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a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)
CHÚC BẠN HỌC TỐT..........
Theo bài ra , ta có :
\(P=\left(\dfrac{x^2}{x^2-y^2}+\dfrac{y}{x-y}\right):\dfrac{x^3-y^3}{x^5-x^4y-xy^4+y^5}\)ĐKXĐ \(x\ne\pm y\)
\(\Leftrightarrow P=\left(\dfrac{x^2}{\left(x-y\right)\left(x+y\right)}+\dfrac{y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x^4\left(x-y\right)-y^4\left(x-y\right)}\)
\(\Leftrightarrow P=\left(\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\right):\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
\(\Leftrightarrow P=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x+y\right)}\times\dfrac{\left(x-y\right)\left(x^4-y^4\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(\Leftrightarrow P=\dfrac{x^4-y^4}{\left(x-y\right)\left(x+y\right)}\)\(\Leftrightarrow P=\dfrac{\left(x^2\right)^2-\left(y^2\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}=x^2+y^2\)(1)
Ta có : \(x+y=5\Rightarrow\left(x+y\right)^2=25\Rightarrow x^2+y^2=25-2xy=25--1=26\)(Vì xy = -1/2)
Thay x2 + y2 = 26 vào (1) ta đk : P = 26
Vậy P = 26 khi x + y = 5 và xy = -1/2
\(P=\left(\dfrac{x^2+y\left(x+y\right)}{\left(x^2-y^2\right)}\right).\left(\dfrac{x^4\left(x-y\right)-y^4\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\right)\\ \)
\(P=\left(\dfrac{x^2+xy+y^2}{\left(x^2-y^2\right)}\right).\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x^2+xy+y^2\right)}\)
\(P=x^2+y^2=\left(x+y\right)^2-2xy=25-2\left(-\dfrac{1}{2}\right)=26\)
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
a) A = ( 6x + 7)( 2x - 3) - ( 4x + 1)( 3x - \(\dfrac{7}{4}\))
A = 12x2 - 18x + 14x - 21 - ( 12x2 - 7x + 3x - \(\dfrac{7}{4}\))
A = \(\dfrac{-77}{4}\)
Vậy biểu thức trên ko phụ thuộc vào biến
b) x2 - 2y2 = xy
⇔ x2 - xy - 2y2 = 0
⇔ x2 + xy - 2xy - 2y2 = 0
⇔ x( x + y) - 2y( x + y) = 0
⇔ ( x - 2y )( x + y ) = 0
Do : x + y # 0
⇒ x - 2y = 0
⇔ x = 2y
Ta có : P = \(\dfrac{x-y}{x+y}\) ( x + y # 0 ; y # 0)
P = \(\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
KL....