Bạn tham khảo cách làm!

`(3x+2)*(x+4)-(3x-1)*(x-5)=0`

\(\Leftrightarrow3x\left(x+4\right)+2\left(x+4\right)-3x\left(x-5\right)+1\left(x-5\right)=0\)

\(\Leftrightarrow3x^2+3x\cdot4+2x+2\cdot4-3x^2+3x\cdot5+x-5=0\)

\(\Leftrightarrow3x^2+12x+2x+8-3x^2+15x+x-5=0\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(12x+2x+15x+x\right)+\left(8-5\right)=0\)

\(\Leftrightarrow30x+3=0\)

\(\Leftrightarrow30x=0-3\)

`=> 30x=-3`

`-> x=-3 \div 30`

`-> x=-1/10 `

cái này mk làm r!!!

đầu tiên bạn có |x-1|+|x-3|+|x-5|+|x-7|+|x-9| là các số lớn hơn bằng 0 nên -6x là số lớn hơn bằng 0

ta có -6x>=0 với mọi x

nên 6x<=0 với mọi x

nên x<=0

vì x<=0 suy ra 

|x-1|=-x+1=1-x

|x-3|=3-x

làm tương tự vs các số còn lại t

sau đó thay vào đề bài ta có 

1-x+3-x+5-x+7-x+9-x=-6x

nên(1+3+5+7+9)-5x=-6x

cậu tự làm nốt nhé!

và kết quả x=-25

\(\dfrac{x+1}{x-1}+\dfrac{x-2}{x+2}+\dfrac{x-3}{x+3}+\dfrac{x+4}{x-4}=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x-4\right)+\left(x-2\right)\left(x-1\right)\left(x+3\right)\left(x-4\right)+\left(x-3\right)\left(x-1\right)\left(x+2\right)\left(x-4\right)+\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow4x^4+20x-96=0\)

\(\Leftrightarrow4\left(x^4+5x-24\right)=0\)

\(\Leftrightarrow x^4+5x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2,45...\\x=1,94...\end{matrix}\right.\)

Vậy: \(S=\left\{-2,45...;1,94...\right\}\)

a)\(-6x^3+18^2+60x\)

\(=6x.-x^2+6x.3x+6x.10\)

=\(6x\left(-x^2+3x+10\right)\)

b)\(x^7+x^2+1\)

=\(\left(x^7-x\right)+\left(x^2+x+1\right)\)

=\(x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right).\left(x^3+1\right)+\left(x^2+x+1\right)\)

=\(x\left(x-1\right).\left(x^2+x+1\right).\left(x^3+1\right)+\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right).\left[x.\left(x-1\right).\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right).\left[\left(x^2-x\right).\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right).\left(x^5+x^4+x^2-x+1\right)\)

Bạn học tốt nhé!

c)\(x^{10}+x^5+1\)

\(=x^{10}+x^9-x^9+x^8-x^8+x^7-x^7+x^6-x^6+\)\(x^5+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+1\)

=\(\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)\)

\(-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)\)

\(+\left(x^2+x+1\right)\)

=\(x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)\)\(-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

\(+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right).\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)

Lộn xộn quá ha, chúc bạn học thật giỏi!

Gửi bạn

\(\left( {\dfrac{1}{7}x - \dfrac{2}{7}} \right)\left( {\dfrac{{ - 1}}{5}x + \dfrac{3}{5}} \right)\left( {\dfrac{1}{3}x + \dfrac{1}{3}} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x - \dfrac{2}{7} = 0\\ \dfrac{{ - 1}}{5}x + \dfrac{3}{5} = 0\\ \dfrac{1}{3}x + \dfrac{1}{3} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \dfrac{1}{7}x = \dfrac{2}{7}\\ - \dfrac{1}{5}x = - \dfrac{3}{5}\\ \dfrac{1}{3}x = - \dfrac{1}{3} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = 3\\ x = - 1 \end{array} \right. \)

C = - 1/2 + 3/10 + 5/50 + 7/170 +9/442 + 11/962

C = - 1/2 + 3/10 + 1/10 + 7/170 + 9/442 + 11/962

C = - 1/37

- Ta có : \(5\left(x+2\right)^3+7=2\)

=> \(5\left(x^3+6x^2+12x+8\right)+7=2\)

=> \(5x^3+30x^2+60x+40+7=2\)

=> \(5x^3+30x^2+60x+40+7-2=0\)

=> \(5x^3+30x^2+60x+45=0\)

=> \(5x^3+15x^2+15x^2+45x+15x+45=0\)

=> \(5x^2\left(x+3\right)+15x\left(x+3\right)+15\left(x+3\right)=0\)

=> \(\left(5x^2+15x+15\right)\left(x+3\right)=0\)

=> \(\left[{}\begin{matrix}x+3=0\\5x^2+15x+15=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\x^2+3x+3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\x^2+2x.\frac{3}{2}+\frac{9}{4}+\frac{3}{4}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2=-\frac{3}{4}\left(VL\right)\end{matrix}\right.\)

=> \(x=-3\)

Vậy phương trình có tập nghiệm là \(S=\left\{-3\right\}\)

5(x + 2)³ + 7 = 2

(x + 2)³ = -1

=> x + 2 = -1

<=> x = -3

\(5.\left(x+2\right)^3+7=2\)

\(\Rightarrow5.\left(x+2\right)^3=2-7\)

\(\Rightarrow5.\left(x+2\right)^3=-5\)

\(\Rightarrow\left(x+2\right)^3=\left(-5\right):5\)

\(\Rightarrow\left(x+2\right)^3=-1\)

\(\Rightarrow\left(x+2\right)^3=\left(-1\right)^3\)

\(\Rightarrow x+2=-1\)

\(\Rightarrow x=\left(-1\right)-2\)

\(\Rightarrow x=-3\)

Vậy \(x=-3.\)

Chúc bạn học tốt!