Tìm x biết (32+x) x 7=329
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a) \(2^4.x-3^2.x=145-\frac{255}{51}\Rightarrow x.\left(2^4-3^2\right)=145-5\)
\(\Rightarrow7.x=140\Rightarrow x=20\)
b) \(\frac{580}{x-24}=329-150.2\Rightarrow\frac{580}{x-24}=329-300\Rightarrow\frac{580}{x-24}=29\Rightarrow x-24=20\)
\(\Rightarrow x=44\)
c) \(7.\left(x-14\right)+35=25+\frac{279}{9}\Rightarrow7.\left(x-14\right)=25+31-35\)
\(\Rightarrow7.\left(x-14\right)=21\Rightarrow x-14=3\Rightarrow x=17\)
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( 32 - x ) + ( -7 + |x| ) - ( |-x| + 25 ) = -45
( 32 - x ) + ( -7 + |x| ) = -45 + ( |-x| + 25 )
( 32 - x ) + ( -7 + |x| ) = -45 + 25 + |-x|
( 32 - x ) + ( -7 + |x| ) = -20 + |-x|
( 32 - x ) = -20 + |-x| - ( -7 ) + |x|
( 32 - x ) = -20 + 7 + |-x| + |x|
( 32 - x ) = -13 + x + x
( 32 - x ) = -13 + 2x
32 - x = -13 + 2x
-2x + ( -x ) = -13 -32
-3x = -45
3x = 45
x = 45 : 3 = 15
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a) (x+1997) + 2003 = 5432
x+ 1997 = 5432 - 2003
x + 1997 = 3429
x = 3429 - 1997
x= 1432
b) 329 - (129 +x) = 14*7
329 - (129 + x) = 98
129 + x= 329 -98
129 + x = 231
x = 231 - 129
x = 102
a. (x+1997) + 2003= 5432
=> x = 5432 -( 2003 + 1997)
=> x = 5432 - 4000
=> x = 1432
b. 329 - (129 + x ) = 14*7
=> 129+ x = 329 - 98
=> 129 + x = 231
=> x = 231 - 129
=> x = 102
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Bài làm:
Ta có: \(290\div\left(x-12\right)=329-75.4\)
\(\Leftrightarrow290\div\left(x-12\right)=329-300\)
\(\Leftrightarrow290\div\left(x-12\right)=29\)
\(\Leftrightarrow x-12=10\)
\(\Rightarrow x=22\)
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− 3 29 − 2 29 ≤ x 29 < 5 29 − 7 29 . ⇒ − 5 29 ≤ x 29 < − 2 29 ⇒ − 5 ≤ x < − 2 ⇒ x ∈ − 5 ; − 4 ; − 3
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Lời giải:
$(x+2)+(x+4)+(x+6)+...+(x+32)=352$
$(x+x+...+x)+(2+4+6+...+32)=352$
$16x+272=352$
$16x=352-272=80$
$x=80:16=5$
{32+x}x7=329
32+x=329:7
32+x=47
x=47-32
x=15