Ta có: \(\left(x-7\right)\left(x^2-9x+20\right)\left(x-2\right)=72\)

\(\Leftrightarrow\left(x^2-9x+20\right)\left(x^2-9x+14\right)=72\)

Đặt \(x^2-9x+17=a\) khi đó:

\(PT\Leftrightarrow\left(a+3\right)\left(a-3\right)=72\)

\(\Leftrightarrow a^2-9-72=0\)

\(\Leftrightarrow a^2=81\Rightarrow\orbr{\begin{cases}a=9\\a=-9\end{cases}}\)

Nếu a = 9 khi đó \(x^2-9x+17=9\)

\(\Leftrightarrow x^2-9x+8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Nếu a = -9 khi đó \(x^2-9x+17=-9\)

\(\Leftrightarrow x^2-9x+26=0\)

\(\Leftrightarrow\left(x^2-9x+\frac{81}{4}\right)+\frac{23}{4}=0\)

\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2=-\frac{23}{4}\left(ktm\right)\)

Vậy \(S=\left\{1;8\right\}\)

( x - 7 )( x² - 9x + 20 )( x - 2 ) = 72

⇔ [ ( x - 7 )( x - 2 ) ]( x² - 9x + 20 ) - 72 = 0

⇔ ( x² - 9x + 14 )( x² - 9x + 20 ) - 72 = 0

Đặt t = x² - 9x + 17

⇔ ( t - 3 )( t + 3 ) - 72

⇔ t² - 9 - 72 = 0

⇔ t² - 81 = 0

⇔ ( t - 9 )( t + 9 ) = 0

⇔ ( x² - 9x + 17 - 9 )( x² - 9x + 17 + 9 ) = 0

⇔ ( x² - 9x + 8 )( x² - 9x + 26 ) = 0

⇔ ( x² - 8x - x + 8 )( x² - 9x + 26 ) = 0

⇔ [ x( x - 8 ) - ( x - 8 ) ]( x² - 9x + 26 ) = 0

⇔ ( x - 8 )( x - 1 )( x² - 9x + 26 ) = 0

⇔ x - 8 = 0 hoặc x - 1 = 0 hoặc x² - 9x + 26 = 0

⇔ x = 8 hoặc x = 1 [ x² - 9x + 26 = ( x² - 9x + 81/4 ) + 23/4 = ( x - 9/2 )² + 23/4 ≥ 23/4 > 0 ∀ x ]

\(\left(x-7\right)\left(x^2-9x+20\right)\left(x-2\right)=72\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)\left(x^2-9x+20\right)-72=0\)

\(\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72=0\)(1)

Đặt \(x^2-9x+17=t\)

Thay \(t=x^2-9x+17\)vào (1) ta được:

\(\left(t-3\right)\left(t+3\right)-72=0\)

\(\Leftrightarrow t^2-9-72=0\)\(\Leftrightarrow t^2-81=0\)

\(\Leftrightarrow\left(t-9\right)\left(t+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-9=0\\t+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=9\\t=-9\end{cases}}\)

TH1: Nếu \(t=9\)\(\Rightarrow x^2-9x+17=9\)

\(\Leftrightarrow x^2-9x+8=0\)\(\Leftrightarrow\left(x^2-x\right)-\left(8x-8\right)=0\)

\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

TH2: Nếu \(t=-9\)\(\Rightarrow x^2-9x+17=-9\)

\(\Leftrightarrow x^2-9x+26=0\)

\(\Leftrightarrow x^2-2.\frac{9}{2}.x+\frac{81}{4}+\frac{23}{4}=0\)

\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2+\frac{23}{4}=0\)

Vì \(\left(x-\frac{9}{2}\right)\ge0\forall x\)\(\Rightarrow\left(x-\frac{9}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\)

\(\Rightarrow\)Với \(t=-9\)thì phương trình vô nghiệm 

Vậy \(x=1\)hoặc \(x=8\)

\(\Leftrightarrow\left[3x+6-23\right]\cdot\left(9-8+20\cdot0\right)=147\)

=>3x=164

hay x=164/3

`- 9 . ( x - 6 ) + 20 = 56`

`=> -9 . ( x - 6 ) = 56-20`

`=> -9 . ( x - 6 ) =36`

`=>x-6=36:(-9)`

`=>x-6=-4`

`=>x=-4+6`

`=>x=2`

`-------------`

` ( 245 - x ) + 7^2=149`

`=>( 245 - x ) +49=149`

`=> 245 - x =149-49`

`=> 245 - x =100`

`=>x=245-100`

`=>x=145`

`------------`

`(2^x-3).7=35`

`=> 2^x-3=35:7`

`=> 2^x-3=5`

`=>2^x=5+3`

`=>2^x=8`

`=>2^x=2^3`

`=>x=3`

<=> (x-18/74 - 1)+(x-20/72 - 1)+(x-22/70 - 1) = 0

<=> x-92/74 + x-92/72 + x-92/70 = 0

<=> (x-92).(1/74+1/72+1/70) = 0

<=> x-92 = 0 ( vì 1/74 + 1/72 + 1/70 > 0 )

<=> x=92

Vậy S = {92}

Tk mk nha

Ta có :

\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=3-3\)

\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)

\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)

Vì \(\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)\ne0\)

\(\Rightarrow\)\(x-92=0\)

\(\Rightarrow\)\(x=92\)

Vậy \(x=92\)

Chúc bạn học tốt 

Ta có : 

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\Leftrightarrow\)\(\left(x-2\right).\frac{2}{9}=\frac{16}{9}\)

\(\Leftrightarrow\)\(x-2=\frac{16}{9}:\frac{2}{9}\)

\(\Leftrightarrow\)\(x-2=\frac{16}{9}.\frac{9}{2}\)

\(\Leftrightarrow\)\(x-2=\frac{8}{1}.\frac{1}{1}\)

\(\Leftrightarrow\)\(x-2=8\)

\(\Leftrightarrow\)\(x=8+2\)

\(\Leftrightarrow\)\(x=10\)

Vậy \(x=10\)

Chúc bạn học tốt ~ 

(x+x+x+x+x+x)-(2/12+2/20+2/30+2/42+2/56+2/72)=16/9
6x-4/9=16/9
6x=20/9=>x=20/7

Trả lời

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)

\(x-2\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{21}{56}+\frac{1}{72}\right)=\frac{16}{9}\)

\(x-2\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)

\(\)\(\Rightarrow2\left(x-2\right)=16\)

\(\Rightarrow x-2=16:2\)

\(\Rightarrow x-2=8\)

\(\Rightarrow x=8+2\)

\(\Rightarrow x=10\)

Vậy x=10