta có: \(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3+27-8x^3+2\)
=29
b: Ta có: \(B=\left(64x^3-1\right)-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-1-64x^3-12x-48x^2+9\)
\(=-12x+8\)
c: Ta có: \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x^2+xy+y^2\right)-3\left(-2xy\right)\)
\(=2x^2+2xy+2y^2+6xy\)
\(=2x^2+8xy+2y^2\)
TN
1
\(\left(4x^4+4x^3-16x^2-16x\right):\left(4x+4\right)\)
\(=[\left(4x^4+4x^3\right)-\left(16x^2+16x\right)]:\left(4x+4\right)\)
\(=[4x^3.\left(x+1\right)-16x.\left(x+1\right)]:\left(4x+4\right)\)
\(=[\left(4x^3-16x\right).\left(x+1\right)]:\left(4x+4\right)\)
\(=[4.\left(x^3-4x\right).\left(x+1\right)]:[4.\left(x+1\right)]\)
\(=\frac{4.\left(x+1\right).\left(x^3-4x\right)}{4.\left(x+1\right)}\)
\(=x^3-4x\)