x-2/5=3-y/7;x+y=3
Mọi ngừ zúp mik vs ạ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
7 tháng 6 2017
( 2 x y + 2/15 ) x 3 = 4/5
( 2 x y + 2/15 ) = 4/5 : 3
( 2 x y + 2/15 ) = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 :2
y = 1/15
DP
7 tháng 6 2017
(2 x y + 2/15) x 3 = 4/5
2 x y + 2/15) = 4/5 : 3
2 x y + 2/15 = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 : 2
y = 1/15
7/9 x (2 - 1/3 x y) = 14/15
(2 - 1/3 x y) = 14/15 : 7/9
(2 - 1/3 x y) = 6/5
2 - y = 6/5 x 1/3
2 - y = 2/5
y = 2/5 + 2
y = 12/5
4/21 + 5 x y - 8/7 = 1/3
4/21 + 5 x y = 1/3 + 8/7
4/21 + 5 x y = 31/21
5 x y = 31/21 - 4/21
5 x y = 9/7
y = 9/7 : 5
y = 9/35
7/12 x y - 3/12 x y = 5
y x (7/12 - 3/12) = 5
y x 1/3 = 5
y = 5 : 1/3
y = 15
HN
30 tháng 5 2021
1)\(\left(x+1\right).\left(y-2\right)=0\) \(\left(x,y\inℤ\right)\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\y-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\y=2\end{cases}}\)
2)\(\left(x-5\right).\left(y-7\right)=1\)
| x-5 | 1 | -1 |
| y-7 | 1 | -1 |
| x | 6 | 4 |
| y | 8 | 6 |
3)\(\left(x+4\right).\left(y-2\right)=2\)
| x+4 | 1 | 2 | -1 | -2 |
| y-2 | 2 | 1 | -2 | -1 |
| x | -3 | -2 | -5 | -6 |
| y | 4 | 3 | 0 | 1 |
4)\(\left(x-4\right).\left(y+3\right)=-3\)
| x-4 | 1 | -1 | 3 | -3 |
| y+3 | -3 | 3 | -1 | 1 |
| x | 5 | 3 | 7 | 1 |
| y | -6 | 0 | -4 | -2 |
5)\(\left(x+3\right).\left(y-6\right)=-4\)
| x+3 | -1 | 1 | -4 | 4 | 2 | -2 |
| y-6 | 4 | -4 | 1 | -1 | -2 | 2 |
| x | -4 | -2 | -7 | 1 | -1 | -5 |
| y | 10 | 2 | 7 | 5 | 4 | 8 |
6)\(\left(x-8\right).\left(y+7\right)=5\)
| x-8 | 1 | 5 | -1 | -5 |
| y+7 | 5 | 1 | -5 | -1 |
| x | 9 | 13 | 7 | 3 |
| y | -2 | -6 | -12 | -8 |
7)\(\left(x+7\right).\left(y-3\right)=-6\)
| x+7 | -1 | 1 | -6 | 6 | -2 | 2 | -3 | 3 |
| y-3 | 6 | -6 | 1 | -1 | 3 | -3 | 2 | -2 |
| x | -8 | -6 | -13 | -1 | -9 | -5 | -10 | -4 |
| y | 9 | -3 | 4 | 2 | 6 | 0 | 5 | 1 |
8)\(\left(x-6\right).\left(y+2\right)=7\)
| x-6 | 1 | 7 | -1 | -7 |
| y+2 | 7 | 1 | -7 | -1 |
| x | 7 | 13 | 5 | -1 |
| y | 5 | -1 | -9 | -3 |
ok :)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 11 2023
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
NN
20 tháng 7 2023
Bài 2:
\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)
\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)
\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)
NT
2
21 tháng 6 2023
a: =2(x-y)^3/(x-y)-7(x-y)^2/(x-y)+(x-y)/(x-y)
=2(x-y)^2-7(x-y)+1
b: =3(x-y)^5/5(x-y)^2-2(x-y)^4/5(x-y)^2+3(x-y)^2/5(x-y)^2
=3/5(x-y)^3-2/5(x-y)^2+3/5
21 tháng 6 2023
\(a,\)
\(\left[2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right]:\left(x-y\right)\)
\(=\left[2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right]:\left(x-y\right)\)
\(=\left\{\left(x-y\right)\left[2\left(x-y\right)^2-7\left(x-y\right)+1\right]\right\}:\left(x-y\right)\)
\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)
\(b,\)
\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:\left[5\left(x-y\right)^2\right]\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
MH
9 tháng 5 2022
Bài 1:
+) \(\dfrac{7}{8}\times y=\dfrac{3}{2}+\dfrac{6}{4}=3\)
\(y=3:\dfrac{7}{8}=\dfrac{24}{7}\)
+) \(\dfrac{1}{y}\times\left(\dfrac{2}{5}+\dfrac{1}{5}\right)=\dfrac{10}{3}\)
\(\dfrac{1}{y}=\dfrac{10}{3}:\dfrac{3}{5}=\dfrac{50}{9}\)
\(y=\dfrac{9}{50}\)
TG
0
Ta có : \(\frac{x-2}{5}=\frac{3-y}{7}=\frac{\left(x-2\right)-\left(3-y\right)}{5-7}=\frac{x-2-3+y}{-2}\) \(=\frac{x+y-5}{-2}=\frac{-2}{-2}=1\)
Nên : \(\frac{x-2}{5}=1\Rightarrow x-2=5\Rightarrow x=7\)
\(\frac{3-y}{7}=1\Rightarrow3-y=7\Rightarrow y=-4\)
Vậy x = 7 ; y = -4
\(\frac{x-2}{5}=\frac{3-y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{x-2}{5}=\frac{3-y}{7}=\frac{x-2-3+y}{5-7}=\frac{3-2-3}{5-7}=\frac{-2}{-2}=1\)
\(\Rightarrow\hept{\begin{cases}x-2=5\\3-y=7\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\y=-4\end{cases}}\)