Tìm X, biết :
[ ( 100 - x ) . 2 - 2 ] = 40
Lưu ý : Dấu " ." là dấu nhân
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Lời giải:
a.
$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$
b.
$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$
$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$
c.
$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$
d.
$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$
$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$
e.
$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$
$\frac{-19}{15}: x=1$
$x=\frac{-19}{15}:1 =\frac{-19}{15}$
f.
$(-\frac{3}{4}+x).2\frac{2}{3}=1$
$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$
$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$
Ta có \(\dfrac{2}{1\cdot5}+\dfrac{2}{5\cdot9}+\dfrac{2}{9\cdot13}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
\(\dfrac{1}{2}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{56}{113}\)
\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{56}{113}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+4}=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=1-\dfrac{112}{113}=\dfrac{1}{113}\)
x + 4 = 113 ⇒ x = 109
\(\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}=\dfrac{56}{113}\)
Xét: \(A=\dfrac{2}{1.5}+\dfrac{2}{5.9}+...+\dfrac{2}{x\left(x+4\right)}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x-4}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+4}\right)\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)\)
Với \(A=\dfrac{56}{113}\) thì
\(\dfrac{1}{2}.\left(1-\dfrac{1}{x+4}\right)=\dfrac{56}{113}\)
\(\left(1-\dfrac{1}{x+4}\right)=\dfrac{112}{113}\)
\(\dfrac{1}{x+4}=\dfrac{1}{113}\)
\(x=109\)
\(b,\left(2x+1\right).\left(39-2\right)=-55\)
\(\Rightarrow\left(2x+1\right).37=-55\)
\(\Rightarrow3x+1=-\frac{55}{37}\)
\(\Rightarrow3x=-\frac{92}{37}\)
\(\Rightarrow x=-\frac{92}{111}\)
\(c,\left(x-7\right)\left(x+3\right)< 0\)
\(\Rightarrow\orbr{\begin{cases}x-7>0;x+3< 0\\x-7< 0;x+3>0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x>7;x< -3\\x< 7;x>-3\end{cases}}\)
\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x}.2+\frac{2}{5}=0,5\)
\(\Rightarrow\frac{2}{5x}+\frac{2}{x}+\frac{2}{5}=\frac{1}{2}\)
\(\Rightarrow2.\left(\frac{1}{5x}+\frac{1}{x}+\frac{1}{5}\right)=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5x}+\frac{5}{5x}+\frac{x}{5x}=\frac{1}{2}:2=\frac{1}{4}\)
\(\Rightarrow\frac{1+5+x}{5x}=\frac{1}{4}\)
\(\Rightarrow4.\left(1+5+x\right)=5x\)
\(\Rightarrow4+20+4x=5x\)
\(\Rightarrow24+4x=5x\)
\(\Rightarrow5x-4x=24\)
\(\Rightarrow x=24\)
a) 5.(x-20) = 35
(x-20) = 35:5
x-20 = 7
x = 27
b) (x+125) -301 = 56
x+125 -301 = 56
x - 176 = 56
x = 56 +176
x= 232
c) 215 + (x-21):2 = 235
(x-21):2 = 235 - 215
(x-21):2 = 20
x-21 = 20 .2
x-21 = 40
x = 61
d) (x:23 +45) .67 = 8911
(x:23 +45) = 8911 : 67
x:23+45 = 133
x:23 = 133-45
x:23 = 88
x = 88.23
x = 2024
a) 5.(x - 20) = 35
⇒ x - 20 = 35 : 5
⇒ x - 20 = 7
⇒ x = 7 + 20
⇒ x = 27
b) (x + 125) - 301 = 56
⇒ x + 125 = 56 + 301
⇒ x + 125 = 357
⇒ x = 357 - 125
⇒ x = 232
c) 215 + (x - 21) : 2 = 235
⇒ (x - 21) : 2 = 235 - 215
⇒ (x - 21) : 2 = 20
⇒ x - 21 = 20 x 2
⇒ x - 21 = 40
⇒ x = 21 + 40
⇒ x = 61
d) (x : 23 + 45).67 = 8911
⇒ x : 23 + 45 = 8911 : 67
⇒ x : 23 + 45 = 133
⇒ x : 23 = 133 - 45
⇒ x : 23 = 88
⇒ x = 88 x 23
⇒ x = 2024
a) \(5+3^{x+1}=86\)
\(=>3^{x+1}=86-5\)
\(=>3^{x+1}=81=3^4\)
\(=>x+1=4\) ( cùng cơ số )
\(=>x=4-1\)
\(=>x=3\)
b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(=>15:\left(x+2\right)=\left(27+3\right):10\)
\(=>15:\left(x+2\right)=30:10=3\)
\(=>x+2=15:3\)
\(=>x+2=5\)
\(=>x=5-2\)
\(=>x=3\)
c) \(\left(9x+2\right).4=80\)
\(=>9x+2=80:4\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:9\)
\(=>x=2\)
d) \(\left(245-x\right)+7^2=14\)
\(=>\left(245-x\right)+14=14\)
\(=>245-x=14-14\)
\(=>245-x=0\)
\(=>x=245-0\)
\(=>x=245\)
a, 7x . (x-10 ) = 0
=> \(\hept{\begin{cases}7x=0\\x-10=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=10\end{cases}}\)
b,17 . (3x-6).(2x-8) = 0
(3x-6).(2x-8) = 0 : 17
(3x-6).(2x-8) = 0
=>\(\hept{\begin{cases}3x-6=0\\2x-8=0\end{cases}}\)=>\(\hept{\begin{cases}x=2\\x=4\end{cases}}\)
Nhớ nha
`@` `\text {Ans}`
`\downarrow`
`[ (100 - x)*2 - 2] = 40`
`=> (100-x)*2 - 2 = 40`
`=> (100 - x)*2 = 40 + 2`
`=> (100 - x)*2 = 42`
`=> 100 - x = 42 \div 2`
`=> 100 - x = 21`
`=> x = 100 - 21`
`=> x = 79`
Vậy, `x=79.`
\(\left[\left(100-x\right).2-2\right]=40\)
\(=>\left(100-x\right).2=40+2\)
\(=>\left(100-x\right).2=42\)
\(=>100-x=42:2\)
\(=>100-x=21\)
\(=>x=100-21\)
\(=>x=79\)
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