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\(Bài.44:\\ a,3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\dfrac{7}{3}\\ b.2x^2+9=0\\ \Leftrightarrow x^2=-\dfrac{9}{2}\left(vô.lí\right)\\ \Rightarrow Không.có.x.thoả.mãn\)
43:
a: \(A=2x\left(x^2-2x-3\right)-6x^2+5x-1+9x^2+3x+3\)
\(=2x^3-4x^2-6x+3x^2+8x+2\)
\(=2x^3-x^2+2x+2\)
b: \(\dfrac{A}{2x-1}=\dfrac{x^2\left(2x-1\right)+2x-1+3}{2x-1}=x^2+1+\dfrac{3}{2x-1}\)
Thương là x^2+1
Dư là 3
c: A chia hết cho 2x-1
=>3 chia hết cho 2x-1
=>2x-1 thuộc {1;-1;3;-3}
=>x thuộc {1;0;2;-1}
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)
Thay x = -2, ta có:
\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)
b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)
Vậy A=0 khi x=0
a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))
A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)
A = 2\(x^3\) + 8\(x\)
b, \(x=-2\) ⇒ A = 2.(-2)3 + 8.(-2) = - 32
A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0
2\(x\left(x^2+4\right)\) = 0
vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\) =0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{5}{2}x+1\) \(B=0,4x-5\)
a) \(A=\frac{5}{2}.\frac{1}{5}+1\) \(B=0,4.\left(-10\right)-5\)
\(A=\frac{1}{2}+1=1\) \(B=-4-5=-9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) P(x) = 7x2 . (x2 – 5x + 2 ) – 5x .(x3 – 7x2 + 3x)
= 7x2 . x2 + 7x2 . (-5x) + 7x2 . 2 – [5x. x3 + 5x . (-7x2) + 5x . 3x]
= 7. (x2 . x2) + [7.(-5)] . (x2 . x) + (7.2).x2 – {5. (x.x3) + [5.(-7)]. (x.x2) + (5.3).(x.x)}
= 7x4 + (-35). x3 + 14x2 – [ 5x4 + (-35)x3 + 15x2 ]
= 7x4 + (-35). x3 + 14x2 - 5x4 + 35x3 - 15x2
= (7x4 – 5x4) + [(-35). x3 + 35x3 ] + (14x2 - 15x2 )
= 2x4 + 0 - x2
= 2x4 – x2
b) Thay x = \( - \dfrac{1}{2}\) vào P(x), ta được:
P(\( - \dfrac{1}{2}\)) = 2. (\( - \dfrac{1}{2}\))4 – (\( - \dfrac{1}{2}\))2 \))
\(\begin{array}{l} = 2.\dfrac{1}{{16}} - \dfrac{1}{4} \\ = \dfrac{1}{8} - \dfrac{{2}}{8} \\ = \dfrac{-1}{8} \end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Khi x=1 thì \(A=\dfrac{x-8}{x-3}=\dfrac{1-8}{1-3}=\dfrac{-7}{-2}=\dfrac{7}{2}\)
Khi x=2/11 thì \(A=\dfrac{\dfrac{2}{11}-8}{\dfrac{2}{11}-3}=\dfrac{-86}{11}:\dfrac{-31}{11}=\dfrac{86}{31}\)
b: Để A là số nguyên thì \(x-8⋮x-3\)
\(\Leftrightarrow x-3-5⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{4;2;8;-2\right\}\)