Wykaż, że jeśli i Wykaż, że jeśli oraz xy=1 i x.... Question from @zysioo

October 2018 2 29 Report

Wykaż, że jeśli $x^2+y^2=2$ i $x+y=1, to xy=-\frac{1}{2}$

Wykaż, że jeśli $x^2+y^2=7$ oraz xy=1 $xy=1$ i x<0, to x+y=-3

irenas

$x^2+y^2=2\\x+y=1\\(x+y)^2=1^2=1\\x^2+y^2+2xy=1\\2+2xy=1\\2xy=-1\\xy=-\frac{1}{2}$

$x^2+y^2=7\\xy=1\\x^2+y^2+2xy=7+2\cdot1=9\\(x+y)^2=9\\(x+y)^2=3\ \ lub\ \ x+y=-3\\x<0\ \ i\ \ xy=1>0\ \ to\ \ y<0\\x+y<0\\x+y=-3$

6 votes Thanks 4

Ingham

$ZAD 1.\\ x+y = 1 / (..)^2\\ (x+y)^2 = 1\\ x^2 + 2xy + y^2 = 1\\ x^2 + y^2 + 2xy = 1\\ 2 + 2xy = 1\\ 2xy = -1\\ xy = -\frac{1}{2}\\ \\ ZAD 2.\\ x^2 + y^2 = 7\\ x^2 + y^2 + 2xy = 7 + 2xy\\ (x+y)^2 = 7 + 2xy\\ (x+y)^2 = 7 + 2*1 = 9\\ (x+y)^2 = 9\\ \\ x + y = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ lub \ \ \ \ x + y = -3\\ to\ odpada\ (bo\ x < 0) \ \ \ \ \ \$

4 votes Thanks 4

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