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Cách 1:
(x3 – 2x2 + x – 1) (3x – 2)
= x3 . (3x – 2) + (-2x2) .(3x – 2) + x .(3x – 2) + (-1) . (3x – 2)
= x3 . 3x + x3 . (-2) + (-2x2). 3x + (-2x2) . (-2) + x . 3x + x. (-2) + (-1). 3x + (-1). (-2)
= 3x4 – 2x3 – 6x3 + 4x2 + 3x2 – 2x – 3x + 2
= 3x4 + (-2x3 -6x3) + (4x2 + 3x2 ) + (-2x – 3x) + 2
= x4 + (-8x3) + 7x2 + (-5x) + 2
= x4 – 8x3 +7x2 – 5x + 2
Cách 2:
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`7,`
`a,`
\(M(x) = - 5x ^ 4 + 3x ^ 5 + x(x ^ 2 + 5) + 14x ^ 4 - 6x ^ 5 - x ^ 3 + x - 1 \)
\(M(x)=-5x^4+3x^5+x^3+5x+14x^4-6x^5-x^3+x-1\)
`M(x)=(3x^5-6x^5)+(-5x^4+14x^4)+(x^3-x^3)+(5x+x)-1`
`M(x)=-3x^5+9x^4+6x-1`
\(N(x)=x ^ 4 (x - 5) - 3x ^ 3 + 3x + 2x ^ 5 - 4x ^ 4 + 3x ^ 3 - 5 \)
\(N(x)=x^5-5x^4-3x^3+3x+2x^5-4x^4+3x^3-5\)
`N(x)=(x^5+2x^5)+(-5x^4-4x^4)+(-3x^3+3x^3)+3x-5`
`N(x)=3x^5-9x^4+3x-5`
`b,`
`H(x)=M(x)+N(x)`
\(H(x)=(-3x^5+9x^4+6x-1)+(3x^5-9x^4+3x-5) \)
`H(x)=-3x^5+9x^4+6x-1+3x^5-9x^4+3x-5`
`H(x)=(-3x^5+3x^5)+(9x^4-9x^4)+(6x+3x)+(-1-5)`
`H(x)=9x-6`
`G(x)=M(x)-N(x)`
\(G(x)=(-3x^5+9x^4+6x-1)-(3x^5-9x^4+3x-5)\)
`G(x)=-3x^5+9x^4+6x-1-3x^5+9x^4-3x+5`
`G(x)=(-3x^5-3x^5)+(9x^4+9x^4)+(6x-3x)+(-1+5)`
`G(x)=-6x^5+18x^4+3x+4`
`c,`
`H(x)=9x-6`
Hệ số cao nhất của đa thức: `9`
Hệ số tự do: `-6`
`G(x)=-6x^5+18x^4+3x+4`
Hệ số cao nhất của đa thức: `-6`
Hệ số tự do: `4`
`d,`
`H(-1)=9*(-1)-6=-9-6=-15`
`H(1)=9*1-6=9-6=3`
`G(1)=-6*1^5+18*1^4+3*1+4`
`G(1)=-6+18+3+4=12+3+4=15+4=19`
`G(0)=-6*0^5+18*0^4+3*0+4=4`
`H(-3/2)=9*(-3/2)-6=-27/2-6=-39/2`
`e,`
Đặt `H(x)=9x-6=0`
`-> 9x=0+6`
`-> 9x=6`
`-> x=6 \div 9`
`-> x=2/3`
Vậy, nghiệm của đa thức là `x=2/3.`
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\(\frac{2x-1}{3x+2}=\frac{3x-3}{5x-2}\)
\(\Rightarrow\left(2x-1\right).\left(5x-2\right)=\left(3x-3\right).\left(3x+2\right)\)
=> (2x - 1).5x - (2x - 1).2 = (3x - 3).3x + (3x - 3).2
=> (10x2 - 5x) - (4x - 2) = (9x2 - 9x) + (6x - 6)
=> 10x2 - 5x - 4x + 2 = 9x2 - 9x + 6x - 6
=> 10x2 - 9x + 2 = 9x2 - 3x - 6
=> 10x2 - 9x - 9x2 + 3x = -6 - 2
=> x2 - 6x = -8
=> x2 - 6x + 8 = 0
=> x2 - 4x - 2x + 8 = 0
=> x.(x - 4) - 2.(x - 4) = 0
=> (x - 4).(x - 2) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x-4=0\\x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=4\\x=2\end{array}\right.\)
Vậy \(x\in\left\{2;4\right\}\)
\(\frac{2x-1}{3x+2}=\frac{3x-3}{5x-2}=\frac{2x-1-3x+3}{3x+2-5x+2}=\frac{-x+2}{-2x+4}=\frac{x+2}{2x+4}=\frac{x+2}{2.\left(x+2\right)}=\frac{1}{2}\)
\(\frac{2x-1}{3x+2}=\frac{1}{2}\Rightarrow4x-2=3x+2\Rightarrow4x-3x=2+2\Rightarrow x=4\)
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A+B=\(\left(x^2y-xy^2+3x^2\right)+\left(x^2y+xy^2-2x^2-1\right)\)
\(=x^2y-xy^2+3x^2+x^2y+xy^2-2x^2-1\)
\(=\left(x^2y+x^2y\right)-\left(xy^2-xy^2\right)+\left(3x^2-2x^2\right)-1\)
\(=2x^2y+x^2-1\)
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Tham khảo:
Cách 1:
P(x) + Q(x) = \(7{x^3} - 8x + 12 + 6{x^2} - 2{x^3} + 3x - 5\)
\(\begin{array}{l} = (7{x^3} - 2{x^3}) + 6{x^2} + ( - 8x + 3x) + (12 - 5)\\ = 5{x^3} + 6{x^2} - 5x + 7\end{array}\)
Cách 2:
\(|x-1|+3x=7\)
\(\Leftrightarrow|x-1|=7-3x\)
Vì \(|x-1|\ge0;\forall x\)
\(\Rightarrow7-3x\ge0\)
\(\Rightarrow3x\le7\)
\(\Rightarrow x\le\frac{7}{3}\)
Ta có: \(|x-1|=7-3x\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=7-3x\\x-1=3x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3x=7+1\\x-3x=-7+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=8\\-2x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2< \frac{7}{3}\left(chon\right)\\x=3>\frac{7}{3}\left(loai\right)\end{cases}}\)
Vậy x=2