32 . x - 42 = - 10 x + 42
42 - 2 . (32 - 2x+1) = 10
\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2\left(32-2^{x+1}\right)=42-10=32\)
\(\Leftrightarrow32-2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=32-16=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
\(\Leftrightarrow x+1=4\)
\(\Rightarrow x=3\)
Tìm x:
42 - 2 . ( 32 - 2x +1) = 10
\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2^{x+1}=16\)
\(\Leftrightarrow2^{x+1}=2^4\)
=> x = 4
\(42-2.\left(32-2^{x+1}\right)=10\)
\(2.\left(32-2^{x+1}\right)=42-10\)
\(2.\left(32-2^{x+1}\right)=32\)
\(32-2^{x+1}=32:2\)
\(32-2^{x+1}=16\)
\(2^{x+1}=32-16\)
\(2^{x+1}=16\)
\(2^{x+1}=2^4\)
\(x+1=4\)
\(x=4-1\)
\(x=3\)
Làm lại :<
\(42-2\left(32-2^{x+1}\right)=10\)
\(\Leftrightarrow2^{x+1}=16\)
\(\Leftrightarrow x^{x+1}=2^4\)
\(\Leftrightarrow x=4-1\)
<=> x = 3
42-2x(32-2^x+1)=10
giúp mình
2x+3x=57:5
7.(42- x) = 53+134
X+18÷32=5×42
32+x=64,m+42+42=168 hay so sanh x va m
cam on cac ban nhieu minh se ket ban tung ban
x = 64 - 32 = 32
m = 168 - 42 = 126
Vì m = 126
x = 32
=> m > x
a) x là bội của 7 và 32< x< 50
b) x là ước của 42 và x>10
a,x-108×42=2378
b,1337+13×x=6745
c,x×10+16×x=972
d,25×x-x=59904:32
Tìm x: (3.x + 11) – 17 = 30
a,x-108×42=2378
b,1337+13×x=6745
c,x×10+16×x=972
d,25×x-x=59904:32
a, x - 108 × 42 = 2378
=> x - 4536 = 2378
=> x = 6914
vậy_
b,1337 + 13 × x = 6745
=> 10 × x = 5408
=> x = 540,8
vậy_
c, x × 10 + 16 × x = 972
=> x × (10 + 16) = 972
=> x× 26 = 972
=> x = \(\frac{486}{13}\)
vậy_
d, 25 × x - x = 59904 : 32
=> x × (25 - 1) = 1872
=> x × 24 = 1872
=> x = 78
vậy_
\(x-108.42=2378\)
\(x-4536=2378\)
\(x=6914\)
Bài 2: Tìm x nguyên
a) 20 – [42 + (x – 6)] = 90
b) (x + 3).(2x – 4) = 0
c) 1000:[30 + (2x – 6)] = 32 + 42 và x ∈ N
\(2,\)
\(a,20-\left[4^2+\left(x-6\right)\right]=90\)
\(\Rightarrow20-16-x+6=90\)
\(\Rightarrow10-x=90\)
\(\Rightarrow x=-80\)
Vậy: \(x=-80\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2^x-6\right)=25\)
\(\Rightarrow24+2^x=40\)
\(\Rightarrow2^x=16\)
\(\Rightarrow x=4\)
Vậy: \(x=4\)