Question

if x=√3-√2/√3+√2 and y=√3+√2/√3-√2 find the valve of x²+y²+xy​

Answer 1

Answer:

sorry i don't know this question again sorry

Answer 2

x=√3-√2/√3+√2=(√3-√2)(√3-√2)/(√3+√2)(√3-√2)=(√3-√2)²/(√3)²-(√2)²

 =(√3-√2)²/3-2=(√3-√2)²=(3-2√6+2)=(5-2√6)

y=√3+√2/√3-√2=(√3+√2)(√3+√2)/(√3-√2)(√3+√2)=(√3+√2)²/(√3)²-(√2)²

 =(√3+√2)²/3-2=(√3+√2)²=(3+2√6+2)=(5+2√6)

xy=(√3-√2)²×(√3+√2)²={(√3)²-2×√3×√2+(√2)²}{(√3)²+2×√3×√2+(√2)²}

   =(3-2√6+2)(3+2√6+2)=(5-2√6)(5+2√6)=5²-(2√6)²=25-24=1

∴x²+y²+xy=​(5-2√6)²+(5+2√6)²+1

                =5²-2×5×2√6+(2√6)²+5²+2×5×2√6+(2√6)²+1

                =25+24+25+24+1=99