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a)\(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\Rightarrow x^2+y^2=a^2-2xy\Rightarrow x^2+y^2=a^2-2b\)
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(\Rightarrow x^5+x^4y+x^3y^2+x^2y^3+y^5-yx^4-x^3y^2-x^2y^3-xy^4-y^5=VP\)
\(\Rightarrow dpcm\)
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(\Rightarrow x^5-x^4y+x^3y^2-x^2y^3+xy^4+yx^4-x^3y^2-xy^4+y^5=VP\)
\(\Rightarrow dpcm\)
c.d làm tương tự
Bài làm
a) Biến đổi vế trái, ta được:
\(VT=\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=\left(x^5-y^5\right)+\left(x^4y-x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5-y^5=VP\left(đpcm\right)\)
b) Biến đổi vế trái, ta có:
\(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=\left(x^5+y^5\right)+\left(-x^4y+x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(-x^2y^3+x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5+y^5=VP\left(đpcm\right)\)
c) Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+a^3b-a^2b^2+ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(-a^3b+a^3b\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)\)
\(=a^4-b^4=VP\left(đpcm\right)\)
d) Đây là hằng đẳng thức, như vế phải hình như bạn viết bị sai, mik sửa là vế phải nha.
\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=\left(a^3+b^3\right)+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(=a^3+b^3=VP\left(đpcm\right)\)
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....
P=(\(\dfrac{x^2}{x^2-y^2}+\dfrac{y\left(x+y\right)}{x^2-y^2}\)):\(\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x^4-y^4\right)}\)
P=\(\dfrac{X^2+xy+y^2}{x^2-y^2}\).\(\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
P=x^2+y^2=(x+y)^2-2xy=5^2-(-1)=26
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) \(x^5+y^5=\left(x+y\right)^5-\left(5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)\)
\(=\left(x+y\right)^5-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x^3+y^3\right)+\left(2x^2y+2xy^2\right)\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-3xy\left(x+y\right)+\left(2xy\left(x+y\right)\right)\right)\)
\(=\left(5\right)^5-5.4\left(\left(\left(5^3-3.4.5\right)+\left(2.4.5\right)\right)\right)\)
\(=3125-20\left(125-65+40\right)\)
\(=3125-20\left(100\right)=3125-2000=1125\)
\(x^2+y^2=\left(x+y\right)^2-2xy=5^2-2\cdot4=25-8=17\\ x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=5^3-3\cdot4\cdot5=125-60=65\\ x^4+y^4 \\ =\left(x+y\right)^4-4xy\left(x^2+y^2\right)-6x^2y^2\\ =5^4-4\cdot4\left[\left(x+y\right)^2-2xy\right]-6\left(xy\right)^2\\ =5^4-4\cdot4\cdot\left(5^2-2\cdot4\right)-6\cdot4^2\\ =625-16\cdot\left(25-8\right)-6\cdot16\\ =625-16\cdot17-96\\ =625-272-96\\ =257\\ x^5+y^5\\ =\left(x+y\right)^5-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\\ =5^5-5\cdot4\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]-10\left(xy\right)^2\cdot5\\ =3125-20\left(5^3-3\cdot4\cdot5\right)-10\cdot4^2\cdot5\\ =3125-20\cdot\left(125-60\right)-10\cdot16\cdot5\\ =3125-20\cdot65-800\\ =3125-1300-800\\ =1025\)