Упр.516 ГДЗ Колягин Ткачёва 9 класс (Алгебра)
1) y(x)=x^3-6x^2+11x-4; 2) y(x)=x^4-9x^3+26x^2-24x;
3) y(x)=(x^2-4x+13)/(x+1); 4) y(x)=v(9+4x-x^2).
$$y(x)=x^3-6x^2+11x-4$$
$$y(0{,}5)=0{,}5^3-6\cdot 0{,}5^2+11\cdot 0{,}5-4=0{,}125\approx 0{,}13$$
$$y(1{,}5)=1{,}5^3-6\cdot 1{,}5^2+11\cdot 1{,}5-4=2{,}375\approx 2{,}38$$
$$y(2{,}5)=2{,}5^3-6\cdot 2{,}5^2+11\cdot 2{,}5-4=1{,}625\approx 1{,}63$$
$$y(3{,}5)=3{,}5^3-6\cdot 3{,}5^2+11\cdot 3{,}5-4=3{,}875\approx 3{,}88$$
$$y(x)=x^4-9x^3+26x^2-24x$$
$$y(0{,}5)=0{,}5^4-9\cdot 0{,}5^3+26\cdot 0{,}5^2-24\cdot 0{,}5=-6{,}5625\approx -6{,}56$$
$$y(1{,}5)=1{,}5^4-9\cdot 1{,}5^3+26\cdot 1{,}5^2-24\cdot 1{,}5=-2{,}8125\approx -2{,}81$$
$$y(2{,}5)=2{,}5^4-9\cdot 2{,}5^3+26\cdot 2{,}5^2-24\cdot 2{,}5=0{,}9375\approx 0{,}94$$
$$y(3{,}5)=3{,}5^4-9\cdot 3{,}5^3+26\cdot 3{,}5^2-24\cdot 3{,}5=-1{,}3125\approx -1{,}31$$
$$y(x)=\frac{x^2-4x+13}{x+1}$$
$$y(0{,}5)=\frac{0{,}5^2-4\cdot 0{,}5+13}{0{,}5+1}=\frac{11{,}25}{1{,}5}=7{,}50$$
$$y(1{,}5)=\frac{1{,}5^2-4\cdot 1{,}5+13}{1{,}5+1}=\frac{9{,}25}{2{,}5}=3{,}70$$
$$y(2{,}5)=\frac{2{,}5^2-4\cdot 2{,}5+13}{2{,}5+1}=\frac{9{,}25}{3{,}5}\approx 2{,}64$$
$$y(3{,}5)=\frac{3{,}5^2-4\cdot 3{,}5+13}{3{,}5+1}=\frac{11{,}25}{4{,}5}=2{,}50$$
$$y(x)=\sqrt{9+4x-x^2}$$
$$y(0{,}5)=\sqrt{9+4\cdot 0{,}5-0{,}5^2}=\sqrt{10{,}75}\approx 3{,}28$$
$$y(1{,}5)=\sqrt{9+4\cdot 1{,}5-1{,}5^2}=\sqrt{12{,}75}\approx 3{,}57$$
$$y(2{,}5)=\sqrt{9+4\cdot 2{,}5-2{,}5^2}=\sqrt{12{,}75}\approx 3{,}57$$
$$y(3{,}5)=\sqrt{9+4\cdot 3{,}5-3{,}5^2}=\sqrt{10{,}75}\approx 3{,}28$$
Ответ
1) $$0{,}13;\ 2{,}38;\ 1{,}63;\ 3{,}88.$$
2) $$-6{,}56;\ -2{,}81;\ 0{,}94;\ -1{,}31.$$
3) $$7{,}50;\ 3{,}70;\ 2{,}64;\ 2{,}50.$$
4) $$3{,}28;\ 3{,}57;\ 3{,}57;\ 3{,}28.$$
