`x^5 : x^3 = 1/16`
`=> x^2 = (1/4)^2`
`=> x = 1/4` hoặc `x = 1/4`
Vậy ...
`x^10 : x^7 = 1/27`
`=> x^3 = (1/3)^3`
`=> x = 1/3`
Vậy ...
`(99^2)/(11^2) = x^2`
`=> (99/11)^2 = x^2`
`=> x^2 = 9^2`
`=> x = 9` hoặc `x = -9`
Vậy ...
`16x^2 = (x+1)^2`
`=> (x+1)^2 = (4x)^2`
`=> x + 1 = 4x` hoặc `x + 1 = -4x`
`=> 3x = 1` hoặc `5x = -1`
`=> x = 1/3` hoặc `x = -1/5`
Vậy ...
Tìm x:
\(a\)) \(\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)
\(b\)) \(\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{27}{8}\right)^3=\dfrac{81}{16}\)
\(c\)) \(\dfrac{1}{2}.2^x+4.2^x=9.2^5\)
\(d\)) \(\text{12 - (2x +1)}^2=-69\)
Câu 1: Tìm x, biết:
a)\(x^2-\dfrac{16}{25}=0\) b)\(\dfrac{2}{5}-\left|\dfrac{1}{2}-x\right|=6\)
C2.Tính giá của biểu thức:
a)\(A=1\dfrac{5}{13}-0,25-\left(2\dfrac{5}{9}+\dfrac{18}{13}-\dfrac{1}{4}\right)\)
b)\(\dfrac{\dfrac{3}{5}.7^2-3.5^6+\dfrac{3}{5}.3^9}{\dfrac{3}{4}.7^2-\dfrac{3}{4}.5^7+\dfrac{3}{4}.3^9}\)
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)
1,\(\dfrac{3}{16}\)- ( x - \(\dfrac{5}{4}\) ) - ( \(\dfrac{3}{4}\) + \(\dfrac{-7}{8}\) - 1 ) = \(2\dfrac{1}{2}\)
2,\(\dfrac{1}{2}\) . ( \(\dfrac{1}{6}\) - \(\dfrac{9}{10}\) ) = \(\dfrac{1}{5}\) - x + ( \(\dfrac{1}{15}\) - \(\dfrac{-1}{5}\) )
Giúp mik nhanh với ạ .
Câu 1: Thực hiện phép tính
a, \(40\dfrac{1}{4}:\dfrac{5}{7}-25\dfrac{1}{4}:\dfrac{5}{7}-\dfrac{1}{2021}\)
b, \(\left|\dfrac{-5}{9}\right|.\sqrt{81}-2021^0.\dfrac{16}{25}\)
Câu 2: Tìm x
\(3\left(x-\dfrac{1}{3}\right)-7\left(x+\dfrac{3}{7}\right)=-2x+\dfrac{1}{3}\)
Tìm x, biết:
a) \(\dfrac{1}{20}\) - (x - \(\dfrac{8}{5}\)) = \(\dfrac{1}{10}\)
b) \(\dfrac{7}{4}\) - (x + \(\dfrac{5}{3}\)) = \(\dfrac{-12}{5}\)
c) x - [\(\dfrac{17}{2}\) - \(\left(\dfrac{-3}{7}+\dfrac{5}{3}\right)\)] = \(\dfrac{-1}{3}\)
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
\(\left(\dfrac{-2}{3}\right)^2.x=\left(\dfrac{-2}{3}\right)^5\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)
\(\dfrac{4}{9}:x=3\dfrac{1}{3}:2,25\)
\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\)
\(b,\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}.x\right)\)
\(c,\dfrac{2x-3}{3}+\dfrac{-3}{2}=\dfrac{5-3x}{6}-\dfrac{1}{3}\)
\(d,\dfrac{2}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)
\(e,2\left(x-1\right)=\left(x-1\right)^2\)
TÍNH
a.\(-\dfrac{5}{4}x^4.\dfrac{8}{15}x\) b.\(-2x\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) c.\(x\left(x-\dfrac{1}{2}\right)-\left(x-2\right)\left(x+3\right)\)
