tìm x biết:
72 : x = 20
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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 ]
\(\Leftrightarrow\left[3x+6-23\right]\cdot\left(9-8+20\cdot0\right)=147\)
=>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 ~
72:x=20
x=72:20
x=3,6
tick nha
72/x=20
x=72/20
x=3,6