Решить 1. (rac{9}{16} +2rac{3}{8})^{2} *4 2. rac{1}{rac{1}{35}-rac{1}{60} - вопрос №1032824019162382421135 от shabishka 20.07.2021 13:51

Question

Алгебра: Решить 1. (rac{9}{16} +2rac{3}{8})^{2} *4 2. rac{1}{rac{1}{35}-rac{1}{60} } 3. rac{1}{rac{1}{63}+rac{1}{72} }

shabishka · Answer

(6x-7)²-(5x+7)(5x-7)=36x²-84x+49-(25x²-49)=36x²-84x+49-25x²+49=11x²-84x+98

y(y+6)²-(y+1)(y-6)²=y(y²+12y+36)-(y+1)(y²+12y+36)=
=y³+12y²+36y-(y³+12y²+36y+y²+12y+36)=y³+12y²+36y-y³-12y²-36y-y²-12y-36=
=-y²-12y-36=-(y²+12y+36)=-(y+6)²

100-140a+49a²=(10-7a)²

x⁴+18x²y+81y²=(x²+9y)²

(x²-4x)²-16 =(x²-4x)²-4²=((x²-4x)+4)((x²-4x)-4)=(x²-4x+4)(x²-4x-4)

9b²-25c²-3b+5c=(9b²-25c²)+(-3b+5c)=(3b+5c)(3b-5c)-(3b-5c)=
=(3b-5c)(3b+5c-1)

$\frac{ x^{2} -y^2}{( x^{2} -xy)^2} = \frac{(x+y)(x-y)}{(x(x-y))^2} =\frac{(x+y)(x-y)}{x^2(x-y)^2} = \frac{x+y}{x^2(x-y)} = \frac{x+y}{x^3-x^2y}$

(a-3b)²=a²-9b²
a²-3ab+9b²=a²-9b²
a²-6ab+9b²-a²+9b²=0
-6ab+18b²=0
-6b(a-3b)=0
a-3b=0
a=3b
значит при любых значениях удовлетворяющих а=3b, исходное равенство будет верным

Решить 1. (\frac{9}{16} +2\frac{3}{8})^{2} *4 2. \frac{1}{\frac{1}{35}-\frac{1}{60} } 3. \frac{1}{\frac{1}{63}+\frac{1}{72} }

Решить
1. (\frac{9}{16} +2\frac{3}{8})^{2} *4

2. \frac{1}{\frac{1}{35}-\frac{1}{60} }

3. \frac{1}{\frac{1}{63}+\frac{1}{72} }