[-3/2]^3+-49/24.3/14-[5/8-27/8]
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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
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x-8:4-(46-23.2+6.3)=0
\(x-8:4-\left(46-23.2+6.3\right)=0\)
\(x-2-\left(46-46+18\right)=0\)
\(x-2-18=0\)
\(x-2=0+18\)
\(x-2=18\)
\(x=18+2\)
\(x=20\)
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
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. 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)
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: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)
Ta có: \(\left(-\dfrac{3}{2}\right)^3+\dfrac{-49}{24}\cdot\dfrac{3}{14}-\left(\dfrac{5}{8}-\dfrac{27}{8}\right)\)
\(=\dfrac{-27}{8}+\dfrac{-7}{2}\cdot\dfrac{1}{8}-\dfrac{-22}{8}\)
\(=\dfrac{-27+22}{8}+\dfrac{-7}{16}\)
\(=\dfrac{-5}{8}+\dfrac{-7}{16}\)
\(=\dfrac{-10}{16}+\dfrac{-7}{16}\)
\(=\dfrac{-17}{16}\)