Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
a)\(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\Rightarrow x^2+y^2=a^2-2xy\Rightarrow x^2+y^2=a^2-2b\)
a,
\(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\cdot\left(-6\right)=1-\left(-12\right)=13\)
\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=1\cdot\left[13-\left(-6\right)\right]=19\)
\(x^5+y^5=\left(x+y\right)\left(x^2+y^2\right)^2-\left(2x^3y^2+xy^4+x^4y+2x^2y^3\right)=169-\left[2\left(xy\right)^2\left(x+y\right)+xy\left(x^3+y^3\right)\right]=169-\left[2\cdot36\cdot1-6\cdot19\right]=211\)
a/\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{54}{2y}\)
\(\Rightarrow2y\cdot y=54\cdot3\Rightarrow2y^2=162\Rightarrow y^2=\dfrac{162}{2}=81\)
Mà y > 0 (gt) => \(y=\sqrt{81}=9\Rightarrow x=\dfrac{54}{9}=6\)
Vậy..............
b/ \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{4}\cdot25=\dfrac{25}{4}\\y^2=\dfrac{1}{4}\cdot9=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\sqrt{\dfrac{25}{4}}=\pm\dfrac{5}{2}\\y=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\end{matrix}\right.\)
Vậy.............
c/ x/2 = y/3 => x/10 = y/15
y/5 = z/7 => y/15 = z/21
=> x/10 = y/15 = z/21
Áp dụng t/c của dãy tỉ số = nhau là ra....
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
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) \(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)
\(=\left(x+y\right)^5-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x^3+y^3\right)+\left(2x^2y+2xy^2\right)\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(2xy\left(x+y\right)\right)\right)\)
\(=\left(5\right)^5-5.4\left(\left(\left(5^3-3.4.5\right)+\left(2.4.5\right)\right)\right)\)
\(=3125-20\left(125-65+40\right)\)
\(=3125-20\left(100\right)=3125-2000=1125\)
\(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2\cdot4=25-8=17\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot4\cdot5=125-60=65\\ x^4+y^4 \\ =\left(x+y\right)^4-4xy\left(x^2+y^2\right)-6x^2y^2\\ =5^4-4\cdot4\left[\left(x+y\right)^2-2xy\right]-6\left(xy\right)^2\\ =5^4-4\cdot4\cdot\left(5^2-2\cdot4\right)-6\cdot4^2\\ =625-16\cdot\left(25-8\right)-6\cdot16\\ =625-16\cdot17-96\\ =625-272-96\\ =257\\ x^5+y^5\\ =\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\\ =5^5-5\cdot4\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-10\left(xy\right)^2\cdot5\\ =3125-20\left(5^3-3\cdot4\cdot5\right)-10\cdot4^2\cdot5\\ =3125-20\cdot\left(125-60\right)-10\cdot16\cdot5\\ =3125-20\cdot65-800\\ =3125-1300-800\\ =1025\)