Rút gọn:
1. (y2 _ 6y + 9) - (3 - y)2
2. ( x - 3)2 - (x-4)(x+4)
3. ( y2 - 6y + 9) - (3 - y)2
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A = x2(x + y) - y(x2 - y) + 2002
A = x2.x + x2.y + (-y).x2 + (-y)(-y) + 2002
A = x3 + x2y - x2y + y2 + 2002
A = x3 + (x2y - x2y) + y2 + 2002
A = x3 + y2 + 2002 (1)
Thay x = 1, y = -1 vào (1), ta có:
A = x3 + y2 + 2002 = 13 + (-1)2 + 2002
= 1 + 1 + 2002
= 2004
B làm tương tự
\(\text{x3 - 19x - 30}=x^3+2x^2-4x-15x-30\)
\(=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x^2-2x-15\right)\left(x+2\right)\)
\(=\left[x^2-5x+3x-15\right]\left(x+2\right)\)
\(=\left[x\left(x-5\right)+3\left(x-5\right)\right]\left(x+2\right)\)
\(=\left(x+3\right)\left(x-5\right)\left(x+2\right)\)
\(x^3-19x+30\)
\(=x^3-9x-10x+30\)
\(=x\left(x^2-9\right)-10\left(x-3\right)\)
\(=x\left(x-3\right)\left(x+3\right)-10\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x-10\right)\)
\(=\left(x-3\right)\left(x^2-2x+5x-10\right)\)
\(=\left(x-3\right)\left[x\left(x-2\right)+5\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-3\right)\left(x+5\right)\)
\(x^3-7x-6=x^3-x-6x-6=x\left(x^2-1\right)-6\left(x-1\right)=x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)=\left(x-1\right)\left(x^2+3x-2x-6\right)=\left(x-1\right)\left(x\left(x+3\right)-2\left(x+3\right)\right)\)
\(=\left(x-1\right)\left(x+3\right)\left(x-2\right)\)
\(\left[\frac{2}{3x}-\frac{2}{x+1}\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}-\frac{2\left(-x-1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\)\(\left[\frac{2}{3x}-\frac{2\left(x+1\right)}{\left(x+1\right).3x}+\frac{2\left(x+1\right)}{x+1}\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\)\(\left(đpcm\right)\)
TL:
1)\(\left(y^2-6y+9\right)-\left(3-y\right)^2=\left(y-3\right)^2-\left(3-y\right)^2\)
\(=\left(y-3+3-y\right)\left(y-3-3+y\right)=0.\left(2y-6\right)=0\)
2)\(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=\left(x-3\right)^2-x^2+16\)
\(=\left(x-3+x\right)\left(x-3-x\right)+16=\left(2x-3\right).\left(-3\right)+16=-6x+9+16\)
\(=-6x+25\)
hc tốt
\(1,\left(y^2-6x+9\right)-\left(3-y\right)^2\)
\(=\left(y-3\right)^2-\left(y-3\right)^2=0\)
\(2,\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16=-6x+21\)
\(3...\)\(< ->1\)