## GIÚP IEM VỚI THÁNKS

Question

GIÚP IEM VỚI
THÁNKS

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3 years 2021-04-16T23:50:00+00:00 1 Answers 25 views 0

1. Giải thích các bước giải:

2.Ta có:

$C(x)=7x^2+\dfrac34x-10+\dfrac92x^2-\dfrac56x$

$\to C(x)=\dfrac{9}{2}x^2+7x^2+\dfrac{3}{4}x-\dfrac{5}{6}x-10$

$\to C(x)=\dfrac{23}{2}x^2-\dfrac1{12}x-10$

Ta có:

$D(x)=-10x^2+\dfrac35x^3+\dfrac25x-10$

$\to -C(x)+D(x)=-(\dfrac{23}{2}x^2-\dfrac1{12}x-10)+(-10x^2+\dfrac35x^3+\dfrac25x-10)$

$\to -C(x)+D(x)=-\dfrac{23}{2}x^2+\dfrac1{12}x+10-10x^2+\dfrac35x^3+\dfrac25x-10$

$\to -C(x)+D(x)=\dfrac{3}{5}x^3-\dfrac{23}{2}x^2-10x^2+\dfrac{2}{5}x+\dfrac{1}{12}x+10-10$

$\to -C(x)+D(x)=\dfrac{3x^3}{5}-\dfrac{43x^2}{2}+\dfrac{29x}{60}$

Ta có:

$C(x)+5D(x)=(\dfrac{23}{2}x^2-\dfrac1{12}x-10)+5(-10x^2+\dfrac35x^3+\dfrac25x-10)$

$\to C(x)+5D(x)=\dfrac{23}{2}x^2-\dfrac1{12}x-10-50x^2+3x^3+2x-50$

$\to C(x)+5D(x)=3x^3+\dfrac{23}{2}x^2-50x^2-\dfrac{1}{12}x+2x-10-50$

$\to C(x)+5D(x)=3x^3-\dfrac{77x^2}{2}+\dfrac{23x}{12}-60$

Để $C(x)+5D(x)=0\to 3x^3-\dfrac{77x^2}{2}+\dfrac{23x}{12}-60=0$

$\to x\approx \:12.90393\dots$