a) \(\dfrac{790^4}{79^4}\)
b) \(\dfrac{3^2}{0,375^2}\)
c) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
d)\(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\)
các bn giup mk vs nha
3) tìm x:
a) \(x^2=\dfrac{1}{16}\)
b)\(x^5\div x^2=\dfrac{-1}{64}\)
c)\(x^3\div x^2=\dfrac{32}{243}\)
d)\(\left(x^2\right)^2=\dfrac{81}{16}\)
(x-2/3)^3=1/27
(X+0,7)^3=-27
(2/3x-1/3)^5=1/243
tìm x thuộc Q bt:
a)\(\left(x-\dfrac{1}{2}\right)^2=0\)
b) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
c) \(\left(2.x-3\right)^3=-8\)
d) \(\left(3.x-2\right)^5=-243\)
e) \(\left(7.x+2\right)^{-1}=3^{-2}\)
f) \(\left(x-1\right)^3=-125\)
g) \(\left(2x-1\right)^4=81\)
h) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
các bn ai bt lm lm ơn lm phước giải ra hộ mk vs mk sẽ tik mừ .... các bn giúp mk vs nhé huhu
Nhờ mọi người giúp mình vs!
Tìm x thuộc Q khi:
a)3x-1 =243
b)\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
c) 3x+3x+2=270
d) \(\left(x-\dfrac{5}{2}\right)^2=144\)
e) \(\left(2x+1\right)^3=\dfrac{215}{27}\)
f) (2x-1)5 = -243
h) (27-1)2 = (2x-1)2
i) (2x+1)2 + (x+1)10 = 0
k) \(\dfrac{x-2}{5}=\dfrac{X+3}{2}\)
l) \(\dfrac{2}{x}=\dfrac{x}{50}\)
m) \(\dfrac{\left(x-3\right)}{4}=\dfrac{4}{x+3}\)
n) \(|x+\dfrac{3}{5}|-|x-\dfrac{7}{3}|\)
(x-1,2)^2=4
(x+1)^3=-125
(x+1,5)^8+(2,7-y)^10=0
3^-1.4^x+ 3.$^x+5/3. 27
9^-x. 27^x= 243
Bài 1:Tính giá trị biểu thức
\(_{1,}\)\(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\)
\(_{2,}\)(\(4.2^5\)):\(\left(2^3.\dfrac{1}{16}\right)\)
\(_{3,}\)\(\left(2^{-1}+3^{-1}\right)+\left(2^{-1}.2^0\right):2^3\)
\(_{4,}\)\(\left(-\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(_{5,}\)[(0,1)\(^2\)]\(^0\)+[(\(\dfrac{1}{7}\))\(^{-1}\)]\(^2\).\(\dfrac{1}{49}.\left[\left(2^2\right)^3:2^5\right]\)
Bài 2:Rút gọn biểu thức
a,\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b,\(\dfrac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)
c,\(\left(\dfrac{2}{5}\right)^7.5^7+\left(\dfrac{9}{4}\right)^3:\left(\dfrac{3}{16}\right)^3\)
\(\overline{2^7.5^2+512}\)
d,\(\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5\)
\(\overline{\left(\dfrac{2}{5}\right)^2}.\left(-\dfrac{5}{12}\right)^3\)
Tìm x
\(1.\left(x+0,7\right)^3=-27\)
\(2.\left(2x-1\right)^{10}=49^5\)
\(3.\left(\frac{2}{3}x-\frac{1}{3}\right)^5=\frac{1}{243}\)
\(4.\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\)
CÁC BẠN GIÚP MÌNH VỚI !!!!!