A(x)= 2x^3-3x^2+3x+8
B(x)= 3x^3+2x^3-5x+1
Tính A(x)-B(x)
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DX
12 tháng 4 2023
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
30 tháng 4 2022
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
6 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`P(x)+Q(x) = (3x^4-2x^3+3x+11)+(3x^2- x^3-5x+3x+4-x+2x^4)`
`= 3x^4-2x^3+3x+11+3x^2- x^3-5x+3x+4-x+2x^4`
`= (3x^4 + 2x^4) + (-2x^3 - x^3) + 3x^2 + (3x + 3x - 5x - x) + (11+4)`
`= 5x^4 - 3x^3 + 3x^2 + 15`
`b,`
` A(x) = P(x) + B(x)`
Thay `B(x) = 2x^3 - 3x^4 - 2`
`A(x) = P(x) + B (x)`
`=> A (x) = (2x^3 - 3x^4 - 2)+(3x^4 - 2x^3 + 3x + 11)`
`= 2x^3 - 3x^4 - 2+ 3x^4 - 2x^3 + 3x + 11`
`= (2x^3 - 2x^3) + (-3x^4 + 3x^4) + 3x + (-2+11) `
`= 3x + 9`
`A(x) = 3x+9 = 0`
`=> 3x = 0-9`
`=> 3x = -9`
`=> x = -9 \div 3`
`=> x = -3`
Vậy, nghiệm của đa thức là `x = -3.`
24 tháng 2 2023
a: \(=\dfrac{2x^4+x^3-5x^2-3x-3}{x^2-3}\)
\(=\dfrac{2x^4-6x^2+x^3-3x+x^2-3}{x^2-3}\)
\(=2x^2+x+1\)
b: \(=\dfrac{x^5+x^2+x^3+1}{x^3+1}=x^2+1\)
c: \(=\dfrac{2x^3-x^2-x+6x^2-3x-3+2x+6}{2x^2-x-1}\)
\(=x+3+\dfrac{2x+6}{2x^2-x-1}\)
d: \(=\dfrac{3x^4-8x^3-10x^2+8x-5}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5}{3x^2-2x+1}\)
\(=x^2-2x-5\)
16 tháng 6 2023
a: \(A\left(x\right)=2x^4-x^3+3x^2+9x-2\)
\(B\left(x\right)=2x^4-5x^3-x+9\)
\(C\left(x\right)=x^4+4x^2+5\)
A(x): bậc 4; hệ số cao nhất là 2; hệ số tự do là -2
B(x): bậc 4; hệ số cao nhất là 4; hệ số tự do là 9
b: M(x)=A(x)+B(x)=4x^4-6x^3+3x^2+8x+7
N(x)=B(x)-A(x)=-4x^3-3x^2-10x+11
c: Q(x)=-N(x)=4x^3+3x^2+10x-11
5 tháng 9 2019
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
2T
5 tháng 9 2019
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
14 tháng 8 2020
a) Ta có: 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)
b) Ta có: \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\)
\(\Leftrightarrow8x=15\)
hay \(x=\frac{15}{8}\)
Vậy: \(x=\frac{15}{8}\)
c) Ta có: \(3x\left(2-x\right)+2x\left(x-1\right)=5x\left(x+3\right)\)
\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)
\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)
\(\Leftrightarrow-6x^2-11x=0\)
\(\Leftrightarrow6x^2+11x=0\)
\(\Leftrightarrow x\left(6x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)
d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)
\(\Leftrightarrow14x^2+18=0\)
\(\Leftrightarrow14x^2=-18\)
mà \(14x^2\ge0\forall x\)
nên \(x\in\varnothing\)
Vậy: \(x\in\varnothing\)
\(A\left(x\right)-B\left(x\right)=\left(2x^3-3x^2+3x+8\right)-\left(3x^3+2x^3-5x+1\right)\\ =2x^3-3x^2+3x+8-3x^3-2x^3+5x-1\\ =\left(2x^3-3x^3+2x^3\right)+\left(-3x^2\right)+\left(3x+5x\right)+\left(8-1\right)\\ =x^3-3x^2+8x+7\)