Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
a) \(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2.4=25-8=17\)
b) \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3.4.5=125-60=65\)
c) \(x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left(\left(x+y\right)^2-2xy\right)^2-2\left(xy\right)^2=\left(5^2-2.4\right)^2-2.4^2\)
\(=\left(25-8\right)^2-2.16=17^2-32=289-32=257\)
d) \(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)
\(=\left(x+y\right)^5-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x^3+y^3\right)+\left(2x^2y+2xy^2\right)\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(2xy\left(x+y\right)\right)\right)\)
\(=\left(5\right)^5-5.4\left(\left(\left(5^3-3.4.5\right)+\left(2.4.5\right)\right)\right)\)
\(=3125-20\left(125-65+40\right)\)
\(=3125-20\left(100\right)=3125-2000=1125\)
\(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2\cdot4=25-8=17\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot4\cdot5=125-60=65\\ x^4+y^4 \\ =\left(x+y\right)^4-4xy\left(x^2+y^2\right)-6x^2y^2\\ =5^4-4\cdot4\left[\left(x+y\right)^2-2xy\right]-6\left(xy\right)^2\\ =5^4-4\cdot4\cdot\left(5^2-2\cdot4\right)-6\cdot4^2\\ =625-16\cdot\left(25-8\right)-6\cdot16\\ =625-16\cdot17-96\\ =625-272-96\\ =257\\ x^5+y^5\\ =\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\\ =5^5-5\cdot4\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-10\left(xy\right)^2\cdot5\\ =3125-20\left(5^3-3\cdot4\cdot5\right)-10\cdot4^2\cdot5\\ =3125-20\cdot\left(125-60\right)-10\cdot16\cdot5\\ =3125-20\cdot65-800\\ =3125-1300-800\\ =1025\)
\(\left(x+y\right)^2-2xy=x^2+y^2=4^2-2.1=14\)
\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=14^2-2=196-2=194\)
\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=4\left(14-1\right)=52\)
\(\left(x^4+y^4\right)\left(x+y\right)=194.4=776\Leftrightarrow x^5+y^5+x^4y+y^4x=\left(x^5+y^5\right)+xy\left(x^3+y^3\right)=\left(x^5+y^5\right)+1.52=\left(x^5+y^5\right)+52=776\Rightarrow x^5+y^5=724\)
\(\left\{{}\begin{matrix}x+y=4\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+2xy+y^2=16\\4xy=4\end{matrix}\right.\Rightarrow x^2+2xy-4xy+y^2=\left(x-y\right)^2=12mà:x>y\Leftrightarrow x-y>0\Rightarrow x-y=\sqrt{12}=2\sqrt{3};x+y=2.2\Rightarrow\left\{{}\begin{matrix}x=\sqrt{3}+2\\y=2-\sqrt{3}\end{matrix}\right.\)
\(x^2-y^2=\left(x-y\right)\left(x+y\right)=4.2\sqrt{3}=8\sqrt{3}\)
\(\left(x^2+y^2\right)\left(x^2-y^2\right)=8\sqrt{3}.14=112\sqrt{3}\Rightarrow x^4-y^4=112\sqrt{3}\)
\(\left(x^3-y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right);x^6-y^6=\left(x^3+y^3\right)\left(x^3-y^3\right)tựlm\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(A=2x^2-2xy-y^2+2xy=2x^2-y^2\)
\(=2\cdot\dfrac{4}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
b: \(B=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)
=1/5-1=-4/5
c \(C=x^3+6x^2+12x+8=\left(x+2\right)^3=\left(-9\right)^3=-729\)
d: \(D=20x^3-10x^2+5x-20x^2+10x+4\)
\(=20x^3-30x^2+15x+4\)
\(=20\cdot5^3-30\cdot5^2+15\cdot2+4=1784\)
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
Ta có
Chọn đáp án A.