2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)

\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)

Đặt \(x^2+5x+3=t\)

\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)

\(\Leftrightarrow t^2-9=280\)

\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)

\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)

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

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

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

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0

\(\Leftrightarrow\) x = 2 hoặc x = - 7

Vậy x = 2 hoặc x = -7.

3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)

\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)

\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)

\(\Leftrightarrow x^3+12x^2+46x+60=0\)

\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)

\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)

\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)

\(\Leftrightarrow x=-6\)

Vậy x = -6.

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

Lúc mới đọc đề tớ tưởng x+\(\frac{3}{2}\)chứ. Nhìn lại thì...

Câu B đây;vừa bị lag

B, \(\frac{x+1}{35}\)+\(\frac{x+3}{33}\)=\(\frac{x+5}{31}\)+\(\frac{x+7}{29}\)

⇔ \(\frac{x+1}{35}\)+1+\(\frac{x+3}{33}\)+1=\(\frac{x+5}{31}\)+1+\(\frac{x+7}{29}\)+1

⇔ \(\frac{x+36}{35}\)+\(\frac{x+36}{33}\)-\(\frac{x+36}{31}\)-\(\frac{x+36}{29}\)=0

⇔ (x+36)(\(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\))=0

Mà \(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\)<0

⇔ x+36=0

⇔ x=-36

Vậy tập nghiệm của phương trình đã cho là:S={-36}

câu C tương tự nhé

Ta có:\(\left|x+5\right|+\left|x-1\right|=\left|x+5\right|+\left|1-x\right|\ge\left|x+5+1-x\right|=6\)

\(\Rightarrow\left|x+5\right|+\left|x+2\right|+\left|x-1\right|=6\Leftrightarrow x=-2\)

* với x ≥ -5

-3x-6=6

⇔ -3x=12

⇔ x=-4 (tm)

*với -5 ≤ x < -2

-x+4=6

⇔ -x=2

⇔ x=-2 (ktm)

* với -2 ≤ x < 1

x+8=6

⇔ x=6-8

⇔ x= -2 (tm)

* với x< 1

3x+4 =6

⇔ 3x=2

⇔ x= \(\dfrac{2}{3}\) (tm)

vậy tập nghiệm của phương trình là S \(\left\{\dfrac{2}{3};-2;-4\right\}\)

\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)+9=0\)

\(\Leftrightarrow\left(x^2-3x+4\right)\left(x^2+3x-10\right)+9=0\)

\(\Leftrightarrow\left(x^2+3x-7+3\right)\left(x^2+3x-7-3\right)+9=0\)

\(x^2+3x-7=0\)

\(x^2+3x=7\)

\(\Rightarrow x^2+2x.\frac{3}{2}+\frac{9}{4}=7+\frac{9}{4}\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2=\frac{37}{4}\)

\(\Rightarrow x+\frac{3}{2}=\pm\sqrt{\frac{37}{4}}\)

\(\Rightarrow x=\frac{-3}{2}-\sqrt{\frac{37}{4}}\)

\(\Rightarrow x=\frac{-3}{2}+\sqrt{\frac{37}{4}}\)

Vậy \(S=\left\{\frac{-3}{2}-\sqrt{\frac{37}{4}};\frac{-3}{2}+\sqrt{\frac{37}{4}}\right\}\)

\(M=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)

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

\(=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi \(x\in\left\{0;-5\right\}\)

Giải PT \(\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}=9\)

\(\Leftrightarrow\frac{x-6}{2010}+\frac{x-603}{471}+\frac{x-1}{403}-9=0\)

\(\Leftrightarrow\left(\frac{x-6}{2010}-1\right)+\left(\frac{x-603}{471}-3\right)+\left(\frac{x-1}{403}-5\right)=0\)

\(\Leftrightarrow\frac{x-2016}{2010}+\frac{x-2016}{471}+\frac{x-2016}{403}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)=0\)

Mà \(\left(\frac{1}{2010}+\frac{1}{471}+\frac{1}{403}\right)\ne0\)

\(\Leftrightarrow x-2016=0\Leftrightarrow x=2016\)

Vậy x=2016

b) \(M=\left(x-1\right)\left(x+2\right).\left(x+3\right)\left(x+6\right)\)

\(M=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right).\left(x+3\right)\right]\)

\(M=\left(x^2+5x-6\right).\left(x^2+5x+6\right)=\left(x^2+5x\right)^2-36\)

Các bạn tự làm tiếp được rồi nhé

           x(x + 2)(x² + 2x + 5) = 6

\(\Leftrightarrow\)(x² + 2x)(x² + 2x + 5) = 6

Đặt    x² + 2x = y; ta có 

           y(y + 5) = 6

\(\Leftrightarrow\)y² + 5y - 6 = 0

\(\Leftrightarrow\)y² - y + 6y - 6 = 0

\(\Leftrightarrow\)y(y - 1)+ 6(y - 1) = 0

\(\Leftrightarrow\)(y - 1)(y + 6) = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}y-1=0\\y+6=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=1\\y=-6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+2x=1\\x^2+2x=-6\end{cases}}\)

Mk lm đc thế thôi, bn lm tiếp nha

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

câu b lm tương tự nha