Matematyka
Szymon500
2

1.     Oblicz  [latex]\frac{10-2\frac{1}{2}:1,5-1,75\cdot(2\frac{2}{7}-5\frac{6}{7}}{2,25\cdot(2,5-2\frac{1}{4})} [/latex] 2.  Oblicz     [latex]\frac{8:(-1\frac{3}{5})-(4\frac{1}{2}\cdot1\frac{2}{3}-3\frac{3}{4})}{\frac{1}{2}\cdot2\frac{1}{2}}[/latex] 3.Oblicz   [latex][(5\frac{1}{2}-3\frac{1}{4}):2\frac{1}{4}+7\frac{5}{8}]:2\frac{7}{8}[/latex] 4.rozwiąż nierówność i przedstaw zbiór rozwiązań na osi liczbowej A.)    [latex](3x+1)^2\leq8x-(2-3x)(2+3x)[/latex] B.)     [latex]4(x-3)^2-7 >(2x+1)^2[/latex] 5. rozwiąż nierówność   [latex](x+2)^2-\frac{2x-1}{3}\geq(x-\sqrt{2})(x+\sqrt{2})-\frac{1}{3}[/latex] 6.Doprowadź do najprostrzej postaci to wyrażenie i oblicz jego wartość dla x=¹¹/₆ [latex](3x-4)^2-2(2x+5)^2-(x-8)(8+x)[/latex]

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(1) Odpowiedź
pumaa3

1.   [latex]\frac{10-2\frac{1}{2}:1,5-1,75\cdot (2\frac{2}{7}-5\frac{6}{7})}{2,25 \cdot (2,5-2\frac{1}{4})}=\frac{10-\frac{5}{2}: \frac{3}{2}-1\frac{3}{4}\cdot (\frac{16}{7}-\frac{41}{7})}{\frac{9}{4} \cdot \frac{1}{4}}=\frac{10-\frac{5}{2}\cdot \frac{2}{3}-\frac{7}{4} \cdot (-\frac{25}{7})}{\frac{9}{16}}=\\ \\ [/latex] [latex]=\frac{10-\frac{5}{3}+\frac{25}{4}}{\frac{9}{16}}=\frac{\frac{120}{12}-\frac{20}{12}+\frac{75}{12}}{\frac{9}{16}}=\frac{\frac{175}{12}}{\frac{9}{16}}=\frac{175}{12} \cdot \frac{16}{9}=\frac{700}{27}=25\frac{25}{27}\\ \\ [/latex]   2. [latex]\frac{8:(-1\frac{3}{5})-(4\frac{1}{2}\cdot 1\frac{2}{3}-3\frac{3}{4})}{\frac{1}{2}\cdot 2\frac{1}{2}} =\frac{8:(-\frac{8}{5})-(\frac{9}{2}\cdot \frac{5}{3}-\frac{15}{4})}{\frac{1}{2}\cdot \frac{5}{2}}=\frac{8\cdot (-\frac{5}{8})-(\frac{15}{2}-\frac{15}{4})}{\frac{5}{4}}=\\ \\ =\frac{-5-\frac{15}{4}}{\frac{5}{4}}=\frac{-\frac{20}{4}-\frac{15}{4}}{\frac{5}{4}}=\frac{-\frac{35}{4}}{\frac{5}{4}}=-\frac{35}{4} \cdot \frac{4}{5}=-7[/latex]   3.   [latex][(5\frac{1}{2}-3\frac{1}{4}):2\frac{1}{4}+7\frac{5}{8}]:2\frac{7}{8}=[(\frac{11}{2}-\frac{13}{4}):\frac{9}{4}+\frac{61}{8}]:\frac{23}{8}=\\ \\ =[\frac{9}{4} \cdot \frac{4}{9}+\frac{61}{8}]\cdot \frac{8}{23}=[1+\frac{61}{8}]\cdot \frac{8}{23}=\frac{69}{8}\cdot \frac{8}{23}=3[/latex]   4. A. [latex](3x+1)^2\le8x-(2-3x)(2+3x)\\ 9x^2+6x+1\le8x-4+9x^2\\ 6x-8x\le-4-1\\ -2x\le-5\\ x \ge \frac{5}{2}\\ \\ x \in <\frac{5}{2}, +\infty)[/latex]                        O-----------------------> --------------5/2---------------------->   B. [latex]4(x-2)^2-7>(2x+1)^2\\ 4(x^2-4x+4)-7>4x^2+4x+1\\ 4x^2-16x+16-7>4x^2+4x+1\\ -16x-4x>1-16+7\\ -20x>-8\\ x<\frac{2}{5}\\ \\ x \in (-\infty, \frac{2}{5})[/latex]   <-----------------o -----------------2/5------------------->     5. [latex](x+2)^2-\frac{2x-1}{3}\ge (x-\sqrt{2})(x+\sqrt{2})-\frac{1}{3}\\ x^2+4x+4-\frac{2x-1}{3}\ge x^2-2-\frac{1}{3}\\ 4x+4-\frac{2x-1}{3} \ge-2-\frac{1}{3} \ \setminus \cdot 3\\ 12x+12-2x+1 \ge-6-1\\ 10x \ge -20\\ x \ge 2\\ \\ x \in <2, +\infty)[/latex]     6. tex](3x-4)^2-2(2x+5)^2-(x-8)(8+x)=\\ =9x^2-24x+16-2(4x^2+20x+25)-(x^2-64)=\\ =9x^2-24x+16-8x^2-40x-50-x^2+64=-64x+30 =\\ = -64\cdot \frac{11}{6}+30 = -\frac{352}{3}+\frac{90}{3}=\frac{-262}{3}=-87\frac{1}{3} [/latex]  

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