\(B=\left(1,6\cdot0,4\right)^2-\dfrac{57}{4}+\dfrac{7}{5}-\dfrac{7}{4}=\dfrac{256}{625}-16+\dfrac{7}{5}=-\dfrac{8869}{625}\)
\(B=\left(1,6\cdot0,4\right)^2-\dfrac{57}{4}+\dfrac{7}{5}-\dfrac{7}{4}=\dfrac{256}{625}-16+\dfrac{7}{5}=-\dfrac{8869}{625}\)
a)\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\)
b)\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
c)\(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
Tính bằng cách hợp lí giá trị của biểu thức.
A = \(\left(3-\dfrac{1}{4} +\dfrac{3}{2}\right)\)- \(\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)\)-\(\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
B =\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
Tính giá trị của các biểu thức sau :
a)\(\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)\)+(0,4 - 5) - \(\left(4\dfrac{1}{4}-1\right)\)
b)\(\dfrac{2}{3}\) - \(\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
c)\(\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right)\):\(\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)
d)3 - \(\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}\)
giúp mình nhé trả lời mình cho tick cảm ơn các bạn !
a,0,4.\(\sqrt{0,25-\sqrt{\dfrac{1}{4}}}\)
b,3/2+2(x-1)=-5\(\dfrac{1}{2}\)
c.7/4|x+4/5|+1/3=8/3
GIÚP MÌNH VỚI
a, x - \(\dfrac{5}{7}\)=\(\dfrac{19}{21}\)
b,\(\dfrac{5}{3}\)-I x - \(\dfrac{1}{5}\)I = \(\dfrac{1}{3}\)
c, (x - \(\dfrac{2}{5}\)) = \(\dfrac{1}{4}\)
d, 5\(\sqrt{x}\) - 30 = 15
Tính bằng cách thuận tiện nhất
B=\(\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}x\dfrac{858585}{313131}x\left(-1\dfrac{14}{17}\right)\)
\(d,\dfrac{\left(13\dfrac{1}{4}-2\dfrac{5}{27}-10\dfrac{5}{6}\right).230\dfrac{1}{25}+46\dfrac{3}{4}}{\left(1\dfrac{3}{7}+\dfrac{10}{3}\right):\left(12\dfrac{1}{3}-14\dfrac{2}{7}\right)}\)
Câu 13. (0,5 điểm): Tính giá trị $\left(2+\dfrac{1}{3}-0,4\right)-\left(7-\dfrac{3}{5}-\dfrac{4}{3}\right)-\left(\dfrac{1}{5}+\dfrac{5}{3}-4\right)$.
a)\([x.\dfrac{1}{2}]^{3}=\dfrac{1}{27}\)
b)\([x+\dfrac{1}{2} ]^{2}=\dfrac{4}{5} \)
c) I 3x-4/5 I = 11/5
d) I 2x - 2I = 0