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1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
\(2,\)
\(a,20-\left[4^2+\left(x-6\right)\right]=90\)
\(\Rightarrow20-16-x+6=90\)
\(\Rightarrow10-x=90\)
\(\Rightarrow x=-80\)
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: \(x=4\)
\(2,\)
\(a,20-\left[42+\left(x-6\right)\right]=90\)
\(\Rightarrow20-42-x+6-90=0\)
\(\Rightarrow x=-106\)
Vậy: \(x=-106\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2x-6\right)=74\)
\(\Rightarrow1000:\left(24+2x\right)=74\)
\(\Rightarrow24+2x=\dfrac{500}{37}\)
\(\Rightarrow2x=-\dfrac{388}{37}\)
\(\Rightarrow x=-\dfrac{194}{37}\)
Mà \(x\in N\)
\(\Rightarrow x\in\varnothing\)
Vậy: \(x\in\varnothing\)
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
\(42-\left(2x+32\right)+12:2=6\)
\(\Rightarrow42-\left(2x+32\right)+12=12\)
\(\Rightarrow42-\left(2x+32\right)=0\)
\(\Rightarrow2x+32=42\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
42-(2x+32)+12:2=6
(2x+32)+6=42-6
(2x-32)+6=36
(2x-32)=36-6
2x-32=30
2x=30+32
2x=62
x=62:2
x=31
Tk mk nha
\(42-\left(2x+32\right)+12:2=6\)
\(\Rightarrow42-\left(2x+32\right)+12=12\)
\(\Rightarrow42-\left(2x+32\right)=0\)
\(\Rightarrow2x+32=42\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
42-(2x+32)+12:2+6
(2x+32)+6=42-6=36
2x+32=36-6=30
2x=30+32=62
x=62:2=31
tk mk nha
42 - (2x + 32) + 12 = 6.2
42 - (2x + 32) + 12 = 12
42 - (2x + 32 ) = 12 - 12
42 - (2x + 32) = 0
2x + 32 = 42 - 0
2x + 32 = 42
2x = 42 - 32
2x = 10
=> x = 10 : 2
=> x = 5
\(42-\left(2x+32\right)=12:2=6\)
\(42-\left(2x+32\right)+6=6\)
\(42-\left(2x+32\right)=6-6=0\)
\(2x+32=42-0=42\)
\(2x=42-32=10\)
\(x=10:2=5\)
42-(2x+32)+12:2=6
42-(2x+32)+6 =6
42-(2x+32) =6-6
42-(2x+32) =0
2x+32 =42-0
2x+32 =42
2x =42-32
2x =10
x =10:2
x =5
nhớ kick cho mình nha
2 . (32 - 2x + 1) = 42
32 - 2x + 1 = 42 : 2
32 - 2x + 1 = 21
2x + 1 = 32 - 21
2x + 1 = 11
2x = 11 -1
2x = 10
x = 10 : 2
x = 5
Vậy...