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a)\(8\sqrt{x}-3\sqrt{\frac{4}{81}}=5,2\)
\(\Rightarrow8\sqrt{x}-3.\frac{2}{9}=5,2\)
\(\Rightarrow8\sqrt{x}-\frac{2}{3}=5,2\)
\(\Rightarrow8\sqrt{x}=5,2+\frac{2}{3}\)
\(\Rightarrow8\sqrt{x}=\frac{40}{3}\)
\(\Rightarrow\sqrt{x}=\frac{40}{3}:8\)
\(\Rightarrow\sqrt{x}=\frac{5}{3}\)
\(\Rightarrow x=\frac{25}{9}\)
b)\(12-3x^2=10+\sqrt{\frac{25}{16}}\)
\(\Rightarrow12-3x^2=10+\frac{5}{4}\)
\(\Rightarrow12-3x^2=11,25\)
\(\Rightarrow3x^2=12-11,25\)
\(\Rightarrow3x^2=0,75\)
\(\Rightarrow x^2=0,25\)
\(\Rightarrow x=\sqrt{0,25}\)
\(\Rightarrow x=0,5\)
a) \(\frac{15^{15}.5^{10}}{9^7.25^{13}}=\frac{3^{15}.5^{15}.5^{10}}{3^{14}.5^{26}}=\frac{3.5^{25}}{5^{26}}=\frac{3}{5}\)
b) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\sqrt{\frac{2^2}{9^2}}:\sqrt{\frac{5^2}{9^2}}-\frac{7}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{9}.\frac{9}{5}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=\frac{-5}{5}=-1\)
c) \(\left(3^2\right)^2-625+64\)
\(=3^4-625+64\)
\(=81-625+64\)
\(=-480\)
d) \(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}\)
\(=\frac{3-39}{7-91}\)
\(=\frac{-36}{-84}\)
\(=\frac{3}{7}\)
Bai 1
a) \(\sqrt{0,36}+\sqrt{0,49}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\frac{2}{3}-\frac{5}{6}\)
=\(-\frac{1}{6}\)
Bài 2
a)\(x^2=81\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
b) \(\left(x-1\right)^2=\frac{9}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{3}{4}\\x-1=\frac{-3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
c) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
a) x = \(\dfrac{-64}{3}\)
b) x = -3,5
c) x = 80
d) x = -1.162
e) x = 0,9436
g) x \(\in\varnothing\)
a) 16/3 : x = -1/4
=> x = 16/3 : (-1/4)
=> x = 16/3 . (-4)
=> x = -64/3
Vậy x= -64/3
b)2x - 13 = -8
=> 2x = (-8) + 1
=> 2x = -7
=> x = -7/2
d) 0,944 - 2x = 3,268
=> 2x = 0,944 - 3,268
=> 2x = -2,324
=> x = (-2,324) : 2
=> x = -1,162
g) \(\sqrt{5^2-3^2}=-\sqrt{81-x}\)
=> \(\sqrt{25-9}\)= \(-\sqrt{81-x}\)
=> \(\sqrt{16}\)=\(-\sqrt{81-x}\)
=> 4=\(-\sqrt{81-x}\)
tới đây mik bí r hk bt lm nữa
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
Bài 1:
a) Ta có: \(\left(0.125\right)\cdot\left(-3\cdot7\right)\cdot\left(-2\right)^3\)
\(=\frac{1}{8}\cdot\left(-21\right)\cdot\left(-8\right)\)
\(=\frac{1}{8}\cdot168\)
\(=21\)
b) Ta có: \(\sqrt{36}\cdot\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{36\cdot\frac{25}{16}}+\frac{1}{4}\)
\(=\sqrt{\frac{225}{4}}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}\)
c) Ta có: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}=-1\)
d) Ta có: \(0,1\cdot\sqrt{225}\cdot\sqrt{\frac{1}{4}}\)
\(=0,1\cdot15\cdot\frac{1}{2}=\frac{3}{4}\)
cảm ơn bạn nhé 
b) \(\left(x+\frac{1}{2}\right)^3:3=-\frac{1}{81}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{81}\right).3\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x+\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{6}\)
Vậy \(x=-\frac{5}{6}.\)
c) \(\frac{x-2}{2}=\frac{8}{x-2}\left(x\ne2\right).\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=8.2\)
\(\Rightarrow\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow x-2=\pm4.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-2\right\}.\)
Chúc bạn học tốt!