Giups mk vs
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Bài 6:
a) Ta có: \(4x-10=0\)
\(\Leftrightarrow4x=10\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)
b) Ta có: \(7-3x=9-x\)
\(\Leftrightarrow-3x+7-9+x=0\)
\(\Leftrightarrow-2x-2=0\)
\(\Leftrightarrow-2x=2\)
\(\Leftrightarrow x=-1\)
Vậy: S={-1}
c) Ta có: \(2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x=4x+12\)
\(\Leftrightarrow7x-3-4x-12=0\)
\(\Leftrightarrow3x-15=0\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy: S={5}
d) Ta có: \(5-\left(6-x\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow x+11-12+8x=0\)
\(\Leftrightarrow9x-1=0\)
\(\Leftrightarrow9x=1\)
\(\Leftrightarrow x=\dfrac{1}{9}\)
Vậy: \(S=\left\{\dfrac{1}{9}\right\}\)
e) Ta có: \(4\left(x+3\right)=-7x+17\)
\(\Leftrightarrow4x+12+7x-17=0\)
\(\Leftrightarrow11x-5=0\)
\(\Leftrightarrow11x=5\)
\(\Leftrightarrow x=\dfrac{5}{11}\)
Vậy: \(S=\left\{\dfrac{5}{11}\right\}\)

1. A
2.A
3.D
4.D
5.C
6.B
7.C
8.A
9.B
10.A
11. Oxit bazo : Fe2O3, CuO , CaO
Những chất còn lại là oxit axit
12.
CTHH sai :
NaO => Na2O , Ca2O => CaO


b: \(B=5+5^2+5^3+5^4+...+5^{99}+5^{100}\)
\(=5\left(1+5\right)+...+5^{99}\left(1+5\right)\)
\(=6\cdot\left(5+...+5^{99}\right)⋮6\)
a) \(A=1+3+3^2+...+3^{99}\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(=40+3^4.40+...+3^{96}.40=40\left(1+3^4+...+3^{96}\right)⋮40\)

nguyen ngoc anh:
\(\frac{3x}{27}=27\)
\(\Rightarrow3x:27=27\)
\(\Rightarrow3x=27\times27\)
\(\Rightarrow3x=729\)
\(x=729:3\)
\(x=243\)
Vậy\(x=27\)
Tham khảo nha!
\(\frac{3x}{27}=27\)<=> 3x=729<=> x=243