Упр.554 ГДЗ Муравин 7 класс (Алгебра)
1) (4у — 5)(3у — 1) — (2у — 7)(6у + 7) + 54 = О;
2) (3у — 6)(4у — 2) — (2у — 9)(6у + 5) — 1 = 0;
3) (2х — 3)^2 — (2х — 1)(2x + 3) = 4;
4) (7 — 3y)^2 — (5у + 1)(2у — 5) + у^2 = 0;
5) (x — 0,2)(х + 0,2) + 2,04 + 24x^2 = (1 — 5x)^2;
6) (3х + 1)^2 — (х + 0,1)(х — 0,1) = 8x^2 — 2,02.
$$\begin{aligned} (4y-5)(3y-1)-(2y-7)(6y+7)+54&=0\\ 12y^2-4y-15y+5-(12y^2+14y-42y-49)+54&=0\\ 12y^2-19y+5-12y^2+28y+49+54&=0\\ 9y+108&=0\\ 9y&=-108\\ y&=-12 \end{aligned}$$
$$\begin{aligned} (3y-6)(4y-2)-(2y-9)(6y+5)-1&=0\\ 12y^2-6y-24y+12-(12y^2+10y-54y-45)-1&=0\\ 12y^2-30y+12-12y^2+44y+45-1&=0\\ 14y+56&=0\\ 14y&=-56\\ y&=-4 \end{aligned}$$
$$\begin{aligned} (2x-3)^2-(2x-1)(2x+3)&=4\\ 4x^2-12x+9-(4x^2+4x-3)&=4\\ -16x+12&=4\\ -16x&=-8\\ x&=\frac{1}{2} \end{aligned}$$
$$\begin{aligned} (7-3y)^2-(5y+1)(2y-5)+y^2&=0\\ 49-42y+9y^2-(10y^2-23y-5)+y^2&=0\\ 49-42y+9y^2-10y^2+23y+5+y^2&=0\\ -19y+54&=0\\ 19y&=54\\ y&=\frac{54}{19} \end{aligned}$$
$$\begin{aligned} (x-0{,}2)(x+0{,}2)+2{,}04+24x^2&=(1-5x)^2\\ x^2-0{,}04+2{,}04+24x^2&=1-10x+25x^2\\ 25x^2+2-1+10x-25x^2&=0\\ 10x+1&=0\\ x&=-\frac{1}{10} \end{aligned}$$
$$\begin{aligned} (3x+1)^2-(x+0{,}1)(x-0{,}1)&=8x^2-2{,}02\\ 9x^2+6x+1-(x^2-0{,}01)&=8x^2-2{,}02\\ 9x^2+6x+1-x^2+0{,}01-8x^2+2{,}02&=0\\ 6x+3{,}03&=0\\ 6x&=-3{,}03\\ x&=-0{,}505 \end{aligned}$$
Ответ
1) $$y=-12$$; 2) $$y=-4$$; 3) $$x=\frac{1}{2}$$; 4) $$y=\frac{54}{19}$$; 5) $$x=-\frac{1}{10}$$; 6) $$x=-0{,}505$$.
