Задачи по математика с полагане

[Гост]

1. (x - 4)(x - 5)(x - 6)(x - 7) = 1680
2. x(x + 1)(x + 2)(x + 3) = 9/16
3. (x + 3)^4 + (x + 5)^4 = 16
4. (3 - x)^4 + (2 - x)^4 = (5 - 2x)^4
Задачите трябва да се решат с полагане!

[ammornil]

1. (x - 4)(x - 5)(x - 6)(x - 7) = 1680

Средно аритметичното на [tex]-4, -5, -6[/tex] и [tex]-7[/tex] е [tex]-\frac{11}{2}[/tex]
[tex]t=x-\frac{11}{2} \Leftrightarrow x=t+\frac{11}{2} \Rightarrow \begin{cases} x-4 = t+\frac{3}{2} \\ x-5 = t+\frac{1}{2} \\ x-6 =t-\frac{1}{2} \\ x-7 = t-\frac{3}{2} \end{cases}[/tex]

[tex]\Rightarrow (x - 4)(x - 5)(x - 6)(x - 7) = 1680 \Leftrightarrow \left( t+\frac{3}{2} \right)\left( t+\frac{1}{2} \right)\left( t-\frac{1}{2} \right)\left( t-\frac{3}{2} \right)=1680 \Leftrightarrow \left( t^{2}-\frac{1}{4} \right)\left( t^{2}-\frac{9}{4} \right)=1680[/tex]

[tex]u=t^{2}-\frac{5}{4}, \left( Du:\> u \ge -\frac{5}{4} \right) \Rightarrow \begin{cases} t^{2}-\frac{1}{4} = u +1 \\ t^{2}-\frac{3}{9}=u-1 \end{cases}[/tex]

[tex]\Rightarrow \left( t^{2}-\frac{1}{4} \right)\left( t^{2}-\frac{9}{4} \right)=1680 \Leftrightarrow (u+1)(u-1)=1680 \Leftrightarrow u^{2}-1=1680 \Leftrightarrow u^{2}=1681 \Rightarrow u_{1,2}=\pm41 \begin{cases} u_{1}=41 \in Du \\ u_{2}=-41 \notin Du \end{cases}[/tex]

[tex]u=41 \rightarrow t^{2}-\frac{5}{4}=41 \Leftrightarrow t^{2}=41+\frac{5}{4} \Leftrightarrow t^{2}=\frac{169}{4} \Rightarrow t_{1,2}=\pm\frac{13}{2}[/tex]

[tex]t_{1}=\frac{13}{2} \Rightarrow x_{1}=\frac{13}{2}+\frac{11}{2}=\frac{24}{2}=12[/tex]
[tex]t_{2}=-\frac{13}{2} \Rightarrow x_{2}=-\frac{13}{2}+\frac{11}{2}=-\frac{2}{2}=1[/tex]

2. x(x + 1)(x + 2)(x + 3) = 9/16

Аналогично на предната: [tex]t=x+\frac{3}{2} \Leftrightarrow x=t-\frac{3}{2} \rightarrow u=t^{2}-\frac{5}{4} \Leftrightarrow t^{2}=u+\frac{5}{4}, \left( Du:\> u \ge -\frac{5}{4} \right)[/tex]

[tex]u_{1,2}=\pm\frac{5}{4} \Rightarrow \begin{cases} t_{1,2}=\pm\sqrt{\frac{5}{2}}=\pm\frac{\sqrt{10}}{2} \Rightarrow \begin{cases} x_{1}=t_{1}-\frac{3}{2}=\frac{\sqrt{10}-3}{2} \\ x_{2}=t_{2}-\frac{3}{2}=\frac{-\sqrt{10}-3}{2} \end{cases} \\ t_{3,4}=0 \Rightarrow x_{3}=x_{4}=-\frac{3}{2} \end{cases}[/tex]

[ammornil]

3. (x + 3)^4 + (x + 5)^4 = 16

[tex]t=x+4 \Leftrightarrow x=t-4 \Rightarrow (t-1)^{4}+(t+1)^{4}=16 \Leftrightarrow t^{4}\cancel{-4t^{3}}+6t^{2}\cancel{-4t}+1+t^{4}\cancel{+4t^{3}}+6t^{2}\cancel{+4t}+1=16 \Leftrightarrow[/tex]

[tex]\Leftrightarrow 2t^{4}+12t^{2}-14=0 \Leftrightarrow t^{4}+6t^{2}-7=0[/tex]

[tex]u=t^{2} (Du:\> u \ge 0) \rightarrow u^{2}+6u-7= 0 \Rightarrow u_{1,2}=\frac{-3\mp \sqrt{3^{2}+7}}{1}=-3 \mp 4 \begin{cases} u_{1}=-7 \notin Du \\ u_{2}=1 \in Du \end{cases}[/tex]

[tex]u=1 \Rightarrow t^{2}=1 \Rightarrow t_{1,2}=\pm1 \Rightarrow \begin{cases} x_{1}=t_{1}-4=-1-4=-5 \\ x_{2}=t_{2}-4=1-4=-3 \end{cases}[/tex]

[ammornil]

4. (3 - x)^4 + (2 - x)^4 = (5 - 2x)^4

[tex]5-2x = (3-x)+(2-x)[/tex]

[tex]u=3-x, v=2-x \Rightarrow (3 - x)^{4} + (2 - x)^{4} = (5 - 2x)^{4} \Leftrightarrow u^{4}+v^{4}-(u+v)^{4}=0 \Leftrightarrow[/tex]

[tex]u^{4}+v^{4}-(u^{4}+4u^{3}v+6u^{2}v^{2}+4uv^{3}+v^{4})=0 \Leftrightarrow \cancel{u^{4}}\cancel{+v^{4}}\cancel{-u^{4}}-4u^{3}v-6u^{2}v^{2}-4uv^{3}\cancel{-v^{4}} =0 \Leftrightarrow[/tex]

[tex]-2uv(2u^{2}+3uv+2v^{2})=0 \Rightarrow u=0 \hspace{2em} \cup \hspace{2em} v=0 \hspace{2em} \cup \hspace{2em} 2u^{2}+3uv+2v^{2}=0[/tex]

[tex]\Rightarrow 3-x=0 \hspace{2em} \cup \hspace{2em} 2-x=0 \hspace{2em} \cup \hspace{2em} 2(3-x)^{2}+3(3-x)(2-x)+2(2-x)^{2}=0 \Leftrightarrow[/tex]

[tex]\Rightarrow x_{1}=3 \hspace{2em} \cup \hspace{2em} x_{2}=2 \hspace{2em} \cup \hspace{2em} 2(9-6x+x^{2})+3(6-3x-2x+x^{2})+2(4-4x+x^{2})=0 \Leftrightarrow[/tex]

[tex]\Rightarrow x_{1}=3 \hspace{2em} \cup \hspace{2em} x_{2}=2 \hspace{2em} \cup \hspace{2em} 2x^{2}-12x+18+3x^{2}-15x+18+2x^{2}-8x+8=0 \Leftrightarrow[/tex]

[tex]\Rightarrow x_{1}=3 \hspace{2em} \cup \hspace{2em} x_{2}=2 \hspace{2em} \cup \hspace{2em} 7x^{2}-35x+44=0 \rightarrow D=-7 \Rightarrow x_{3,4} \notin R[/tex]

[Евва]

1 зад. Ето още една идея .
Полагаме а=(х-4)(х-7) =[tex]x^{2 }[/tex]-11х+28 ; b=(x-5)(x-6) =[tex]x^{2 }[/tex]-11x+30
[tex]\begin{array}{|l} ab = 1 680 \\ b = a+2 \end{array}[/tex]
a(a+2)=1 680
[tex]a^{2 }[/tex]+2a-1680=0 ;k= 1 ; D=1+1680 =1681=[tex]41^{2 }[/tex]

[tex]a_{1,2 }[/tex]=[tex]\frac{- 1 \pm 41}{1}[/tex] ;получихме [tex]а_{1 }[/tex]=40 ,или [tex]а_{2 }[/tex]= -42

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Към полагането

1 случай а=40
[tex]x^{2 }[/tex]-11х+28=40
[tex]x^{2 }[/tex]-11х-12=0
D=121+48=169=[tex]13^{2 }[/tex]

[tex]x_{1,2 }[/tex]=[tex]\frac{11 \pm 13}{2}[/tex]
[tex]x_{1 }[/tex]=12 ,[tex]x_{2 }[/tex]= -1
___________________________________
2 случай а= -42
[tex]x^{2 }[/tex]-11x+28= -42
[tex]x^{2 }[/tex]-11x+70=0
D=121-280 <0 [tex]\Rightarrow[/tex] няма решение в R
краен отговор 12 и -1