a) A = 110 - (-761) + 296 + 1454 - (-813 + 1077)

        = 110 + 761 + 296 + 1454 - 264

        = 871 + 1750 - 264

        = 2631 - 264

        = 2357

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

\(\Leftrightarrow\sqrt{x+3}=3\)

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

1) \(125^5:25^7\)

\(=\left(5^3\right)^5:\left(5^2\right)^7\)

\(=5^{15}:5^{14}\)

= 5

2) \(27^8:9^9\)

\(=\left(3^3\right)^8:\left(3^2\right)^9\)

\(=3^{24}:3^{18}\)

\(=3^6\)

3) \(36^5:6^8\)

\(=\left(6^2\right)^5:6^8\)

\(=6^{10}:6^8\)

\(=6^2\)

4) \(49^6:7^{10}\)

\(=\left(7^2\right)^6:7^{10}\)

\(=7^{12}:7^{10}=7^2\)

5) \(7^{20}:49^9\)

\(=7^{20}:\left(7^2\right)^9\)

\(=7^{20}:7^{18}=7^2\)

6) \(\frac{1}{2^{10}}:\frac{1}{8^3}\)

\(=\frac{1}{2^{10}}:\frac{1}{\left(2^3\right)^3}\)

\(=\frac{1}{2^{10}}:\frac{1}{2^9}=\frac{1}{2^{10}}.\frac{2^9}{1}=\frac{1}{2}\)

7) \(\left(-\frac{1}{2}\right)^{21}:\frac{1}{4^{10}}\)

\(=\frac{\left(-1\right)^{21}}{2^{21}}:\frac{1}{\left(2^2\right)^{10}}\)

\(=-\frac{1}{2^{21}}:\frac{1}{2^{20}}=-\frac{1}{2^{21}}.\frac{2^{20}}{1}\)

\(=-\frac{1}{2}\)

8) \(\frac{1}{16^5}:\left(-\frac{1}{2}\right)^{18}\)

\(=\frac{1}{\left(2^4\right)^5}:\frac{\left(-1\right)^{18}}{2^{18}}\)

\(=\frac{1}{2^{20}}:\frac{1}{2^{18}}\)

\(=\frac{1}{2^{20}}.\frac{2^{18}}{1}=\frac{1}{4}\)

9) \(\frac{1}{5^{30}}:\frac{1}{25^{14}}\)

\(=\frac{1}{5^{30}}:\frac{1}{\left(5^2\right)^{14}}\)

\(=\frac{1}{5^{30}}:\frac{1}{5^{28}}=\frac{1}{25}\)

240-[23+(13+24.3-x)]=132

240-[23+(13+168-x)]=132

240-[23+(181-x)]=132

|

x-8:4-(46-23.2+6.3)=0

\(x-8:4-\left(46-23.2+6.3\right)=0\)

\(x-2-\left(46-46+18\right)=0\)

\(x-2-18=0\)

\(x-2=0+18\)

\(x-2=18\)

\(x=18+2\)

\(x=20\)

\(240-\left[23+\left(13+24.3-x\right)\right]=132\)

\(240\left[23+\left(13+82-x\right)\right]=132\)

\(23+\left(95-x\right)=132:240\)

\(23+\left(95-x\right)=\frac{132}{240}=\frac{11}{30}\)

\(23+\left(95-x\right)=\frac{11}{30}\)

\(95-x=\frac{11}{30}-23\)

\(95-x=\frac{11}{30}-\frac{690}{30}\)

\(95-x=-\frac{679}{30}\)

\(x=95+\frac{679}{30}\)

\(x=\frac{2850}{30}+\frac{679}{30}\)

\(x=\frac{3529}{30}=\)tự rút gọn ( nếu có thể )

\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)

\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)

\(C=\frac{2}{3}\)

Bạn Cô nàng Thiên Bình làm đúng hết òi =.= 

a=7.[1/8+1/27-1/49]

   ------------------------

11.[1/8+1/27-1/49]

=7/11

cau b,c tuong tu nha h mk   

a)\(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)

\(A=\frac{7.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}{11.\left(\frac{1}{8}+\frac{1}{27}-\frac{1}{49}\right)}\).

\(A=\frac{7}{11}\)

b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)

\(B=\frac{\frac{8}{9}.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}{4.\left(1-\frac{1}{3}-\frac{1}{9}+\frac{1}{27}\right)}\)

\(B=\frac{8}{9}:4=\frac{2}{9}\)

c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)\(C=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{239}\right)}\)

C=\(\frac{2}{3}\)

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

a: =-2/49-5/3+2/49=-5/3

b: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)