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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) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10
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.
Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a) -4/5 + 5/2x = -3/10
5/2x = -3/10 + 4/5
5/2x = 1/5
5/2x = 1/2
x = 1/2 : 5/2
x = 1/5
b) 4/3 + 5/8 : x = 1/12
5/8x = 1/12 - 4/3
5/8x = -5/4
5 = -5/4.8x
5 = -10x
5/-10 = x
-1/2 = x
x = -1/2
c) (x - 1/3)(x - 2/5) = 0
x - 1/3 = 0 hoặc x - 2/5 = 0
x = 0 + 1/3 x = 0 + 2/5
x = 1/3 x = 2/5
a)\(\frac{9}{4}\cdot\left|x\right|-\frac{5}{2}=\frac{8}{3}\)\(\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{8}{3}+\frac{5}{2}\Rightarrow\frac{9}{4}\cdot\left|x\right|=\frac{31}{6}\)
\(\Rightarrow\left|x\right|=\frac{31}{6}:\frac{9}{4}\Rightarrow\left|x\right|=\frac{31}{6}\cdot\frac{4}{9}\Rightarrow\left|x\right|=\frac{62}{27}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{62}{27}\\x=-\frac{62}{27}\end{cases}}\)
b)\(\frac{1}{2}\cdot\left|x\right|+\frac{3}{4}=\frac{2}{3}\Rightarrow\frac{1}{2}\cdot\left|x\right|=\frac{2}{3}-\frac{3}{4}\Rightarrow\frac{1}{2}\cdot\left|x\right|=-\frac{1}{12}\)
\(\Rightarrow\left|x\right|=-\frac{1}{12}:\frac{1}{2}\Rightarrow\left|x\right|=-\frac{1}{12}\cdot2\Rightarrow\left|x\right|=-\frac{1}{6}\)
Ta có\(\left|x\right|\ge0\)mà \(-\frac{1}{6}\le0\)
Do đó ko có giá trị của x thỏa mãn