a.

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

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

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

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

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

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

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

*Với a = -5 ta được:

\(x^2+x-1=-5\)

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

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

b)(ĐKXĐ:x khác 2;3;4;5;6)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-6\right)\left(x-2\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow x^2-8x-20=0\)

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

\(\Leftrightarrow x=-2\) or x=10(đều thỏa)

Vậy ...

.

.

.

\(\dfrac{30}{x}-\dfrac{1}{2}=\dfrac{30}{x}+5+\dfrac{1}{2}\)

`<=> 60/(2x) - x/(2x) = 60/(2x) + (10x)/(2x) + x/(2x)`

`=> 60-x=60+10x +x`

`<=> 60-60=10x+x+x`

`<=>0 = 13x`

`<=>13x=0`

`<=>x=0`

Vậy phương trình có nghiệm `x=0`

\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)

Chọn A

mik nghĩ A

\(\dfrac{7}{16}+\dfrac{-1}{8}+\dfrac{9}{32}=\dfrac{14}{32}-\dfrac{4}{32}+\dfrac{9}{32}=\dfrac{19}{32}\)

\(\dfrac{5}{4}-\dfrac{25}{30}\times\dfrac{37}{44}+\dfrac{-25}{30}\times\dfrac{13}{44}+\dfrac{-25}{30}\times\dfrac{-6}{44}\)

\(=-\dfrac{25}{30}\times\left(\dfrac{5}{4}+\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)\)

\(=-\dfrac{25}{30}\times\left(\dfrac{55}{44}+\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)\)

\(=-\dfrac{25}{30}\times\dfrac{99}{44}\)

\(=-\dfrac{5}{6}\times\dfrac{9}{4}\)

\(=-\dfrac{15}{8}\)

\(\left(\dfrac{7}{16}+\dfrac{-1}{8}+\dfrac{9}{32}\right):\dfrac{5}{4}\)

\(=\left(\dfrac{14}{32}+\dfrac{-4}{32}+\dfrac{9}{32}\right):\dfrac{5}{4}\)

\(=\dfrac{19}{32}:\dfrac{5}{4}\)

\(=\dfrac{19}{32}.\dfrac{4}{5}=\dfrac{19.4}{32.5}=\dfrac{19}{40}\)

\(\dfrac{-25}{30}.\dfrac{37}{44}+\dfrac{-25}{30}.\dfrac{13}{44}+\dfrac{-25}{30}.\dfrac{-6}{44}\)

\(=\dfrac{-25}{30}.\left(\dfrac{37}{44}+\dfrac{13}{44}+\dfrac{-6}{44}\right)\)

\(=\dfrac{-25}{30}.\dfrac{44}{44}=\dfrac{-5}{6}.1=-\dfrac{5}{6}\)

a) x.31+(1+2+3+4+.....+30)=1240

x.31+465=1240

x.31=1240-465

x.31=775

x=775:31

x=25

b) (1+x).x:2=210

(1+x).x=210.2

(1+x).x=420

Vì 1+x và x là 2 số nhiên liên tiếp và 420=20.21

=> x=20

Ta có: \(\dfrac{30}{x}-\dfrac{30}{x+3}=\dfrac{1}{2}\)

\(\Leftrightarrow60\left(x+3\right)-60x=x\left(x+3\right)\)

\(\Leftrightarrow x^2+3x-180=0\)

\(\Leftrightarrow x^2+15x-12x-180=0\)

\(\Leftrightarrow\left(x+15\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-15\left(nhận\right)\\x=12\left(nhận\right)\end{matrix}\right.\)

Bài làm:

1) đk: \(x\ne0;x\ne-5\)

Ta có: \(\frac{30}{x}-\frac{30}{x+5}=1\)

\(\Leftrightarrow\frac{30\left(x+5\right)-30x}{x\left(x+5\right)}=1\)

\(\Leftrightarrow x^2+5x=150\)

\(\Leftrightarrow x^2+5x-150=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)

2) đk: \(x\ne0;x\ne-2\)

Ta có: \(\frac{60}{x}-\frac{60}{x+2}=1\)

\(\Leftrightarrow\frac{60\left(x+2\right)-60x}{x\left(x+2\right)}=1\)

\(\Leftrightarrow x^2+2x=120\)

\(\Leftrightarrow x^2+2x-120=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)

\(\frac{30}{x}-\frac{30}{x+5}=1\)( ĐKXĐ : \(x\ne0;x\ne-5\))

<=> \(30\left(\frac{1}{x}-\frac{1}{x+5}\right)=1\)

<=> \(30\left(\frac{x+5}{x\left(x+5\right)}-\frac{x}{x\left(x+5\right)}\right)=1\)

<=> \(30\left(\frac{5}{x\left(x+5\right)}\right)=1\)

<=> \(\frac{5}{x\left(x+5\right)}=\frac{1}{30}\)

<=> \(5\cdot30=x\left(x+5\right)\)

<=> \(x^2+5x-150=0\)

<=> \(x^2+15x-10x-150=0\)

<=> \(x\left(x+15\right)-10\left(x+15\right)=0\)

<=> \(\left(x-10\right)\left(x+15\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-15\end{cases}}\)( tmđk )

Vậy S = { 10 ; -15 }

\(\frac{60}{x}-\frac{60}{x+2}=1\)( ĐKXĐ : \(x\ne0;x\ne-2\))

<=> \(60\left(\frac{1}{x}-\frac{1}{x+2}\right)=1\)

<=> \(60\left(\frac{x+2}{x\left(x+2\right)}-\frac{x}{x\left(x+2\right)}\right)=1\)

<=> \(60\left(\frac{2}{x\left(x+2\right)}\right)=1\)

<=> \(\frac{2}{x\left(x+2\right)}=\frac{1}{60}\)

<=> \(2\cdot60=x\left(x+2\right)\)

<=> \(x^2+2x-120=0\)

<=> \(x^2+12x-10x-120=0\)

<=> \(x\left(x+12\right)-10\left(x+12\right)=0\)

<=> \(\left(x-10\right)\left(x+12\right)=0\)

<=> \(\orbr{\begin{cases}x-10=0\\x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-12\end{cases}}\)

Vậy S = { 10 ; -12 }

a, \(\frac{30}{x}-\frac{30}{x+5}=1\)                                                                           b, \(\frac{60}{x}-\frac{60}{x+2}=1\)

ĐKXĐ: x khác 0                                                                                                   ĐKXĐ:     x khác 0

            x khác -5                                                                                                                  x khác -2

Ta có:                                                                                                                 Ta có:

\(\frac{30}{x}-\frac{30}{x+5}=1\)                                                                                          \(\frac{60}{x}-\frac{60}{x+2}=1\)

<=>\(\frac{30\left(x+5\right)}{x\left(x+5\right)}-\frac{30x}{x\left(x+5\right)}=\frac{x\left(x+5\right)}{x\left(x+5\right)}\)                                <=>\(\frac{60\left(x+2\right)}{x\left(x+2\right)}-\frac{60x}{x\left(x+2\right)}=\frac{x\left(x+2\right)}{x\left(x+2\right)}\)

=>30(x+5)-30x=x(x+5)                                                                                     =>60(x+2)-60x=x(x+2)

<=>30x+150-30x=x²+5x                                                                                 <=>60x+120-60x=x²+2x

<=>150=x²+5x                                                                                                <=>120=x²+2x

<=>0=x²+5x-150                                                                                             <=>0=x²+2x-120

<=>0=x²-10x+15x-150                                                                                     <=>x²-10x+12x-120

<=>0=(x²-10x)+(15x-150)                                                                                 <=>(x²-10x)+(12x-120)

<=>0=x(x-10)+15(x-10)                                                                                     <=>x(x-10)+12(x-10)

<=>0=(x-10)(x+15)                                                                                            <=>(x-10)(x+12)

<=>x-10=0 hoặc x+15=0                                                                                    <=>x-10=0 hoặc x+12=0

1, x-10=0          2,x+15=0                                                                                      1, x-10=0                         2, x+12=0

<=>x=10             <=>x=-15                                                                                      <=>x=10                          <=>x=-12

( thỏa mãn ĐKXĐ)                                                                                                                   (thỏa mãn ĐKXĐ)

Vậy tâp nghiệm của PT là S={10;-15}                                                                            Vậy tập ngiệm của PT là S={10;-12}