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a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
`@` `\text {dnammv}`
`a,`
`4x(x^2-x-1)-(x^2-2)(x+3)`
`= 4x^3-4x^2-4x- [x^2(x+3)-2(x+3)]`
`= 4x^3-4x^2-4x- (x^3+3x^2-2x-6)`
`= 4x^3-4x^2-4x-x^3-3x^2+2x+6`
`= 3x^3 - 7x^2-2x+6`
`b,`
`(x+5)(x+7)-7x(x+3)`
`= x(x+7)+5(x+7)-7x^2-21x`
`= x^2+7+5x+35-7x^2-21x`
`= -6x^2-16x+35`
`c,`
`x(x^2-x-2)-(x+5)(x-1)`
`= x^3-x^2-2x- [x(x-1)+5(x-1)]`
`= x^3-x^2-2x- (x^2-x+5x-5)`
`= x^3-x^2-2x - x^2 + x -5x+5`
`= x^3-2x^2- 4x+5`
`d,`
`(x+5)(x+7)-(x-4)(x+3)`
`= x(x+7)+5(x+7)- [x(x+3)-4(x+3)]`
`= x^2+7x+5x+35 - (x^2+3x-4x-12)`
`= x^2+12x+35 - x^2+x+12`
`= 13x+47`
Dễ
Thế
Mà
Cũnhoir
Dc
Ạ
Chịu
Chắc
Phải
Ngu
Lamqs
Mới
Hỏi
Câu
Này
a: \(P\left(x\right)=2x^3-x^3+x^2+3x-2x+2=x^3+x^2+x+2\)
\(Q\left(x\right)=3x^3-4x^3-4x^2+5x^2+3x-4x+1=-x^3+x^2-x+1\)
b: M(x)=P(x)+Q(x)
\(=x^3+x^2+x+2-x^3+x^2-x+1=2x^2+3\)
N(x)=P(x)-Q(x)
\(=x^3+x^2+x+2+x^3-x^2+x-1=2x^3+2x+1\)
c: Vì \(2x^2+3>0\forall x\)
nên M(x) vô nghiệm
a, \(P\left(x\right)=x^3+x^2+x+2\)
\(Q\left(x\right)=-x^3+x^2-x+1\)
b, \(M\left(x\right)=x^3+x^2+x+2-x^3+x^2-x+1=2x^2+3\)
\(N\left(x\right)=x^3+x^2+x+2+x^3-x^2+x-1=2x^3+2x+1\)
c, giả sử \(M\left(x\right)=2x^2+3=0\)( vô lí )
vì 2x^2 >= 0 ; 2x^2 + 3 > 0
Vậy giả sử là sai hay đa thức M(x) ko có nghiệm
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
a: \(=-4x^2+20x+2x-10=-4x^2+22x-10\)
b: =x^2-9
c: =x^3+27
d: \(=-2x^2-6x+x+3=-2x^2-5x+3\)
e: =8a^3+1
f: =(3-x)(x+1)(x+2)
=(3-x)(x^2+3x+2)
=3x^2+9x+6-x^3-3x^2-2x
=-x^3+7x+6
\(A\left(x\right)⋮x-1\)
=>\(mx^2-mx+\left(m-n-1\right)x-m+n+1+m-n-1-3⋮x-1\)
=>m-n-4=0(2)
\(A\left(x\right)⋮x+1\)
=>\(mx^2+mx-\left(m+n+1\right)x-\left(m+n+1\right)+m+n-2⋮x+1\)
=>m+n-2=0(1)
Từ (1),(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}m+n=2\\m-n=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m=6\\m+n=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m=3\\n=2-3=-1\end{matrix}\right.\)
b: \(B\left(x\right)⋮x-2\)
=>\(\left(m+1\right)x^2-\left(2m+2\right)x+\left(2m+6\right)x-2\left(2m+2\right)+2\left(2m+2\right)+3⋮x-2\)
=>2(2m+2)+3=0
=>4m+7=0
=>\(m=-\dfrac{7}{4}\)
c: \(C\left(x\right)⋮x-1\)
=>\(\left(2n-3\right)x^2-\left(n+2\right)x-9⋮x-1\)
=>\(\left(2n-3\right)x^2-\left(2n-3\right)+\left(2n-3-n-2\right)x-9⋮x-1\)
=>\(\left(n-5\right)x-9⋮x-1\)
=>\(x\left(n-5\right)-\left(n-5\right)+n-5-9⋮x-1\)
=>n-14=0
=>n=14
d: \(D\left(x\right)⋮x-3\)
=>\(5x^2-15x+\left(-3n-1+15\right)x+7⋮x-3\)
=>\(\left(-3n+14\right)x+7⋮x-3\)
=>\(\left(-3n+14\right)x-3\left(-3n+14\right)+3\left(-3n+14\right)+7⋮x-3\)
=>3(-3n+14)+7=0
=>-9n+49=0
=>-9n=-49
=>\(n=\dfrac{49}{9}\)