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Do A = x183y chia cho 2 và 5 đều dư 1 nên y = 1. Ta có A = x183y
Vì A = x183y chia cho 9 dư 1
→ x183y - 1 chia hết cho 9
→ x183y chia hết cho 9
↔ x + 1 + 8 + 3 + 0 chia hết cho 9 ↔ x + 3 chia hết cho 9, mà x là chữ số nên x = 6
Vậy x = 6; y = 1
a)
= 48 + 288 : ( x - 3 )2 = 50
288 : ( x - 3 )2 = 50 - 48
288: ( x - 3 )2= 2
(x - 3 )2= 288 : 2
(x - 3)2= 144
(x - 3)2 = 122
x - 3 = 12
x = 12 + 3 = 15
\(8.6+288:\left(x-3\right)^2=50\)
\(48+288:\left(x-3\right)^2=50\)
\(288:\left(x-3\right)^2=50-48\)
\(288:\left(x-3\right)^2=2\)
\(\left(x-3\right)^2=288:2\)
\(\left(x-3\right)^2=144\)
\(\left(x-3\right)^2=12^2=\left(-12\right)^2\)
\(=>\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.=>\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Vậy x = 15 hoặc x = -9
\(f\)) \(32^{-x}.16^x=1024\)
\(\left(2\right)^{-5x}.2^{4x}=2^{10}\)
\(\Leftrightarrow2^{4x-5x}=2^{10}\)
\(\Leftrightarrow2^{-x}=2^{10}\)
\(\Leftrightarrow-x=10\)
\(\Leftrightarrow x=-10\)
\(g\)) \(3^{x-1}.5+3^{x-1}=162\)
\(3^{x-1}.\left(5+1\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=162:6\)
\(3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
\(h\)) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
\(\Leftrightarrow\left(2x-1\right)^6-\left(2x-1\right)^6.\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[1-\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)^6=0\\1-\left(2x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=1\\\left(2x-1\right)^2=\left(1,-1\right)^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x-1=-1\\2x-1=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\2x=0\\2x=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=0\\x=1\end{cases}}\)
\(i\)) \(5^x+5^{x+2}=650\)
\(5^x.\left(1+5^2\right)=650\)
\(5^x.26=650\)
\(5^x=650:26\)
\(5^x=25\)
\(\Leftrightarrow5^x=5^2\)
\(\Leftrightarrow x=2\)
a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)
\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)
\(10x-70-8x-40=-6\)
\(10x-8x=\left(-6\right)+70+40\)
\(2x=104\)
\(x=104\div2\)
\(x=52\)
b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)
\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)
\(8x-16-21-7x=6\)
\(8x-7x=6+16+21\)
\(x=43\)
Bài 5 :
Ta có : \(x+3⋮x+2\)
\(\Leftrightarrow x+2+1⋮x+2\)
\(\Leftrightarrow1⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1\right\}\)
Vậy ...
Bài 6 :
Ta có : \(2x+7⋮x+1\)
\(\Leftrightarrow2\left(x+1\right)+5⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;-6;4\right\}\)
Vậy ...
(x+x+x+x+x+...+x)+(1+3+5+...+99)=0
50x + 2500 = 0
50x=0- 2500
50x =-2500
x=-2500:50
x=-50
Vậy x=-50
288:(x-3)2=2
=> (x-3)2=144
=>x-3=12(vì x thuộc N)
=> x=15
\(8.6+288:\left(x-3\right)^2=50\)
\(\Rightarrow48+288:\left(x-3\right)^2=50\)
\(\Rightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Rightarrow\left(x-3\right)^2=288:2=144\)
Mà \(\left(x-3\right)^2=144=12^2\)
\(\Rightarrow x-3=12\)
\(\Rightarrow x=12+3\)
\(\Rightarrow x=15\)