Предмет: Алгебра,
автор: BJIADA
1.408,1.410,1.411 под буквами "а"
Приложения:
Ответы
Автор ответа:
0
решение во вложенииииииииииииииииииииии
11 добавлю
добавила
11 добавлю
добавила
Приложения:
Автор ответа:
0
1.408a
![f(x)= frac{1}{x+1} + frac{2}{x-1}
\
f(x)=f(a), a=3
\
frac{1}{x+1} + frac{2}{x-1} =frac{1}{3+1} + frac{2}{3-1}
\
frac{1}{x+1} + frac{2}{x-1} =frac{1}{4} + frac{2}{2}
\
frac{1}{x+1} + frac{2}{x-1} =frac{5}{4} (x neq pm1)
\
4(x-1)+8(x+1)=5(x^2-1)
\
4x-4+8x+8=5x^2-5
\
5x^2-12x-9=0
\
D_1=(-6)^2-5cdot(-9)=36+45=81
\
x_1= frac{6+9 }{5} =3
\
x_2= frac{6-9 }{5} =-0.6](https://tex.z-dn.net/?f=f%28x%29%3D+frac%7B1%7D%7Bx%2B1%7D+%2B+frac%7B2%7D%7Bx-1%7D+%0A%5C%0Af%28x%29%3Df%28a%29%2C++a%3D3%0A%5C%0Afrac%7B1%7D%7Bx%2B1%7D+%2B+frac%7B2%7D%7Bx-1%7D+%3Dfrac%7B1%7D%7B3%2B1%7D+%2B+frac%7B2%7D%7B3-1%7D+%0A%5C%0Afrac%7B1%7D%7Bx%2B1%7D+%2B+frac%7B2%7D%7Bx-1%7D+%3Dfrac%7B1%7D%7B4%7D+%2B+frac%7B2%7D%7B2%7D+%0A%5C%0Afrac%7B1%7D%7Bx%2B1%7D+%2B+frac%7B2%7D%7Bx-1%7D+%3Dfrac%7B5%7D%7B4%7D++%28x+neq+pm1%29%0A%5C%0A4%28x-1%29%2B8%28x%2B1%29%3D5%28x%5E2-1%29%0A%5C%0A4x-4%2B8x%2B8%3D5x%5E2-5%0A%5C%0A5x%5E2-12x-9%3D0%0A%5C%0AD_1%3D%28-6%29%5E2-5cdot%28-9%29%3D36%2B45%3D81%0A%5C%0Ax_1%3D+frac%7B6%2B9+%7D%7B5%7D+%3D3%0A%5C%0Ax_2%3D+frac%7B6-9+%7D%7B5%7D+%3D-0.6 "f(x)= frac{1}{x+1} + frac{2}{x-1}
\
f(x)=f(a), a=3
\
frac{1}{x+1} + frac{2}{x-1} =frac{1}{3+1} + frac{2}{3-1}
\
frac{1}{x+1} + frac{2}{x-1} =frac{1}{4} + frac{2}{2}
\
frac{1}{x+1} + frac{2}{x-1} =frac{5}{4} (x neq pm1)
\
4(x-1)+8(x+1)=5(x^2-1)
\
4x-4+8x+8=5x^2-5
\
5x^2-12x-9=0
\
D_1=(-6)^2-5cdot(-9)=36+45=81
\
x_1= frac{6+9 }{5} =3
\
x_2= frac{6-9 }{5} =-0.6")
Ответ: 3 и -0,6
1.410а
![phi(x)=x-3+ frac{1-4x}{x} =x-3+ frac{1}{x} -4=x+ frac{1}{x} -7, x neq 0
\
phi(2.5)=2.5+ frac{1}{2.5} -7=-4.5+ frac{1}{5/2}=-4.5+ frac{2}{5}=-4.5+0.4=-4.1
\
phi(0.4)=0.4+ frac{1}{0.4} -7=-6.6+ frac{1}{2/5}=-6.6+ frac{5}{2}=-6.6+2.5=-4.1
\
Rightarrow phi(2.5)=phi(0.4)](https://tex.z-dn.net/?f=phi%28x%29%3Dx-3%2B+frac%7B1-4x%7D%7Bx%7D+%3Dx-3%2B+frac%7B1%7D%7Bx%7D+-4%3Dx%2B+frac%7B1%7D%7Bx%7D+-7%2C++x+neq+0%0A%5C%0Aphi%282.5%29%3D2.5%2B+frac%7B1%7D%7B2.5%7D+-7%3D-4.5%2B+frac%7B1%7D%7B5%2F2%7D%3D-4.5%2B+frac%7B2%7D%7B5%7D%3D-4.5%2B0.4%3D-4.1%0A%5C%0Aphi%280.4%29%3D0.4%2B+frac%7B1%7D%7B0.4%7D+-7%3D-6.6%2B+frac%7B1%7D%7B2%2F5%7D%3D-6.6%2B+frac%7B5%7D%7B2%7D%3D-6.6%2B2.5%3D-4.1%0A%5C%0ARightarrow+phi%282.5%29%3Dphi%280.4%29 "phi(x)=x-3+ frac{1-4x}{x} =x-3+ frac{1}{x} -4=x+ frac{1}{x} -7, x neq 0
\
phi(2.5)=2.5+ frac{1}{2.5} -7=-4.5+ frac{1}{5/2}=-4.5+ frac{2}{5}=-4.5+0.4=-4.1
\
phi(0.4)=0.4+ frac{1}{0.4} -7=-6.6+ frac{1}{2/5}=-6.6+ frac{5}{2}=-6.6+2.5=-4.1
\
Rightarrow phi(2.5)=phi(0.4)")
1.411a
![y_1=x^2+ frac{1}{x-4} , x neq -4
\
y_2= frac{4x-15}{x-4} = frac{4x-16+1}{x-4} =frac{4(x-4)+1}{x-4} =4+frac{1}{x-4} , x neq -4
\
y_1=y_2
\
x^2+ frac{1}{x-4} =4+ frac{1}{x-4}
\
x^2=4
\
x=pm 2](https://tex.z-dn.net/?f=y_1%3Dx%5E2%2B+frac%7B1%7D%7Bx-4%7D+%2C++x+neq+-4%0A%5C%0Ay_2%3D+frac%7B4x-15%7D%7Bx-4%7D+%3D+frac%7B4x-16%2B1%7D%7Bx-4%7D+%3Dfrac%7B4%28x-4%29%2B1%7D%7Bx-4%7D+%3D4%2Bfrac%7B1%7D%7Bx-4%7D+%2C++x+neq+-4%0A%5C%0Ay_1%3Dy_2%0A%5C%0Ax%5E2%2B+frac%7B1%7D%7Bx-4%7D+%3D4%2B+frac%7B1%7D%7Bx-4%7D+%0A%5C%0Ax%5E2%3D4%0A%5C%0Ax%3Dpm+2 "y_1=x^2+ frac{1}{x-4} , x neq -4
\
y_2= frac{4x-15}{x-4} = frac{4x-16+1}{x-4} =frac{4(x-4)+1}{x-4} =4+frac{1}{x-4} , x neq -4
\
y_1=y_2
\
x^2+ frac{1}{x-4} =4+ frac{1}{x-4}
\
x^2=4
\
x=pm 2")
Ответ: 3 и -0,6
1.410а
1.411a
Приложения:
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