Vô nghiệm rồi bạn

\(x^3+6x^2-13x-42=0\)

\(\Leftrightarrow\left(x^3-3x^2\right)+\left(9x^2-27x\right)+\left(14x-42\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)+\left(x^2+9x+14\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+7x+2x+14\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[x\left(x+7\right)+2\left(x+7\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\\x+7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\\x=-7\end{array}\right.\)

x³ + 6x² - 13x - 42 = 0

=> x³ - 3x² + 9x² - 27x + 14x - 42 = 0

=> x² ( x - 3 ) + 9x ( x - 3 ) + 14 ( x - 3 ) = 0

=> ( x - 3 ) ( x² + 9x + 14) = 0

=> ( x - 3 ) ( x² + 2x + 7x + 14 ) = 0

=> ( x - 3 ) [ x ( x + 2 ) + 7 ( x + 2 ) ] = 0

=> ( x - 3 ) ( x + 2 ) ( x + 7 ) = 0

=> x - 3 = 0 => x = 3

=> x + 2 = 0 => x = -2

=> x + 7 = 0 => x = -7

\(x^3+6x^2-13x-42\)

\(=x^2\left(x+7\right)-x\left(x+7\right)-6\left(x+7\right)\)

\(=\left(x^2-x-6\right)\left(x+7\right)\)

\(=\left(x-3\right)\left(x+2\right)\left(x+7\right)\)

x3+6x2−13x−42

x3+6x2−13x−42

=(x+7)(x−3)(x+2)

$ĐKXĐ:x \neq -4;-5;-6;-7$

$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$

$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$

$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$

$⇔x^2+11x+28=54$

$⇔x^2+11x-26=0$

$⇔x^2-2x+13x-26=0$

$⇔(x-2)(x+13)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)

Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$

Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

 x²+13x+42=0 

Two solutions were found :

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

 1.1     Factoring  x²+13x+42 

The first term is,  x²  its coefficient is  1 .
The middle term is,  +13x  its coefficient is  13 .
The last term, "the constant", is  +42 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 42 = 42 

Step-2 : Find two factors of  42  whose sum equals the coefficient of the middle term, which is   13 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  6  and  7 
                     x² + 6x + 7x + 42

Step-4 : Add up the first 2 terms, pulling out factors :
                    x • (x+6)
              Add up the last 2 terms, pulling out common factors :
                    7 • (x+6)
Step-5 : Add up the four terms of step 4 :
                    (x+7)  •  (x+6)
             Which is the desired factorization

Equation at the end of step  1  :

(x + 7) • (x + 6) = 0

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