Cho các biểu thức sau
A=2/5x(x^2-5x+15)
B=x(x-20)+(3-x)(3+x)
C=(x-40)^2-2(x-5)(x+4)+(x-5)^2
a)Rút gọn A,B và C
b) Tính giá trị biểu thức B tại x=5
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a) Ta có: \(B=\dfrac{x^2}{5x+25}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x+5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10\left(x+5\right)^2}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+100x+250+250+25x}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+125x+500}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+5x^2+5x^2+25x+100x+500}{5x\left(x+5\right)}\)
\(=\dfrac{x^2\left(x+5\right)+5x\left(x+5\right)+100\left(x+5\right)}{5x\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x^2+5x+100\right)}{5x\left(x+5\right)}\)
\(=\dfrac{x^2+5x+100}{5x}\)
b) Thay x=-2 vào biểu thức \(B=\dfrac{x^2+5x+100}{5x}\), ta được:
\(B=\dfrac{\left(-2\right)^2+5\cdot\left(-2\right)+100}{-5\cdot2}=\dfrac{4+100-10}{-10}=\dfrac{94}{-10}=-\dfrac{94}{10}=\dfrac{-47}{5}\)
Vậy: Khi x=-2 thì \(B=-\dfrac{47}{5}\)
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
a) A = (x - 5)(x² + 5x + 25) - (x - 2)(x + 2) + x(x² + x + 4)
= x³ - 125 - x² + 4 + x³ + x² + 4x
= (x³ + x³) + (-x² + x²) + 4x + (-125 + 4)
= 2x³ + 4x - 121
b) Tại x = -2 ta có:
A = 2.(-2)³ + 4.(-2) - 121
= 2.(-8) - 8 - 121
= -16 - 129
= -145
c) x² - 1 = 0
x² = 1
x = -1; x = 1
*) Tại x = -1 ta có:
A = 2.(-1)³ + 4.(-1) - 121
= 2.(-1) - 4 - 121
= -2 - 125
= -127
*) Tại x = 1 ta có:
A = 2.1³ + 4.1 - 121
= 2.1 + 4 - 121
= 2 - 117
= -115
\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)