42-2x(32-2^x+1)=10
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\(M=343-8x^3-64+8x^3=279\\ N=8x^3-1-1+8x^3=16x^3=16\cdot1000=16000\)
`|2x+1|-3=x+4`
`<=>|2x+1|=x+4+3=x+7(x>=-7)`
`**2x+1=x+7`
`<=>x=7-1=6(tm)`
`**2x+1=-x-7`
`<=>3x=-6`
`<=>x=-2(tm)`
`|3x-5|=1-3x(x<=1/3)`
`**3x-5=1-3x`
`<=>6x=6`
`<=>x=1(l)`
`**3x-5=3x-1`
`<=>-5=-1` vô lý
`|2x+2|+|x-1|=10`
Nếu `x>=1`
`pt<=>2x+2+x-1=10`
`<=>3x+1=10`
`<=>3x=9`
`<=>x=3(tm)`
Nếu `x<=-1`
`pt<=>-2x-2+1-x=10`
`<=>-1-3x=10`
`<=>-11=3x`
`<=>x=-11/3(tm)`
Nếu `-1<=x<=1`
`pt<=>2x+2+1-x=10`
`<=>x+3=10`
`<=>x=7(l)`
Vậy `S={3,-11/3}`
\(3^{200}=9^{100}>4^{100}\\ 5^{200}=25^{100}< 64^{100}=4^{300}\\ 6^{50}=36^{25}>7^{25}\\ 8^{40}=64^{20}>10^{20}\\ 16^{20}=256^{10}>32^{10}\)
tick mik nha!!
3200=9100>41005200=25100<64100=4300650=3625>725840=6420>10201620=25610>3210
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: \(x\in\varnothing\)
Ta có:
97 x 32 ⋮ 8
8 ⋮ 8
⇒ 97 x 32 + 8 ⋮ 8
Ta có:
2020 x 30 ⋮ 10
8 x 5 = 40 ⋮ 10
⇒ 2020 x 30 + 8 x 5 ⋮ 10