35

В тетради можно записать так:

№ 35.

\[ \begin{aligned} \text{а)}\;&2x^2-5x-7=(2x-7)(x+1);\\ \text{б)}\;&-5y^2-2y+3=(3-5y)(y+1);\\ \text{в)}\;&x^2+8x-9=(x+9)(x-1);\\ \text{г)}\;&\frac16x^2+\frac12x+\frac13=\frac16(x+1)(x+2);\\ \text{д)}\;&-5x^2-10x+15=-5(x+3)(x-1);\\ \text{е)}\;&y^2-16y+15=(y-1)(y-15);\\ \text{ж)}\;&-x^2+12x-24 =-\bigl(x-6-2\sqrt3\bigr)\bigl(x-6+2\sqrt3\bigr);\\ \text{з)}\;&-2x^2+5x-3=-(2x-3)(x-1);\\ \text{и)}\;&2x^2-2x+\frac12=2\left(x-\frac12\right)^2;\\ \text{к)}\;&-16a^2-24a-9=-(4a+3)^2;\\ \text{л)}\;&m^2-5m+6=(m-2)(m-3);\\ \text{м)}\;&0{,}25m^2-2m+4=(0{,}5m-2)^2. \end{aligned} \]

Если требуется решение с пояснениями, напиши, и я покажу, как оформить один пункт по шагам.

Решение № 35

Разложим выражения на множители.

а)
\[ 2x^2-5x-7=2x^2+2x-7x-7 \] \[ =2x(x+1)-7(x+1)=(x+1)(2x-7). \]

б)
\[ -5y^2-2y+3=-5y^2-5y+3y+3 \] \[ =-5y(y+1)+3(y+1)=(y+1)(3-5y). \]

в)
\[ x^2+8x-9=x^2+9x-x-9 \] \[ =x(x+9)-(x+9)=(x+9)(x-1). \]

г)
\[ \frac16x^2+\frac12x+\frac13 =\frac16(x^2+3x+2) \] \[ =\frac16(x+1)(x+2). \]

д)
\[ -5x^2-10x+15=-5(x^2+2x-3) \] \[ =-5(x+3)(x-1). \]

е)
\[ y^2-16y+15=y^2-y-15y+15 \] \[ =y(y-1)-15(y-1)=(y-1)(y-15). \]

ж)
\[ -x^2+12x-24=-(x^2-12x+24) \] \[ =-\bigl((x-6)^2-12\bigr) \] \[ =-\bigl((x-6)^2-(2\sqrt3)^2\bigr) \] \[ =-(x-6-2\sqrt3)(x-6+2\sqrt3). \]

з)
\[ -2x^2+5x-3=-2x^2+2x+3x-3 \] \[ =-2x(x-1)+3(x-1)=(x-1)(3-2x). \]

и)
\[ 2x^2-2x+\frac12 =2\left(x^2-x+\frac14\right) \] \[ =2\left(x-\frac12\right)^2. \]

к)
\[ -16a^2-24a-9=-(16a^2+24a+9) \] \[ =-(4a+3)^2. \]

л)
\[ m^2-5m+6=m^2-2m-3m+6 \] \[ =m(m-2)-3(m-2)=(m-2)(m-3). \]

м)
\[ 0{,}25m^2-2m+4=(0{,}5m)^2-2\cdot0{,}5m\cdot2+2^2 \] \[ =(0{,}5m-2)^2. \]

Да, вот решение через дискриминант.

Для квадратного трёхчлена \(At^2+Bt+C\):

\[ D=B^2-4AC,\qquad t_{1,2}=\frac{-B\pm\sqrt D}{2A}, \]

\[ At^2+Bt+C=A(t-t_1)(t-t_2). \]

№ 35

а)
\[ D=(-5)^2-4\cdot2\cdot(-7)=81. \]

\[ x_{1,2}=\frac{5\pm9}{4}=\frac72;\,-1. \]

\[ 2x^2-5x-7=2\left(x-\frac72\right)(x+1) =(2x-7)(x+1). \]

б)
\[ D=(-2)^2-4\cdot(-5)\cdot3=64. \]

\[ y_{1,2}=\frac{2\pm8}{-10}=-1;\,\frac35. \]

\[ -5y^2-2y+3=-5(y+1)\left(y-\frac35\right) =(y+1)(3-5y). \]

в)
\[ D=8^2-4\cdot1\cdot(-9)=100. \]

\[ x_{1,2}=\frac{-8\pm10}{2}=1;\,-9. \]

\[ x^2+8x-9=(x-1)(x+9). \]

г)
\[ D=\left(\frac12\right)^2 -4\cdot\frac16\cdot\frac13 =\frac1{36}. \]

\[ x_{1,2}= \frac{-\frac12\pm\frac16}{\frac13} =-1;\,-2. \]

\[ \frac16x^2+\frac12x+\frac13 =\frac16(x+1)(x+2). \]

д)
\[ D=(-10)^2-4\cdot(-5)\cdot15=400. \]

\[ x_{1,2}=\frac{10\pm20}{-10}=-3;\,1. \]

\[ -5x^2-10x+15=-5(x+3)(x-1). \]

е)
\[ D=(-16)^2-4\cdot1\cdot15=196. \]

\[ y_{1,2}=\frac{16\pm14}{2}=15;\,1. \]

\[ y^2-16y+15=(y-15)(y-1). \]

ж)
\[ D=12^2-4\cdot(-1)\cdot(-24)=48. \]

\[ x_{1,2}=\frac{-12\pm4\sqrt3}{-2} =6-2\sqrt3;\,6+2\sqrt3. \]

\[ -x^2+12x-24 =-(x-6+2\sqrt3)(x-6-2\sqrt3). \]

з)
\[ D=5^2-4\cdot(-2)\cdot(-3)=1. \]

\[ x_{1,2}=\frac{-5\pm1}{-4}=1;\,\frac32. \]

\[ -2x^2+5x-3 =-2(x-1)\left(x-\frac32\right) =(x-1)(3-2x). \]

и)
\[ D=(-2)^2-4\cdot2\cdot\frac12=0. \]

\[ x_0=\frac{-(-2)}{2\cdot2}=\frac12. \]

\[ 2x^2-2x+\frac12 =2\left(x-\frac12\right)^2. \]

к)
\[ D=(-24)^2-4\cdot(-16)\cdot(-9)=0. \]

\[ a_0=\frac{-(-24)}{2\cdot(-16)}=-\frac34. \]

\[ -16a^2-24a-9 =-16\left(a+\frac34\right)^2 =-(4a+3)^2. \]

л)
\[ D=(-5)^2-4\cdot1\cdot6=1. \]

\[ m_{1,2}=\frac{5\pm1}{2}=3;\,2. \]

\[ m^2-5m+6=(m-3)(m-2). \]

м)
\[ D=(-2)^2-4\cdot0{,}25\cdot4=0. \]

\[ m_0=\frac{-(-2)}{2\cdot0{,}25}=4. \]

\[ 0{,}25m^2-2m+4 =0{,}25(m-4)^2 =(0{,}5m-2)^2. \]

Да, понял: нужно оформить как в образце:

\[ \text{выражение}=0\;\longrightarrow\; a,b,c\;\longrightarrow\; D\;\longrightarrow\;\text{корни}\;\longrightarrow\;\text{разложение}. \]

Формула:

\[ D=b^2-4ac,\qquad x_{1,2}=\frac{-b\pm\sqrt D}{2a}, \] \[ ax^2+bx+c=a(x-x_1)(x-x_2). \]

№ 35

а)

\[ 2x^2-5x-7=0 \]

\[ a=2,\quad b=-5,\quad c=-7 \]

\[ D=(-5)^2-4\cdot2\cdot(-7)=81 \]

\[ x_1=\frac{5-9}{4}=-1,\qquad x_2=\frac{5+9}{4}=\frac72 \]

\[ 2x^2-5x-7=2(x+1)\left(x-\frac72\right) =(x+1)(2x-7). \]

б)

\[ -5y^2-2y+3=0 \]

\[ a=-5,\quad b=-2,\quad c=3 \]

\[ D=(-2)^2-4\cdot(-5)\cdot3=64 \]

\[ y_1=\frac{2-8}{-10}=\frac35,\qquad y_2=\frac{2+8}{-10}=-1 \]

\[ -5y^2-2y+3 =-5\left(y-\frac35\right)(y+1) =(3-5y)(y+1). \]

в)

\[ x^2+8x-9=0 \]

\[ a=1,\quad b=8,\quad c=-9 \]

\[ D=8^2-4\cdot1\cdot(-9)=100 \]

\[ x_1=\frac{-8-10}{2}=-9,\qquad x_2=\frac{-8+10}{2}=1 \]

\[ x^2+8x-9=(x+9)(x-1). \]

г)

\[ \frac16x^2+\frac12x+\frac13=0 \]

\[ a=\frac16,\quad b=\frac12,\quad c=\frac13 \]

\[ D=\left(\frac12\right)^2 -4\cdot\frac16\cdot\frac13 =\frac14-\frac29=\frac1{36} \]

\[ x_1=\frac{-\frac12-\frac16}{\frac13}=-2,\qquad x_2=\frac{-\frac12+\frac16}{\frac13}=-1 \]

\[ \frac16x^2+\frac12x+\frac13 =\frac16(x+2)(x+1). \]

д)

\[ -5x^2-10x+15=0 \]

\[ a=-5,\quad b=-10,\quad c=15 \]

\[ D=(-10)^2-4\cdot(-5)\cdot15=400 \]

\[ x_1=\frac{10+20}{-10}=-3,\qquad x_2=\frac{10-20}{-10}=1 \]

\[ -5x^2-10x+15=-5(x+3)(x-1). \]

е)

\[ y^2-16y+15=0 \]

\[ a=1,\quad b=-16,\quad c=15 \]

\[ D=(-16)^2-4\cdot1\cdot15=196 \]

\[ y_1=\frac{16-14}{2}=1,\qquad y_2=\frac{16+14}{2}=15 \]

\[ y^2-16y+15=(y-1)(y-15). \]

ж)

\[ -x^2+12x-24=0 \]

\[ a=-1,\quad b=12,\quad c=-24 \]

\[ D=12^2-4\cdot(-1)\cdot(-24)=48 \]

\[ \sqrt D=4\sqrt3 \]

\[ x_1=\frac{-12+4\sqrt3}{-2}=6-2\sqrt3, \]

\[ x_2=\frac{-12-4\sqrt3}{-2}=6+2\sqrt3 \]

\[ -x^2+12x-24 =-(x-6+2\sqrt3)(x-6-2\sqrt3). \]

з)

\[ -2x^2+5x-3=0 \]

\[ a=-2,\quad b=5,\quad c=-3 \]

\[ D=5^2-4\cdot(-2)\cdot(-3)=1 \]

\[ x_1=\frac{-5+1}{-4}=1,\qquad x_2=\frac{-5-1}{-4}=\frac32 \]

\[ -2x^2+5x-3 =-2(x-1)\left(x-\frac32\right) =(x-1)(3-2x). \]

и)

\[ 2x^2-2x+\frac12=0 \]

\[ a=2,\quad b=-2,\quad c=\frac12 \]

\[ D=(-2)^2-4\cdot2\cdot\frac12=0 \]

\[ x_0=\frac{2}{4}=\frac12 \]

\[ 2x^2-2x+\frac12 =2\left(x-\frac12\right)^2. \]

к)

\[ -16a^2-24a-9=0 \]

\[ A=-16,\quad B=-24,\quad C=-9 \]

\[ D=(-24)^2-4\cdot(-16)\cdot(-9)=0 \]

\[ a_0=\frac{24}{-32}=-\frac34 \]

\[ -16a^2-24a-9 =-16\left(a+\frac34\right)^2 =-(4a+3)^2. \]

л)

\[ m^2-5m+6=0 \]

\[ a=1,\quad b=-5,\quad c=6 \]

\[ D=(-5)^2-4\cdot1\cdot6=1 \]

\[ m_1=\frac{5-1}{2}=2,\qquad m_2=\frac{5+1}{2}=3 \]

\[ m^2-5m+6=(m-2)(m-3). \]

м)

\[ 0{,}25m^2-2m+4=0 \]

\[ a=0{,}25,\quad b=-2,\quad c=4 \]

\[ D=(-2)^2-4\cdot0{,}25\cdot4=0 \]

\[ m_0=\frac{2}{0{,}5}=4 \]

\[ 0{,}25m^2-2m+4 =0{,}25(m-4)^2 =(0{,}5m-2)^2. \]