Bài 1:
a) 4x2 +9x +5
b) x2 + 8x +15
c) -x4 + 4x3 + 9
d) -4x4 + 2x3 + 2x2 - 8x + 7
K
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27 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
Thu gọn:
`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)
`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`
`= -x^5 + 5x^4 + 2x^2 + 2x - 4`
`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)
`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`
`= x^5 - x^4 - x^3 - x^2 + 7x - 2`
`@` Tổng:
`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`
`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`
`= 4x^4 - x^3 + x^2 + 9x - 6`
`@` Hiệu:
`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`
`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`
`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`
`b)`
`@` Thu gọn:
\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)
`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`
`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`
`= x^4 - 2x^3 - x^2 + 15x + 10`
\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)
`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`
`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`
`= x^4 + 3x^3 + 2x - 4`
`@` Tổng:
`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)
`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`
`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`
`= 2x^4 + x^3 - x^2 + 17x + 6`
`@` Hiệu:
`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)
`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`
`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`
`= -5x^3 - x^2 + 13x + 14`
`@` `\text {# Kaizuu lv u.}`
8 tháng 8 2021
c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)
\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)
\(=\left(2x-y+2\right)^2\)
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=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)
DH
6 tháng 6 2017
a,\(x^3-7x+6\)
\(=x^3-2x^2+2x^2-4x-3x+6\)
\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)
\(=x^2.\left(x-2\right)+2x.\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)
\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)
b,\(x^3-9x^2+6x+16\)
\(=x^3-8x^2-x^2+8x-2x+16\)
\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)
\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)
\(=\left(x-8\right).\left(x^2-x-2\right)\)
\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)
\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)
\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)
c,\(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)
\(=\left(x-5\right).\left(x^2-x-6\right)\)
\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)
\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)
\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)
\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)
Chúc bạn học tốt!!!
DH
6 tháng 6 2017
d,\(2x^3-x^2+5x+3\)
\(=2x^3+x^2-2x^2-x+6x+3\)
\(=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2.\left(2x+1\right)-x.\left(2x+1\right)+3.\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x^2-x+3\right)\)
e, \(27x^3-27x^2+18x-4\)
\(=27x^3-9x^2-18x^2+6x+12x-4\)
\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)
\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)
\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)
Chúc bạn học tốt!!!
6 tháng 6 2017
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
18 tháng 7 2021
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
18 tháng 7 2021
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
DX
1
31 tháng 10 2021
b: \(\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
c: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
\(=5x^3+14x^2+12x+8\)
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