Lí do: 2(x - 1) - x = 4

Áp dụng tính chất sau:

a.b + c.b = b (a + c)

Áp dụng tính chất trên:

2(x - 1) = 2.x - 1.2 = 2x - 2 

Như vậy: 2(x-1) - x = 2x - 2 - x = 4

2(x - 1) - x = 4

=> 2x - 2 - x = 4

=> 2x - x = 4 + 2

=> x = 6

(x-4)(2x+6)=0

=>\(\left[{}\begin{matrix}x-4=0\\2x+6=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

trình bày rõ đc không ạ?

1) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(\Rightarrow C=30+2^4.30...+2^{96}.30\)

\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)

mà \(30=5.6\)

\(\Rightarrow C⋮5\left(dpcm\right)\)

1,

Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)

=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)

=> \(3^x\).\(13\) = \(351\)

=> \(3^x\) = \(27\)

=> \(x\) = \(3\)

2,

C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)

2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)

2C - C = \(2^{101}\) - \(2\)

C = \(2^{101}\) - \(2\)

C = \(2\).\(\left(2^{100}-1\right)\)

C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)

Có \(2^5\) \(-1\) \(⋮\) 5

=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5

=> C \(⋮\) 5

3,

Xét \(\overline{abcdeg}\)

= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)

= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)

Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)

=> \(\overline{abcdeg}⋮9\)

4,

S = \(3^0+3^2+3^4+...+3^{2002}\)

9S = \(3^2+3^4+3^6+...+3^{2004}\)

9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))

8S = \(3^{2004}-1\)

=> 8S \(< 3^{2004}\)

2^x+84=116

2^x=116-84

2^x=32

2^x=2^5

=>x=5

Ta có:

 \(\frac{x^2}{x^2+x}=\frac{x.x}{x\left(x+1\right)}=\frac{x}{x+1}\)

Ta thấy: \(\frac{x^2}{x^2+x}=\frac{x\cdot x}{x\cdot\left(x+1\right)}=\frac{1\cdot x}{1\cdot\left(x+1\right)}=\frac{x}{x+1}\) nên \(\frac{x^2}{x^2+x}=\frac{x}{x+1}\) ( đpcm )

Do x là số tự nhiên => 2x + 13 > x + 2

=> 3^a > 3^b

\(\Rightarrow3^a⋮3^b\Leftrightarrow\left(2x+13\right)⋮\left(x+2\right)\)

\(\RightarrowĐPCM\)

\(x^2+x+35=x\left(x+1\right)+35\)

mà \(\left\{{}\begin{matrix}x\left(x+1\right)⋮2\\35⋮̸2\end{matrix}\right.\)

\(\Rightarrow x\left(x+1\right)+35⋮̸2\)

\(\Rightarrow dpcm\)

\(x^2+x+35=x\left(x+1\right)+35\)

mà \(x\left(x+1\right)\) là 2 số liên tiếp nên chia hết cho 2

      35 là số lẻ không chia hết cho 2

\(\Rightarrow x\left(x+1\right)+35\) không chia hết cho 2

\(\Rightarrow dpcm\)

Vì em xài máy tính cầm tay=))

-135 nha bạn