Пусть a - сторона, тогда
площадь = a*a = 100 => a = 10
диагональ по теореме пифагора
![d=\sqrt{a^2+a^2}=a\sqrt{2}=10\sqrt2](https://tex.z-dn.net/?f=d%3D%5Csqrt%7Ba%5E2%2Ba%5E2%7D%3Da%5Csqrt%7B2%7D%3D10%5Csqrt2)
(a-b)²=a²-2ab+b²
(a+b)²=a²+2ab+b²
a²-b²=(a-b)(a+b)
(11-√17)(11+√17)
=11²-(√17)²=121-17=104
<span>(√43-5)²+ 10(√43-0,2)=66
1) </span>(√43-5)²=(√43)² - 2*5*√43 + 5²=43-10√43+25=68<span>-10√43
2) </span><span><span>10(√43-0,2)=10</span></span><span>√43 - 10*0.2 =</span><span>10<span>√43-2</span>
3) </span><span>68<span>-10√43 +</span></span><span>10√43-2=68-2=66
14(1+√31)+(7-√31)²=94
1) </span><span>14(1+√31)=14+</span><span>14√31
2) </span><span><span>(7-√31)²=7² - 2*7*√31 + (²√31)</span>=49-14√43+31=80-14√43
3) </span><span><span>14+14√31</span>+</span><span><span>80-14√43=14+80=94
</span> (√2-√15)²+(√6+√5)²</span>=28
1) <span>(√2-√15)²=(√2)² - 2*√2*√15 + (√15)²=2-2√30+15=17-4√30
2) </span><span><span>(√6+√5)²=(√6)² + 2*√6*√5 </span>+ (√5)²=6+2√30+5=11+2√30
3) </span><span>17-2√30</span>+11+2√30=17+11=28
90+45=135
......................
Первая длина = х, вторая = у; система уравнений: х - у = 15 , х + у = 40
х = 27,5; у = 12,5
1.
3y=-6x+11
y= (-6x+11):3
y=-2x+3,(6)
k=-2
m=+3,(6)
2.
y=3x+2
x = -1
y=3*(-1)+2
y=-1