delta --> to ten trójkąt, nie wiedziałam jak go wstawić, więc napisałam słownie
a)
[tex](2x - 3)(2x + 3) = ( {x + 5})^{2} - 9[/tex]
[tex]4 {x}^{2} - 9 = {x}^{2} + 10x + 25 - 9[/tex]
[tex]3 {x}^{2} - 10x - 25 = 0[/tex]
[tex]delta = 100 - 4 \times 3 \times ( - 25) = 100 + 300 = 400[/tex]
[tex] \sqrt{delta} = \sqrt{400} = 20[/tex]
[tex]x \frac{}{1} = \frac{10 - 20}{6} = \frac{ - 10}{6} = - 1 \frac{4}{6} = - 1 \frac{2}{3} [/tex]
[tex]x \frac{}{2} = \frac{10 + 20}{6} = \frac{30}{6} = 5[/tex]
Odp. x należy do zbioru {-1 2/3, 5}
b)
[tex]2x(x - 3) - (x + 1)(x + 2) = ( {2x - 3})^{2} [/tex]
[tex]2 {x}^{2} - 6x - ( {x}^{2} + 2x + x + 2) = 4 {x}^{2} - 12x + 9[/tex]
[tex]2 {x}^{2} - 6x - {x}^{2} - 2x - x - 2 = 4 {x}^{2} - 12x + 9[/tex]
[tex] - 3 {x}^{2} + 3x - 11 = 0[/tex]
[tex]delta = 9 - 4 \times ( - 3) \times ( - 11) = 9 - 132 = - 123 < 0[/tex]
równanie nie ma rozwiązań
c)
[tex]( {2x - 1})^{2} - (4x + 1)(4x - 1) = 4x(1 - 2x)[/tex]
[tex]4 {x}^{2} - 4x + 1 - (16 {x}^{2} - 1) = 4x - 8 {x}^{2} [/tex]
[tex]4 {x}^{2} - 4x + 1 - 16 {x}^{2} + 1 = 4x - 8 {x}^{2} [/tex]
[tex] - 4 {x}^{2} - 8x + 2 = 0[/tex]
[tex]delta = 64 - 4 \times ( - 4) \times 2 = 64 + 32 = 96[/tex]
[tex] \sqrt{delta} = \sqrt{96} = \sqrt{16 \times 6} = 4 \sqrt{6} [/tex]
[tex]x \frac{}{1} = \frac{8 - 4 \sqrt{6} }{ - 8} = \frac{4(2 - \sqrt{6} )}{ - 8} = \frac{2 - \sqrt{6} }{ - 2} = - \frac{2 - \sqrt{6} }{2} [/tex]
[tex]x \frac{}{2} = \frac{8 + 4 \sqrt{6} }{ - 8} = \frac{4(2 + \sqrt{6}) }{ - 8} = \frac{2 + \sqrt{6} }{ - 2} = - \frac{2 + \sqrt{6} }{2} [/tex]
Odp. x należy do zbioru {[tex]{{ - \frac{2 - \sqrt{6} }{2} }, - \frac{2 + \sqrt{6} }{2}}[/tex]}
d)
[tex] {(2x + 1)}^{2} - 9 {(x - 1)}^{2} = 4 - x[/tex]
[tex]4 {x}^{2} + 4x + 1 - 9( {x}^{2} - 2x + 1) = 4 - x[/tex]
[tex]4 {x}^{2} + 4x + 1 - 9 {x}^{2} + 18x - 9 = 4 - x[/tex]
[tex] - 5 {x}^{2} + 23x - 12 = 0[/tex]
[tex]delta =529 - 4 \times ( - 5) \times ( - 12) = 529 - 240 = 289[/tex]
[tex] \sqrt{delta} = \sqrt{289} = 17[/tex]
[tex]x \frac{}{1} = \frac{ - 23 - 17}{ - 10} = \frac{ - 40}{ - 10} = 4[/tex]
[tex]x \frac{}{2} = \frac{ - 23 + 17}{ - 10} = \frac{ - 6}{ - 10} = \frac{6}{10} = \frac{3}{5} [/tex]
Odp. x należy do zbioru {4, 3/5}
e)
[tex] {(x - 1)}^{2} + 2( {x - 3)}^{2} = 18 - 10x[/tex]
[tex] {x}^{2} - 2x + 1 + 2( {x}^{2} - 6x + 9) = 18 - 10x[/tex]
[tex] {x}^{2} - 2x + 1 + 2 {x}^{2} - 12x + 18 = 18 - 10x[/tex]
[tex]3 {x}^{2} - 4x + 1 = 0[/tex]
[tex]delta = 16 - 4 \times 3 \times 1 = 16 - 12 = 4[/tex]
[tex] \sqrt{delta} = \sqrt{4} = 2[/tex]
[tex]x \frac{}{1} = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3} [/tex]
[tex]x \frac{}{2} = \frac{4 + 2}{6} = \frac{6}{6} = 1[/tex]
Odp. x należy do zbioru {1/3, 1}
f)
[tex] {(2x + 5)}^{2} - {(2x - 5)}^{2} = 3 {x}^{2} + 41x - 4[/tex]
[tex]4 {x}^{2} + 20x + 25 - 4 {x}^{2} + 20x - 25 = 3 {x}^{2} + 41x - 4[/tex]
[tex] - 3 {x}^{2} - x + 4 = 0[/tex]
[tex]delta = 1 - 4 \times ( - 3) \times 4 = 1 + 48 = 49[/tex]
[tex] \sqrt{delta} = \sqrt{49} = 7[/tex]
[tex]x \frac{}{1} = \frac{1 - 7}{ - 6} = \frac{ - 6}{ - 6} = 1[/tex]
[tex]x \frac{}{2} = \frac{1 + 7}{ - 6} = \frac{8}{ - 6} = - \frac{4}{3} = - 1 \frac{1}{3} [/tex]
Odp. x należy do zbioru {1, -1 1/3}
