Упр.936 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

а) $$(3x-2y)^2-(2x-y)^2$$ при $$x=2{,}35,\ y=-1{,}65$$;

б) $$(2m-n)^2+(m+2n)^2$$ при $$m=3{,}2,\ n=-3{,}4$$;

в) $$(6a-1)^2-\bigl((10a+3)(10a-3)-(8a+1)^2\bigr)$$ при $$a=-0{,}05$$;

г) $$\bigl((k+4)^2-(k+3)^2\bigr)^2-4(k-3)(k+10)$$ при $$k=1{,}375$$;

д) $$5mn(m+5n)-9n^3-\bigl(4mn^2-(m+n)(5m-3n)^2\bigr)$$ при $$m=-0{,}2,\ n=\frac{1}{2}$$;

е) $$(x-2y)(4x-3y)^2-(57xy-2y)(28x^2+9y^2)$$ при $$x=-0{,}5,\ y=\frac{1}{19}$$;

ж) $$(4a-3b)^2(b-a)-\bigl(9b^3-a(4a-5b)^2\bigr)$$ при $$a=-0{,}4,\ b=\frac{1}{2}$$;

з) $$(2x-9y)^2(x+y)-\bigl(y(9y+2{,}5x)^2+x^2(4x+1{,}75y)\bigr)$$ при $$x=-5,\ y=0{,}1$$.

Подробный ответ

а)
$$\begin{aligned} (3x-2y)^2-(2x-y)^2&=((3x-2y)-(2x-y))((3x-2y)+(2x-y))\\ &=(x-y)(5x-3y). \end{aligned}$$
Подставим $$x=2{,}35,$$ $$y=-1{,}65$$:
$$\begin{aligned} (x-y)(5x-3y)&=(2{,}35-(-1{,}65))(5\cdot 2{,}35-3\cdot(-1{,}65))\\ &=4\cdot 16{,}7=66{,}8. \end{aligned}$$
б)
$$\begin{aligned} (2m-n)^2+(m+2n)^2&=4m^2-4mn+n^2+m^2+4mn+4n^2\\ &=5m^2+5n^2=5(m^2+n^2). \end{aligned}$$
Подставим $$m=3{,}2,$$ $$n=-3{,}4$$:
$$\begin{aligned} 5(m^2+n^2)&=5(3{,}2^2+(-3{,}4)^2)\\ &=5(10{,}24+11{,}56)=5\cdot 21{,}8=109. \end{aligned}$$
в)
$$\begin{aligned} (6a-1)^2-\bigl((10a+3)(10a-3)-(8a+1)^2\bigr) &=36a^2-12a+1-(100a^2-9-(64a^2+16a+1))\\ &=36a^2-12a+1-(36a^2-16a-10)\\ &=4a+11. \end{aligned}$$
При $$a=-0{,}05$$:
$$4\cdot(-0{,}05)+11=-0{,}2+11=10{,}8.$$
г)
$$\begin{aligned} \bigl((k+4)^2-(k+3)^2\bigr)^2-4(k-3)(k+10) &=\bigl((k+4-k-3)(k+4+k+3)\bigr)^2-4(k^2+7k-30)\\ &=(2k+7)^2-4k^2-28k+120\\ &=169. \end{aligned}$$
Значение не зависит от $$k$$.
д)
$$\begin{aligned} 5mn(m+5n)-9n^3-\bigl(4mn^2-(m+n)(5m-3n)^2\bigr) &=5m^2n+25mn^2-9n^3\\ &\quad-\bigl(4mn^2-(m+n)(25m^2-30mn+9n^2)\bigr)\\ &=5m^2n+25mn^2-9n^3\\ &\quad-\bigl(4mn^2-25m^3+30m^2n-9mn^2-25m^2n+30mn^2-9n^3\bigr)\\ &=25m^3. \end{aligned}$$
При $$m=-0{,}2$$:
$$25\cdot(-0{,}2)^3=25\cdot(-0{,}008)=-0{,}2.$$
е)
$$\begin{aligned} (x-2y)(4x-3y)^2-(57xy-2y)(28x^2+9y^2) &=(x-2y)(16x^2-24xy+9y^2)\\ &\quad-(1596x^3y+513xy^3-56x^2y-18y^3)\\ &=16x^3+57xy^2-1596x^3y-513xy^3. \end{aligned}$$
Подставим $$x=-0{,}5=-\frac12,$$ $$y=\frac1{19}$$:
$$\begin{aligned} 16x^3+57xy^2-1596x^3y-513xy^3 &=8\frac{331}{722}. \end{aligned}$$
ж)
$$\begin{aligned} (4a-3b)^2(b-a)-\bigl(9b^3-a(4a-5b)^2\bigr) &= (16a^2-24ab+9b^2)(b-a)-\bigl(9b^3-a(16a^2-40ab+25b^2)\bigr)\\ &= -8ab^2. \end{aligned}$$
При $$a=-0{,}4=-\frac25,$$ $$b=\frac12$$:
$$-8\cdot\left(-\frac25\right)\cdot\left(\frac12\right)^2=\frac45.$$
з)
$$\begin{aligned} (2x-9y)^2(x+y)-\bigl(y(9y+2{,}5x)^2+x^2(4x+1{,}75y)\bigr) &=(4x^2-36xy+81y^2)(x+y)\\ &\quad-\bigl(y(81y^2+45xy+6{,}25x^2)+4x^3+1{,}75x^2y\bigr)\\ &=-40x^2y. \end{aligned}$$
При $$x=-5,$$ $$y=0{,}1$$:
$$-40\cdot(-5)^2\cdot 0{,}1=-100.$$