câu 1
tính a) 1^5+2^5+3^5+.....+10^5
b) (x-y)*(x^3+x*y+y^3)
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CG
1
NM
Nguyễn Minh Quang
Giáo viên
12 tháng 8 2021
a. ta có : \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2\times\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3\times\left(-6\right)\times1=19\)
\(x^5+y^5=\left(x+y\right)\left[x^4-x^3y+x^2y^2-xy^3+y^4\right]\)
\(=\left(x+y\right)\left[\left(x^2+y^2\right)^2-x^2y^2-xy\left(x^2+y^2\right)\right]=1.\left(13^2-\left(-6\right)^2-\left(-6\right).13\right)=211\)
b.\(x^2+y^2=\left(x-y\right)^2+2xy=1+2\times6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+6.3.1=19\)
\(x^5-y^5=\left(x-y\right)\left[\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\right]\)
\(=\left(x-y\right)\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=1.\left(13^2-6^2+6.13\right)=211\)
19 tháng 9 2021
\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)
\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)
\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
Thay vào BT
\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2023
a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)
y = \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)
y = \(\dfrac{4}{3}\)
b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)
y - 0,5 + 0,5 = \(\dfrac{3}{4}\)
y = \(\dfrac{3}{4}\)
c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2
0,8 - 0,4y = 0,2
0,4y = 0,8 - 0,2
0,4y = 0,6
y = 1,5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8 2023
d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)
y + \(\dfrac{3}{4}\) = \(\dfrac{14}{9}\)
y = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)
y = \(\dfrac{29}{36}\)
e, y : \(\dfrac{5}{4}\) = \(\dfrac{9}{5}\) + \(\dfrac{1}{2}\)
y : \(\dfrac{5}{4}\) = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{10}\)
y = \(\dfrac{23}{8}\)
f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y = \(\dfrac{4}{5}\)
y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\)) = \(\dfrac{4}{5}\)
2y = \(\dfrac{4}{5}\)
y = \(\dfrac{2}{5}\)
A
5 tháng 9 2020
a, \(\left(x+2\right)^2-\left(x+3\right)\left(x-3\right)+10=x^2+4x+4-x^2+9+10=4x+23\)
b, \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)=25-10x+x^2+x^2+10x+25-2x^2+50=100\)
5 tháng 9 2020
a) ( x + 2 )2 - ( x + 3 )( x - 3 ) + 10
= x2 + 4x + 4 - ( x2 - 9 ) + 10
= x2 + 4x + 4 - x2 + 9 + 10
= 4x + 23
b) ( x + 1 )2 + ( x - 2 )( x + 3 ) - 4x
= x2 + 2x + 1 + x2 + x - 6 - 4x
= 2x2 - 2x - 5
c) ( x - 2 )( x + 2 ) - ( x - 3 )( x + 1 )
= x2 - 4 - ( x2 - 2x - 3 )
= x2 - 4 - x2 + 2x + 3
= 2x - 1
d) ( x + 4 )2 + ( x + 5 )( x - 5 ) - 2x( x + 1 )
= x2 + 8x + 16 + x2 - 25 - 2x2 - 2x
= 6x - 9
e) ( 5 - x )2 + ( x + 5 )2 - ( 2x + 10 )( x - 5 )
= 25 - 10x + x2 + x2 + 10x + 25 - ( 2x2 - 50 )
= 2x2 + 50 - 2x2 + 50
= 100
f) ( x - 2 )2 + ( x + 1 )2 + 2( x - 2 )( -1 - x )
= x2 - 4x + 4 + x2 + 2x + 1 + 2( -x2 + x + 2 )
= 2x2 - 2x + 5 - 2x2 + 2x + 4
= 9
g) ( 3x - 5 )2 - 2( 3x - 5 )( 3x + 5 ) + ( 3x + 5 )2
= [ ( 3x - 5 ) - ( 3x + 5 ) ]2
= ( 3x - 5 - 3x - 5 )2
= ( -10 )2 = 100
h) ( y - 3 )( y + 3 )( y2 + 9 ) - ( y2 + 2 )( y2 - 2 )
= ( y2 - 9 )( y2 + 9 ) - [ ( y2 )2 - 4 ]
= [ ( y2 )2 - 81 ] - y4 + 4
= y4 - 81 - y4 + 4
= -77
29 tháng 6 2015
a)
A=\(x^2+y^2=\left(x^2+2xy+y^2\right)-2xy=\left(x+y\right)^2-2xy=a^2-2b\)
\(B=x^3+y^3=\left(x^3+3x^2y+3xy^2+y^3\right)-3x^2y-3xy^2=\left(x+y\right)^3-3xy\left(x+y\right)=a^3-3ab\)
\(C=x^5+y^5=\left(x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4\right)-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)
\(=\left(x+y\right)^5-5xy\left(x^3+2xy^2+2x^2y+y^3\right)=\left(x+y\right)^5-5xy\left(x^3+3xy^2+3x^2y+y^3-xy^2-x^2y\right)\)
\(=\left(x+y\right)^5-5xy\left(\left(x+y\right)^3-xy\left(x+y\right)\right)=a^5-5b\left(a^3-ab\right)\)
môn Tin lớp 8 đó nha mấy bn