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Задание 1.
============== 1 ==============
$\frac{y^2-11y-26}{9y+18}\\ y^2-11y-26=0\\ D=b^2-4ac=(-11)^2-4\cdot1\cdot(-26) =121+104=225=15^2\\ y_1= \frac{11+15}{2}= 13\\ y_2= \frac{11-15}{2}= -2\\ y^2-11y-26=(y-13)(y+2)\\ \frac{y^2-11y-26}{9y+18}= \frac{(y-13)(y+2)}{9(y+2)}= \frac{y-13}{9} \\ y=-5 \rightarrow \frac{-5-13}{9}= \frac{-18}{9}=-2\\ y=31 \rightarrow \frac{31-13}{9}= \frac{18}{9}=2\\ y=112 \rightarrow \frac{112-13}{9}= \frac{99}{9}=11\\$
============== 2 ==============
$\frac{x^2-18x+80}{5x-50} \\ x^2-18x+80=0\\ D=b^2-4ac=(-18)^2-4\cdot1\cdot80=324-320=4=2^2\\ x_1= \frac{18+2}{2} =10\\ x_2= \frac{18-2}{2} =8\\ x^2-18x+80=(x-10)(x-8)\\ \frac{x^2-18x+80}{5x-50}= \frac{(x-10)(x-8)}{5(x-10)}= \frac{x-8}{5} \\ x=-12 \rightarrow \frac{-12-8}{5}= \frac{-20}{5}= -4\\ x=8.5 \rightarrow \frac{8.5-8}{5}= \frac{0.5}{5}= 0.1\\ x=48 \rightarrow \frac{48-8}{5}= \frac{40}{5}= 8\\$
============== 3 ==============
$\frac{x^2-8x-33}{10x+30} \\ x^2-8x-33 = 0\\ D=b^2-4ac=(-8)^2-4\cdot1\cdot(-33)=64+132=196=14^2\\ x_1= \frac{8+14}{2} =11\\ x_2= \frac{8-14}{2} =-3\\ x^2-8x-33 =(x-11)(x+3)\\ \frac{x^2-8x-33}{10x+30} = \frac{(x-11)(x+3)}{10(x+3)} = \frac{x-11}{10}\\ x=-9 \rightarrow \frac{-9-11}{10}= \frac{-20}{5}= -4\\ x=12 \rightarrow \frac{12-11}{10}= \frac{1}{10}\\ x=111 \rightarrow \frac{111-11}{10}= \frac{100}{5}= 20\\$
============== 4 ==============
$\frac{8y-56}{y^2-27y+140}\\ y^2-27y+140=0\\ D=b^2-4ac=(-27)^2-4\cdot1\cdot140=729-560=169=13^2\\ y_1= \frac{27+13}{2}= \frac{40}{2}=20\\ y_2= \frac{27-13}{2}= \frac{14}{2}=7\\ y^2-27y+140= (y-20)(y-7)\\ \frac{8y-56}{y^2-27y+140}= \frac{8(y-7)}{(y-20)(y-7)}= \frac{8}{y-20}\\ y = -4 \rightarrow \frac{8}{-4-20}= \frac{8}{-24}= - \frac{1}{3}\\ y = 22.5 \rightarrow \frac{8}{22.5-20}= \frac{8}{2.5}= 3.2\\ y = 24 \rightarrow \frac{8}{24-20}= \frac{8}{4}= 2\\$

Задание 2.
============== 1 ==============
$\frac{2.5+3.48}{6.34}= \frac{5.98}{6.34} =5 \frac{98}{100}:6 \frac{34}{100}= 5 \frac{49}{50}:6 \frac{17}{50}= \frac{299}{50} \cdot \frac{50}{317}= \frac{299}{317}$
============== 2 ==============
$\frac{2.5\cdot3.48}{6.34}= \frac{8.7}{6.34} =8 \frac{7}{10}:6 \frac{34}{100}= \frac{87}{10} \cdot \frac{50}{317}= \frac{87 \cdot5}{317}= \frac{435}{317} =1 \frac{118}{317}$
============== 3 ==============
$\frac{2.5}{6.34+3.48}= \frac{2.5}{9.82} =2 \frac{5}{10}:9 \frac{82}{100}= 2 \frac{1}{2}:9 \frac{41}{50}= \frac{5}{2} \cdot \frac{50}{491} = \frac{5\cdot25}{491}= \frac{125}{491}$
============== 4 ==============
$\frac{2.5}{6.34\cdot3.48}= \frac{2 \frac{5}{10} }{6 \frac{34}{100}\cdot 3 \frac{48}{100} } =\frac{2 \frac{1}{2} }{6 \frac{17}{50}\cdot 3 \frac{12}{25} } =\frac{ \frac{5}{2} }{ \frac{317}{50}\cdot \frac{87}{25} } = \frac{5}{2}\cdot \frac{50 \cdot 25}{317\cdot87} = \frac{3125}{27579}$
============== 5 ==============
$2.7(18^2-5^2)=2.7(18-5)(18+5)=2.7 \cdot 13 \cdot 23 = 807.3$
============== 6 ==============
<img src="https://tex.z-dn.net/?f=%281%2B+%5Csqrt%7B+%5Cfrac%7Ba-x%7D%7Ba%2Bx%7D+%7D+%29%281-+%5Csqrt%7B+%5Cfrac%7Ba-x%7D%7Ba%2Bx%7D+%7D+%29%3D1%5E2-+%28%5Csqrt%7B+%5Cfrac%7Ba-x%7D%7Ba%2Bx%7D+%7D%29%5E2%3D1-+%5Cfrac%7Ba-x%7D%7Ba%2Bx%7D+%7D%3D+%5Cfrac%7Ba%2Bx-a%2Bx%7D%7Ba%2Bx%7D+%3D%5C%5C%0A%3D+%5Cfrac%7B2x%7D%7Ba%2Bx%7D+%5C%5C%0Aa%3D5%2C+x%3D4+%5Crightarrow++%5Cfrac%7B2x%7D%7Ba%2Bx%7D+%3D+%5Cfrac%7B2%5Ccdot4%7D%7B5%2B4%7D%3D+%5Cfrac%7B8%7D%7B9%7D+" id="TexFormula10" title="(1+ \sqrt{ \frac{a-x}{a+x} } )(1- \sqrt{ \frac{a-x}{a+x} } )=1^2- (\sqrt{ \frac{a-x}{a+x} })^2=1- \frac{a-x}{a+x} }= \frac{a+x-a+x}{a+x} =\\ = \frac{2x}{a+x} \\ a=5, x=4 \rightarrow \frac{2x}{a+x} = \frac{2\cdot4}{5+4}= \frac{8}{9} " alt="(1+ \sqrt{ \frac{a-x}{a+x} } )(1- \sqrt{ \frac{a-x}{a+x} } )=1^2- (\sqrt{ \frac{a-x}{a+x} })^2=1- \frac{a-x}{a+x} }= \frac{a+x-a+x}{a+x} =\\ = \frac{2x}{a+x}

paradiseva_zn (39.4k баллов) 17 Май, 18