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