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giá trị của : f(0) + f(1) + f(2) + f(3) + f(4) + f(5) + f(6) +f(7) + f(8)
= -3-3-2+1+8+23+54+117+244
= 439
a, \(P=\left(\dfrac{2}{x+2}-\dfrac{x}{2-x}-\dfrac{x^2}{x^2-4}\right):\dfrac{4-4x}{x^2+2x}\)
\(=\left(\dfrac{2}{x+2}+\dfrac{-x}{x-2}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4-4x}{x^2+2x}\)
\(=\left(\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4-4x}{x^2+2x}\)
\(=\left(\dfrac{2\left(x-2\right)-x\left(x+2\right)-x^2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{4-4x}{x^2+2x}\)
\(=\left(\dfrac{2x-4+x^2+2x-x^2}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2+2x}{4-4x}\)
\(=\dfrac{4x-4}{\left(x-2\right)\left(x+2\right)}.\dfrac{-x\left(x+2\right)}{4x-4}\)
\(=-\dfrac{x}{x-2}\)
b, Để P có nghĩa
\(\Leftrightarrow x-2\ne0\)
\(\Leftrightarrow x\ne2\)
Thay x= -8 vào biểu thức P ,có :
\(-\dfrac{-8}{-8-2}=-\dfrac{-8}{-10}=\dfrac{8}{10}=-\dfrac{4}{5}\)
Vậy tại x = -8 giá trị của P là
c, Để P có giá trị nguyên
\(\Leftrightarrow-x⋮x-2\)
\(\Leftrightarrow-x+2-2⋮x-2\)
\(\Leftrightarrow-\left(x-2\right)-2⋮x-2\)
\(\Leftrightarrow2⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;2;-1;-2\right\}\)
| \(x-2\) | 1 | 2 | -1 | -2 |
| x | 3 | 4 | 1 | 0 |
Vậy \(x\in\left\{0;1;3;4\right\}\) thì P có giá trị nguyên
Giá trị của phân thức \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) được xác định với điều kiện ( x + 1 )( 2x - 6 )\(\ne\) 0
<=> 2( x + 1 )( x - 3 ) \(\ne\) 0
<=> x + 1 \(\ne\) 0 và x - 3 \(\ne\) 0
+, x + 1 \(\ne\) 0
<=> x \(\ne\) -1
+, x - 3 \(\ne\) 0
<=> x \(\ne\) 3
Vậy ĐKXĐ : x \(\ne\) -1; 3
Ta có : \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\)
\(=\frac{3x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}\)
\(=\frac{3x}{2\left(x-3\right)}\)
Giá trị của biểu thức \(\frac{3x^2+3x}{\left(x+1\right)\left(2x-6\right)}\) bằng 1
\(\Leftrightarrow\frac{3x}{2\left(x-3\right)}=1\)
\(\Rightarrow3x=2x-6\)
\(\Rightarrow3x-2x=-6\)
\(\Rightarrow x=-6\)
Vậy x = -6
a) \(Q=\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y\right)^2\)
\(Q=\left(x-y\right)^2-2\cdot\left(x-y\right)\cdot2\left(x+2y\right)+\left[2\left(x+2y\right)\right]^2\)
\(Q=\left[\left(x-y\right)-2\left(x+2y\right)\right]^2\)
\(Q=\left(x-y-2x-4y\right)^2\)
\(Q=\left(-x-5y\right)^2\)
b) \(A=\left(xy+2\right)^3-6\left(xy+2\right)^2+12\left(xy+2\right)-8\)
\(A=\left(xy+2\right)^3-3\cdot2\cdot\left(xy+2\right)^2+3\cdot2^2\cdot\left(xy+2\right)-2^3\)
\(A=\left[\left(xy+2\right)-2\right]^3\)
\(A=\left(xy+2-2\right)^3\)
\(A=\left(xy\right)^3\)
\(A=x^3y^3\)
c) \(\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)
\(=\left(x^3+6x^2+12x+8\right)+\left(x^2-6x^2+12x-8\right)-\left(2x^3+24x\right)\)
\(=x^3+6x^2+12x+8+x^2-6x^2+12x-8-2x^3-24x\)
\(=\left(x^3+x^3-2x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x-24x\right)+\left(8-8\right)\)
\(=0\)
a: =(x-y)^2-2(x-y)(2x+4y)+(2x+4y)^2
=(x-y-2x-4y)^2=(-x-5y)^2=x^2+10xy+25y^2
b: =(xy+2-2)^3=(xy)^3=x^3y^3
c: =x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x(x^2+12)
=24x+2x^3-2x^3-24x
=0
B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)
\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)
b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)
\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))
\(\Leftrightarrow x>-1\).
-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).
a: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)
b: \(=3x^2-6x-5x+5x^2-8x^2+24\)
=-11x+24
a) \(A=\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right).\left(x^2+4\right)\)
\(=x^4+4x^2+4-\left(x^2-4\right)\left(x^2+4\right)\)
\(=x^4+4x^2+4-\left(x^4-16\right)\)
\(=x^4+4x^2+4-x^4+16\)
\(=4x^2+20\)
b) Nếu x = -2 thì \(A=4.\left(-2\right)^2+20=36\)
Nếu x = 0 thì \(A=4.0^2+20=20\)
Nếu x = 2 thì \(A=4.2^2+20=36\)
c) Ta có: \(4x^2=\left(2x\right)^2\ge0\left(\forall x\in Z\right)\)
\(\Rightarrow A=4x^2+20\ge20\left(\forall x\in Z\right)\)
Vậy A luôn đạt giá trị dương với mọi giá trị của x