\(-2\dfrac{3}{4}\) (-0,4) + \(1\dfrac{2}{5}\) . 2,75 - (-1,2) : \(\dfrac{4}{11}\)
\(-2\dfrac{3}{4}\) . (-0,4) + \(1\dfrac{3}{5}\) . 2,75 - (1,2) : \(\dfrac{4}{11}\)
\(-2\dfrac{3}{4}.\left(-0,4\right)+1\dfrac{3}{5}.2,75-1,2:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}:\dfrac{4}{11}\)
\(=\dfrac{11}{4}.\dfrac{-2}{5}+\dfrac{8}{5}.\dfrac{11}{4}-\dfrac{6}{5}.\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(\dfrac{-2}{5}+\dfrac{8}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{11}{4}.0\)
\(=0\)
a,\(\dfrac{2}{3}+5\dfrac{3}{7}:2+\dfrac{9}{49}.\left(-7\right)+\dfrac{1}{12}.\left(-2\right)^4\)
b,\(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
a: \(=\dfrac{2}{3}+\dfrac{38}{7}\cdot\dfrac{1}{2}-\dfrac{9}{7}+\dfrac{1}{12}\cdot16\)
\(=\dfrac{2}{3}+\dfrac{4}{3}+\dfrac{19}{7}-\dfrac{9}{7}=2+\dfrac{10}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{4}\cdot\dfrac{-2}{5}-\dfrac{11}{4}\cdot\dfrac{8}{5}+\dfrac{11}{4}\cdot\dfrac{-6}{5}\)
\(=\dfrac{11}{4}\cdot\dfrac{-16}{5}=\dfrac{-44}{5}\)
Giúp mk vs !!!
Bài 1. Tính giá trị các biểu thức sau:
A = -1,7 . 2,3 + 1,7 . (-3,7) - 1,7 . 3 - 0,17 : 0,1
B = \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{2}{3}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
C = \(\dfrac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
tính:
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
Cái này bn lầy máy tính ra tính tí là xong thôi
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
1. Thực hiện phép tính
a. \(25\%-\dfrac{5}{4}+1\dfrac{5}{6}\)
b. \(75\%:2\dfrac{1}{5}-\left(0,5\right)^2.\left(-7\right)+2.5\left(7\dfrac{2}{3}-5\dfrac{2}{3}\right)\)
c. \(45:2\dfrac{4}{7}+50\%-1,25\)
d. \(350\%:\dfrac{105}{24}+4\dfrac{5}{6}:2-\left(0,5\right)^2.30\%\)
e. \(4\dfrac{2}{5}.0.5-1\dfrac{3}{7}.14\%+\left(-0,8\right)\)
f. \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right);\dfrac{4}{11}\)
Bài 4: Tính
a) 0,(3) + \(\dfrac{10}{3}\) + 0,4(2) b) \(\dfrac{4}{9}\) + 1,2(31) - 0,(13)
c) 2,(4) x \(\dfrac{3}{11}\) d) -0,(3) + \(\dfrac{1}{3}\)
a: Ta có: \(0,\left(3\right)+\dfrac{10}{3}+0,4\left(2\right)\)
\(=\dfrac{1}{3}+\dfrac{10}{3}+\dfrac{4}{9}\)
\(=\dfrac{33}{9}+\dfrac{4}{9}=\dfrac{37}{9}\)
b: Ta có: \(\dfrac{4}{9}+1.2\left(31\right)-0,\left(13\right)\)
\(=\dfrac{4}{9}+\dfrac{1219}{990}-\dfrac{13}{99}\)
\(=\dfrac{440}{990}+\dfrac{1219}{990}-\dfrac{130}{990}\)
\(=\dfrac{139}{90}\)
c: Ta có: \(2,\left(4\right)\cdot\dfrac{3}{11}\)
\(=\dfrac{22}{9}\cdot\dfrac{3}{11}\)
\(=\dfrac{2}{3}\)
d: Ta có: \(-0,\left(3\right)+\dfrac{1}{3}\)
\(=-\dfrac{1}{3}+\dfrac{1}{3}\)
=0
1. Tìm x, biết : \(\sqrt{x^2}=0\)
2. Tính :
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}\right).\)\(\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(B=\dfrac{\left(1+2+3+...+100\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
Help me !!! Nhanh lên nha chiều mk phải nộp rồi
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
Tính giá trị của các biểu thức sau :
\(A=27+46+79+34+53\)
\(B=-377-\left(98-277\right)\)
\(C=-1,7.2,3+1,7.\left(-3,7\right)-1,7.3-0,17:0,1\)
\(D=2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
\(E=\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
A = 27 + 46 + 79 + 34 + 53 = 239
B = - 377 - (98 - 277)
= - 377 - 98 + 277
= -198
C = - 1,7× 2,3 + 1,7.× (- 3,7) - 1,7×3 - 0,17:0,1
= 1,7. (-2,3 – 3,7 – 3 – 1)
= 1,7 . (-10)
= -17.
D =
\(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)