Vì dãy trên có 2012 số 

Nên chữ số tân jcungf là : 2²⁰¹² = 2^4.503 = (....6)

Vậy chữ số tận cùng là 6 

tận cùng 6 nhe bé 

12¹² = 12^4.503 = (....6)

Tận cùng là 6

1. \(3-|2x+1|=-5\)

\(\Rightarrow|2x+1|=8\)

\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)

2.\(12+|3-x|=9\)

\(\Rightarrow|3-x|=-3\)

Mà \(|3-x|\ge0\forall x\)

\(\Rightarrow\)Vô lí

Vậy không có x

3.\(|x+9|=12+\left(-9\right)+2\)

\(\Rightarrow|x+9|=5\)

\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Vậy \(x\in\left\{-4;-14\right\}\)

4.\(5x-16=40+x\)

\(\Rightarrow5x-x=40+16\)

\(\Rightarrow4x=56\)

\(\Rightarrow x=14\)

Vậy \(x=14\)

5.\(5x-7=-21-2x\)

\(\Rightarrow5x+2x=-21+7\)

\(\Rightarrow7x=-14\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

6.\(\left(2x-1\right)\left(y-2\right)=12\)

Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)

\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Ta có bảng : (em tự xét bảng nhé)

<=> 7.( 12 - x ) = 9. ( 15 - x )

<=> 84 - 7x = 135 - 9x 

<=> 84 - 7x - ( 135 - 9x ) =0

<=> 84 - 7x - 135 + 9x = 0

<=> 2x - 51 = 0 

<=> 2x = 51 <=> x = 51/2 = 25,5

7/15 - x = 9/12 - x

So sánh 7/15 và 9/12

7/15 < 9/12 => ta có :

9/12 - 7/15 = 17/60 

=> Khi 7/15 trừ đi số nào thì 17/60 cộng thêm số đó

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

<=> 1/2 * x - 2/3 * x = 5/12

<=> x *(1/2 - 2/3) = 5/12

<=> x * (-1/6) = 5/12

<=> x = 5/12 : (-1/6)

<=> x = -5/2 (tm)

Câu 2: 

\(\dfrac{7}{9}\cdot\dfrac{3}{35}=\dfrac{1}{5}\cdot\dfrac{1}{3}=\dfrac{1}{15}\)

\(\dfrac{9}{22}\cdot55=\dfrac{9\cdot55}{22}=\dfrac{9\cdot5}{2}=\dfrac{45}{2}\)

A\(\frac{7}{8}+\frac{7}{12}+\frac{9}{8}+\frac{5}{12}\)

\(=\left(\frac{7}{12}+\frac{5}{12}\right)+\left(\frac{7}{8}+\frac{9}{8}\right)\)

\(=1+2\)

\(=3\)

\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)

\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)

\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)

\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)

\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)

\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\)         ( 4^3 = 8^2 ; 5^4 = 25^2 )

\(=\frac{1}{25\times7}=\frac{1}{175}\)

\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)

\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)

\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)

\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)

a) \(\frac{-9}{12}=\frac{-9\div3}{12\div3}=\frac{-3}{4}\)

\(\frac{-18}{-24}=\frac{-18\div\left(-6\right)}{-24\div\left(-6\right)}=\frac{3}{4}\)

\(\frac{-35}{70}=\frac{-35\div35}{70\div35}=\frac{-1}{2}\)

\(\frac{-9}{27}=\frac{-9\div9}{27\div9}=\frac{-1}{3}\)

b) \(\frac{1313}{4242}=\frac{1313\div101}{4242\div101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{-353535\div\left(-70707\right)}{-424242\div\left(-70707\right)}=\frac{5}{6}\)

c)\(\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{\left(2^3\right)^2\times\left(5^2\right)^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^6\times5^6\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^3\times2^3\times5^2\times5^4\times7}\)= \(\frac{2^3\times2^3}{2^3\times5^2\times7}\)= \(\frac{8}{25\times7}\)= \(\frac{8}{175}\)