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a)\(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\Rightarrow x^2+y^2=a^2-2xy\Rightarrow x^2+y^2=a^2-2b\)
Giải:
a) \(\left(x-5\right)^2-16\)
\(=\left(x-5-4\right)\left(x-5+4\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
b) \(25-\left(3-x\right)^2\)
\(=\left(5-3+x\right)\left(5+3-x\right)\)
\(=\left(2+x\right)\left(8-x\right)\)
c) \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
d) \(11x+11y-x^2-xy\)
\(=11\left(x+y\right)-x\left(x+y\right)\)
\(=\left(11-x\right)\left(x+y\right)\)
e) \(x^2-xy-8x+8y\)
\(=x\left(x-y\right)-8\left(x-y\right)\)
\(=\left(x-8\right)\left(x-y\right)\)
Vậy ...
\(\left(x-5\right)^2-16\)
\(=\left(x-5\right)^2-4^2\)
\(=\left(x-5-4\right)\left(x-5+4\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(25-\left(3-x\right)^2\)
\(=5^2-\left(3-x\right)^2\)
\(=\left(5+3-x\right)\left(5-3+x\right)\)
\(=\left(8-x\right)\left(2+x\right)\)
\(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=7^2\left(y-4\right)^2-3^2\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
a) /x-5/=/2x+1/
=>x-5=2x+1 hoặc x-5=-(2x+1)
Th1 x-5=2x+1
-5-1=2x-x
x=-6
Thử lại thấy đúng
Th2: x-5=-(2x+1)
x-5=-2x-1
x+2x=-1+5
3x=4
x=4/3
Thử lại thấy đúng
Vậy x=-6 hoặc x=4/3
Các câu còn lại liên quan đến giá trị tuyệt đối thì làm tương tự
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Chọn B
Ta có: x - y 2 = x 2 -2xy+ y 2 = ( x 2 + y 2 ) - 2xy = 26 - 2.5=16