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\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)
Bài 1:
\(P=2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2+4xy-7y^2\)
Ta có
\(C=\left(x^2+2.x.2y+\left(2y\right)^2\right)-\left(2x+4y\right)+10\)
\(\Rightarrow C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(\Rightarrow C=5^2-2.5+10\)
\(\Rightarrow C=25-10+10=25\)
\(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left[x^2+2.x.2y+\left(2y\right)^2\right]-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2\right)+10\)
\(=5^2-2.5+10\)
\(=5^2-10+10\)
\(=25-10+10\)
\(=25\)
CHÚC BẠN HỌC TỐT !!!
\(x^2+4y^2-5x+10y-4xy+20\)
\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)
\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)
\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được :
\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)
\(B=x^2-2xy-2x+2y+y^2\)
\(=x^2-2xy+y^2-2\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được :
\(=1-2=-1\)
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-1}{x-2}\)
b: Khi x=1/2 thì \(B=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Khi x=-1/2 thì B=2/5
c: Để B nguyên thì \(x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
a, đk : x khác -2 ; 2
\(B=\left(\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{1}{2-x}\)
b, Ta có \(\left|x\right|=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2}\)
Với x = 1/2 ta được \(B=\dfrac{1}{2-\dfrac{1}{2}}=\dfrac{2}{3}\)
Với x = -1/2 ta được \(B=\dfrac{1}{2+\dfrac{1}{2}}=\dfrac{2}{5}\)
c, \(\dfrac{1}{2-x}\Rightarrow2-x\inƯ\left(1\right)=\left\{\pm1\right\}\)
| 2-x | 1 | -1 |
| x | 1 | 3 |
A=(x^2+4xy+4y^2)-(2x+4y)+10
A=(x+2y)^2-2(x+2y)+1+9
A=(x+2y-1)^2+9
A=(5-1)^2+9=16+9=25
(x²+4xy+4y²)-(2x+4y)+10=(x+2y)²-2(x+2y)+10=5²-10+10=25 :333
:333 là biểu cảm nhé