1.<span><span>(<span>a−b</span>)^</span>2</span><span>(<span>a+b</span>)</span>=<span>(<span><span>a^2</span>−2ab+<span>b^2)</span></span></span><span>(<span>a+b</span>)</span>=<span>a^3</span>+<span>a^2</span>b−2<span>a^2</span>b−2a<span>b^2</span>+<span>b^2</span>a+<span>b^3</span>=<span>a^3</span>−<span>a^2</span>b−a<span>b^2</span>+<span>b^3;
2.</span><span><span>(<span>a+2b</span>)^</span>2</span><span>(<span>a−2b</span>)</span>=<span>(<span><span>a^2</span>+4ab+4<span>b^2</span></span>)</span><span>(<span>a−2b</span>)</span>=<span>a^3-2</span><span>a^2</span>b+4<span>a^2</span>b−8a<span>b^2</span>+4<span>b^2</span>a−8<span>b^3</span>=<span>a^3</span>+2<span>a^2</span>b−4a<span>b^2</span>−8<span>b^3;
3.</span>7c<span>(<span>4c+2</span>)</span>−<span><span>(<span>7+c</span>)^</span>2</span>=28<span>c^2</span>+14c−49−14c−<span>c^2</span>=27<span>c^2</span>−<span>49;
4.</span><span><span>(<span>b+4</span>)^</span>2</span>−2b<span>(<span>5b+4</span>)</span>=<span>b^2</span>+8b+16−10<span>b^2</span>−8b=−9<span>b^2</span>+<span>16;
5.</span><span><span>(<span>b−2</span>)^</span>2</span>−2b<span>(<span>6b−2</span>)=</span><span>b^2</span>−4b+4−12<span>b^2</span>+4b=<span>−11<span>b^2</span>+4;
6.</span>−3c<span>(<span>6c+2</span>)</span>−<span><span>(<span>−3+c</span>)^</span>2</span>=−18<span>c^2</span>−6c−9+6c−<span>c^2</span>=−19<span>c^2</span>−9;7.(−8x+5y)^2−16x(−8x−5y)=64x^2−80xy+25y^2+128x^2+80xy= 192x^2+25y^2;
8.y(7y+4x)−(2x+y)^2=7y^2+4xy−4x^2−4xy−y^2=6y^2−4x^2.
(log1331):(log2)/(log11):log(2)=3
ΔABC,AB=BC
cos<A=0,2
BH_|_AC
<ABH=90-arccos(0,2)
sin<ABH=sin(90-arccos0,2)=cos(arccos0,2)=0,2⇒<ABH=arcsin0,2
<B=2<ABH=2arcsin0,2
Ответ: (x-1)(x-3)=x^2-4x+3, переведем в квадратное уравнение
x^2-4x+3 = 0
По теореме обратной теореме виета
x1= 3
x2= 1
Это нули функции, т.е где y = 0, этому соответствует рисунок номер 1
