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Bài 2:
a: Để A là số nguyên thì \(3n^3+10n^2-5⋮3n+1\)
\(\Leftrightarrow3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
\(\Leftrightarrow3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow n\in\left\{0;-1;1\right\}\)(do n là số nguyên)
b: Để B là số nguyên thì \(n^3-4n^2+5n-1⋮n-3\)
\(\Leftrightarrow n^3-3n^2-n^2+3n+2n-6+5⋮n-3\)
\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{4;2;8;-2\right\}\)
a: \(\Leftrightarrow3x^3-2x^2+6x^2-4x-3x+2+a-2⋮3x-2\)
=>a-2=0
=>a=2
b: \(\Leftrightarrow3x^3-2x^2+6x^2-4x-3x+2+3⋮3x-2\)
=>\(3x-2\in\left\{1;-1;3;-3\right\}\)
mà x là số nguyên
nên x=1
c: \(\Leftrightarrow x^2+x-3x-3-a+3⋮x+1\)
=>3-a=0
=>a=3
Bài 2:Tìm x biết
\\(\\left(4x+3\\right)^3+\\left(5-7x\\right)^3+\\left(3x-8\\right)^3=0\\)
\\(\\Leftrightarrow\\left[\\left(4x\\right)^3+3.\\left(4x\\right)^2.3+3.4x.3^2+3^3\\right]+\\left[5^3-3.5^2.7x+3.5.\\left(7x\\right)^2-\\left(7x\\right)^3\\right]+\\left[\\left(3x\\right)^3-3.\\left(3x\\right)^2.8+3.3x.8^2-8^3\\right]=0\\)
\\(\\Leftrightarrow64x^3+144x^2+108x+27+125-525x+735x^2-343x^3+27x^3-216x^2+576x-512=0\\)
\\(\\Leftrightarrow-252x^3+663x^2+159x-360=0\\)
\\(\\Leftrightarrow3\\left(-84x^3+221x^2+53x-120\\right)=0\\)
Câu 1:
\(\dfrac{A}{B}=\dfrac{4x^{n+1}y^2}{3x^3y^{n-1}}=\dfrac{4}{3}x^{n-2}y^{2-n+1}=\dfrac{4}{3}x^{n-2}y^{3-n}\)
Để A chia hết cho B thì \(\left\{{}\begin{matrix}n-2>=0\\3-n>=0\end{matrix}\right.\Leftrightarrow2\le n\le3\)
Bài 2:
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\left(x-y\right)+3\left(x+y\right)^2}{x+y}\)
\(=x^2-xy+y^2-2\left(x-y\right)+3\left(x+y\right)\)
\(=x^2-xy+y^2-2x+2y+3x+3y\)
\(=x^2-xy+y^2+x+5y\)
Câu 1 :
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:
\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)
\(\Leftrightarrow\left(x+2\right)^2=-2\)
Ta có:
\(\left(x+2\right)^2\ge0\forall x\)
Vậy pt vô nghiệm
Vậy:ko......
Câu 3:
\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)
\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)
Câu 4:
\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)
\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)
câu 6,
Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)
\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)
\(\Rightarrow-4x+3=8\)
\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)
Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)
\(\Rightarrow7x^2=-x\)
\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).
a: \(x^3+x^2-2x+a⋮x+1\)
\(\Leftrightarrow x^3+x^2-2x-2+a+2⋮x+1\)
=>a+2=0
hay a=-2
b: \(2x^3-4x^2-3a⋮2x-3\)
\(\Leftrightarrow2x^3-3x^2-x^2+1.5x-1.5x+2.25-3a-2.25⋮2x-3\)=>-3a-2,25=0
=>-3a=2,25
hay a=-0,75
c: \(4x^4+3x^2-ax+3⋮x+3\)
\(\Leftrightarrow4x^4+12x^3-12x^3-36x^2+39x^2+117x-ax+3⋮x+3\)
\(\Leftrightarrow-ax+3⋮x+3\)
\(\Leftrightarrow-ax-3a+3+3a⋮x+3\)
=>3a+3=0
hay a=-1
a)=\(-\left(x^2+2x+1\right).\left(x-3\right)\)
=\(-\left(x+1\right)^2.\left(x-3\right)\)
Câu 2:
a: Để f(x) chia hết cho g(x) thì \(2x^3+3x^2-x+4⋮2x+1\)
\(\Leftrightarrow2x^3+x^2+2x^2+x-2x-1+5⋮2x+1\)
\(\Leftrightarrow2x+1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{0;-1;2;-3\right\}\)
b: Để f(x) chia hết cho g(x) thì \(3x^3-x^2+6x⋮3x-1\)
\(\Leftrightarrow3x^3-x^2+6x-2+2⋮3x-1\)
\(\Leftrightarrow3x-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)