Tính
a)\(\left[6\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
b)
tính
a,\(6\dfrac{4}{5}-\left(1\dfrac{2}{3}+3\dfrac{4}{5}\right)\)
b,\(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
c,\(6\dfrac{7}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
a: \(=6+\dfrac{4}{5}-1-\dfrac{2}{3}-3-\dfrac{4}{5}\)
\(=2-\dfrac{2}{3}=\dfrac{4}{3}\)
b: \(=7+\dfrac{5}{9}-2-\dfrac{3}{4}-3-\dfrac{5}{9}=2-\dfrac{3}{4}=\dfrac{5}{4}\)
c: =6+7/7-1-3/4-2-5/7
=3+2/7-3/4
=84/28+8/28-21/28
=84/28-13/28
=71/28
1.tính
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
b)\(\left(\dfrac{-5}{24}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(\dfrac{-5}{16}\right)\)
c)\(\dfrac{6}{7}+\dfrac{5}{4}:\left(-5\right)-\dfrac{-1}{28}.\left(-2\right)^2\)
ai giải đc mik sẽ tick
a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)
\(=1-\dfrac{-5}{2}\)
\(=\dfrac{2}{2}-\dfrac{-5}{2}\)
\(=\dfrac{7}{2}\)
dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!
Giải:
a)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
b)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
c)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7-1/4-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
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.\(x\left(x-\dfrac{1}{2}\right)-\left(x-2\right)\left(x+3\right)\)
a. \(\dfrac{-5}{4}\) x4 . \(\dfrac{8}{15}\) x = \(\dfrac{-40}{60}\) x5 = \(\dfrac{-2}{3}\) x5
b. -2x\(\left(\dfrac{3}{4}x^2-x+\dfrac{1}{2}\right)\) = -\(\dfrac{-3}{2}\) x3 + 2x3 - x
c. \(x\left(x-\dfrac{1}{2}\right)\) - (x - 2)(x + 3)
= x2 - \(\dfrac{1}{2}\) x - x2 - 3x - 2x - 6
Tính:
a/\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(\dfrac{-1}{6}\right):\left(-3\right)\)
b/\(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
\(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-\dfrac{2}{4}-\dfrac{1}{4}\right).\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{3}\right)\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{1}{5}\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{3}{20}+\dfrac{1}{48}-\dfrac{1}{18}=\dfrac{108}{720}+\dfrac{15}{720}-\dfrac{40}{720}=\dfrac{83}{720}\)
bài 1:tính
a)\(\left(\dfrac{-7}{15}-\dfrac{27}{70}\right)-\left(\dfrac{8}{15}+\dfrac{43}{70}\right)\)
b)\(\dfrac{3}{7}+\left(\dfrac{-1}{5}+\dfrac{-3}{7}\right)\)
c)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)
d)\(\left(\dfrac{-5}{24}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(-\dfrac{5}{16}\right)\)
e)\(\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}\)
g)\(\dfrac{6}{7}+\dfrac{5}{4}:\left(-5\right)-\dfrac{-1}{28}.\left(-2\right)^2\)
Giải:
a)(-7/15-27/70)-(8/15+43/70)
=-7/15-27/70-8/15-43/70
=(-7/15-8/15)+(-27/70-43/70)
=-1+(-1)
=-2
b)3/7+(-1/5+-3/7)
=3/7-1/5+-3/7
=(3/7+-3/7)-1/5
=0-1/5
=-1/5
c)(4-12/5).25/8-2/5:-4/25
=8/5.25/8-(-5/2)
=5+5/2
=15/2
d)(-5/24+3/4-7/12):(-5/16)
=-1/24:(-5/16)
=2/15
e)-5/7.2/11+-5/7.9/11
=-5/7.(2/11+9/11)
=-5/7.1
=-5/7
g)6/7+5/4:(-5)-(-1/28).(-2)2
=6/7+(-1/4)-(-1/28).4
=6/7+(-1/4)-(-1/7)
=6/7-1/4+1/7
=(6/7+1/7)-1/4
=1-1/4
=3/4
Chúc bạn học tốt!
Bài 27: Tính
a) \(\left(\dfrac{-1}{3}\right)^4\)
b)\(\left(\dfrac{-9}{4}\right)^3\)
c) \(\left(-0,2\right)^2\)
d) \(\left(-5,3\right)^0\)
a) \(=\dfrac{\left(-1\right)^4}{3^4}=\dfrac{1}{81}\)
b) \(=\dfrac{\left(-9\right)^3}{4^3}=\dfrac{-729}{64}\)
c) \(=\left(-\dfrac{2}{10}\right)^2=\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\)
d) \(=1\)
\(a,=\dfrac{1}{81}\\ b,=\dfrac{729}{64}\\ c,=0,04\\ d,=1\)
Tính
a, \(\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^4+\dfrac{1}{6}x^2y+\dfrac{1}{9}y^2\right)\)
\(\left(\dfrac{1}{2}x^2-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^4+\dfrac{1}{6}x^2y+\dfrac{1}{9}y^2\right)\\ =\left(\dfrac{1}{2}x^2\right)^3-\left(\dfrac{1}{3}y\right)^3\\ =\dfrac{1}{8}x^6-\dfrac{1}{27}y^3.\)
Bài 1 : Thực hiện phép tính
a , \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
b , \(\dfrac{-5}{11}.\dfrac{4}{13}+\dfrac{-5}{11}.\dfrac{9}{13}+3\dfrac{5}{11}\)
c , \(3^2-12.\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
Bài 2 : Cho đường thẳng xy , lấy điểm O thuộc đường thẳng xy . Trên tia Ox , lấy 2 điểm A , B sao cho OA = 3 cm , AB = 2 cm
a, Trong 3 điểm O , A , B điểm nào nằm giữa 2 điểm còn lại.
b, Có tất cả bao nhiêu tia ? nêu tên ?
c, Tính độ dài OB ?
Bài 3 : Tính A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{2021.2022}\)
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)