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\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
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Tìm X
a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)
\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)
\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)
\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)
\(\Leftrightarrow\dfrac{5}{32}-x=0\)
\(\Leftrightarrow x=\dfrac{5}{32}\)
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1. So sánh
a) \(25^{50}\) và \(2^{300}\)
\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)
\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25< 64\) nên \(25^{50}< 64^{50}\)
Vậy \(25^{50}< 2^{300}\)
b) \(625^{15}\) và \(12^{45}\)
\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)
\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)
Vậy \(625^{15}< 12^{45}\)
1.So sánh
a)\(25^{50}\) và \(2^{300}\)
Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)
b)\(625^{15}\) và \(12^{45}\)
Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)
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1) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)
=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)
=> \(\frac{25}{12}.x=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{25}{12}\)
=> \(x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\).
3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)
=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)
=> \(\frac{29}{12}.x=\frac{29}{24}\)
=> \(x=\frac{29}{24}:\frac{29}{12}\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\).
4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)
=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)
=> \(4x+\frac{3}{4}=\frac{13}{4}\)
=> \(4x=\frac{13}{4}-\frac{3}{4}\)
=> \(4x=\frac{5}{2}\)
=> \(x=\frac{5}{2}:4\)
=> \(x=\frac{5}{8}\)
Vậy \(x=\frac{5}{8}\).
5) 2x + 2x+3 = 144
⇔ 2x + 2x . 23 = 144
⇔ 2x . (1 + 23) = 144
⇔ 2x . 9 = 144
⇔ 2x = 144 : 9
⇔ 2x = 16
⇔ 2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
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1)
a. \(\left(3x^2-50\right)^2=5^4\)
\(\Leftrightarrow3x^4-50=625\)
\(\Leftrightarrow3x^4=675\)
\(\Leftrightarrow x^4=225\)
\(\Leftrightarrow x=\sqrt{15}\)
2)
a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)
\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)
b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)
\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)
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a) (2x)5 : 43 = 815 => 25x = 815.43 = (23)15.(22)3 = 245.26 = 251 => 5x = 51 => x = 10,2
b) (32)x .93 = 2439 => 32x = 2439 : 93 = (35)9 : (32)3 = 345 : 36 = 339 => 2x = 39 => x = 19,5
c) (1/125)3.5x = 255 => 5x = 255 : (1/125)3 = (52)5 : (1/53)3 = 510 : (5-3)3 = 510 : 5-9 = 519 => x = 19
d) 1/81 : 3x = 1/729 => 3x = 1/81 : 1/729 = 1/34.729 = 3-4.36 = 32 => x = 2
e) (5x - 2)4 = 168 = (162)4 = 2564
=> 5x - 2 = -256 ; 256 => 5x = -254 ; 258 => x = -50,8 ; 51,6
P/S : Thay x = 10,2 vào câu a , x = 19,5 vào câu b sẽ thấy điều hư cấu : 210,2 và 919,5.Ko thể tính được giá trị của 2 lũy thừa này.
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Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
\(\left(-\dfrac{3}{5}\right)^2-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\\ \Rightarrow\dfrac{9}{25}-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\\ \Rightarrow x-\dfrac{1}{3}=\dfrac{9}{25}-\dfrac{4}{25}\\ \Rightarrow x-\dfrac{1}{3}=\dfrac{5}{25}\\ \Rightarrow x=\dfrac{1}{5}+\dfrac{1}{3}\\ \Rightarrow x=\dfrac{3}{15}+\dfrac{5}{15}\\ \Rightarrow x=\dfrac{8}{15}\)
Vậy `x=8/15`
(-3/5)^2 - (x - 1/3) = 4/25
9/25 - (x - 1/3) = 4/25
x - 1/3 = 9/25 - 4/25 = 1/5
x = 1/5 + 1/3 = 8/15
Vậy x = 8/15