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Potrzebuje podpunktów d i e z góry dziękuje :)

Potrzebuje Podpunktów D I E Z Góry Dziękuje class=

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Krysiaviz

[tex]d)\\\\x= \frac{1}{2}\\\\(- 9\frac{3}{7}x^3+ 3\frac{5}{7}x^2+ 1\frac{3}{7}x^3+x- \frac{2}{7}x^2) (4x-1)^3 =(- 8x^3+ 3\frac{3}{7}x^2 +x ) (4x-1)^3 =\\\\=(- 8 *(\frac{1}{2})^3+ \frac{24}{7}*(\frac{1}{2})^2 + \frac{1}{2}) (4*\frac{1}{2}-1)^3 =(- \not{8}^1 * \frac{1}{\not{8}^1} + \frac{\not{24}^6}{7}* \frac{1}{\not{4}^1} + \frac{1}{ 2}) (\not{4}^2*\frac{1}{\not{2}^1}-1)^3=\\\\ =(-1+\frac{6}{7}+\frac{1}{2})(2-1)^3=(- \frac{14}{14}+\frac{12}{14}+\frac{7}{14})*1^3=\frac{5}{14}*1=\frac{5}{14}[/tex]

[tex]e)\\\\ x=0,25=\frac{25}{100}=\frac{1}{4}\\\\(0,6x-1,4+7,4x^3+3,4x+8,6x^3+1,15)(3-8x)^2=(16x^3+4x- 0,25)(3-8x)^2=[/tex]

[tex]=(16 *(\frac{1}{4})^3+ *\frac{1}{\not{4}^1}- \frac{1}{4})(3-\not{8}^2*\frac{1}{\not{4}^1} )^2=(\not{16}^1\frac{1}{\not{64}^4}+1-\frac{1}{4}) (3-2)^2=\\\\=(\frac{1}{4}+1-\frac{1}{4})*1^2=1*1=1[/tex]

ZbiorJviz

[tex]zad.d\\\\(-9\frac{3}{7} x^{3} +3\frac{5}{7} x^{2} +1\frac{3}{7} x^{3} +x-\frac{2}{7} x^{2} )\cdot (4x-1)^{3} =(-8x^{3} +3\frac{3}{7} x^{2}+x)\cdot (4x-1)^{3} =(-8x^{3} +\frac{24}{7} x^{2}+x)\cdot (4x-1)^{3}\\\\gdy~~x=\frac{1}{2} \\\\(-8\cdot (\frac{1}{2}) ^{3} +\frac{24}{7}\cdot (\frac{1}{2}) ^{2}+\frac{1}{2} )\cdot (4\cdot \frac{1}{2}-1)^{3}=(-8\cdot \frac{1}{8}+\frac{24}{7}\cdot \frac{1}{4}+\frac{1}{2})\cdot (2-1)^{3} =[/tex]

[tex]=(-1+\frac{6}{7}+\frac{1}{2})\cdot 1^{3} =(-1+\frac{12}{14}+\frac{7}{14})\cdot 1=-1+\frac{19}{14} =-1+1\frac{5}{14} =\frac{5}{14}[/tex]

[tex]zad.e\\\\(0,6x-1,4+7,4x^{3} +3,4x+8,6x^{3} +1,15)\cdot (3-8x)^{2} =(16x^{3} +4x-0,25) \cdot (3-8x)^{2}\\\\gdy~~x=0,25=\frac{1}{4} \\\\(16\cdot ( \frac{1}{4})^{3} +4\cdot \frac{1}{4} - \frac{1}{4})\cdot (3-8\cdot \frac{1}{4}) ^{2} =(16\cdot \frac{1}{64}+1-\frac{1}{4})\cdot (3-2)^{2} =(\frac{1}{4} +1-\frac{1}{4})\cdot 1^{2} =1\cdot 1 = 1[/tex]

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