Bài 1: Tính
a)\(\left(\frac{9}{25}-2\cdot18\right):\left(3\frac{4}{5}+0,2\right)\)
b)\(\frac{3}{8}\cdot19\frac{1}{3}-\frac{3}{8}\cdot33\frac{1}{3}\)
c)\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
d)\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
LÀM NHANH GIÚP MIK ,MIK TICK CHO Ạ
Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
a) \(\frac{21}{47}\) + \(\frac{9}{45}\) + \(\frac{26}{47}\) + \(\frac{4}{5}\)
b) 12 x \(\left(-\frac{2}{3}\right)\)2 +\(\frac{4}{3}\)
c) 12,5 x \(\left(-\frac{5}{7}\right)\) + 1,5 x \(\left(-\frac{5}{7}\right)\)
d) \(\frac{4}{5}\) x \(\left(\frac{7}{2}+\frac{1}{4}\right)\)2
a) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{4}{5}\right)\)
\(=\frac{47}{47}+\left(\frac{1}{5}+\frac{4}{5}\right)\)
\(=1+1=2\)
b) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
\(=12.\frac{4}{9}+\frac{4}{3}\)
\(=\frac{16}{3}+\frac{4}{3}\)
\(=\frac{20}{3}\)
c) \(12,5.\left(-\frac{5}{7}\right)+15.\left(-\frac{5}{7}\right)\)
\(=\left(-\frac{5}{7}\right).\left(12,5+15\right)\)
\(=\left(-\frac{5}{7}\right).27,5\)
\(=\left(-\frac{5}{7}\right).\frac{55}{2}\)
\(=-\frac{275}{14}\)
d) \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
a)\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
=\(\frac{21}{47}+\frac{1}{5}+\frac{26}{47}+\frac{4}{5}\)
=\(\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)
=\(\frac{47}{47}+\frac{5}{5}=1+1=2\)
b)\(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
=\(12.\frac{4}{9}+\frac{4}{3}\)
=\(\frac{12}{1}.\frac{4}{9}+\frac{4}{3}=\frac{48}{9}+\frac{4}{3}\)
=\(\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)
c)\(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
=\(\left(-\frac{5}{7}\right).\left(12,5+1,5\right)\)
=\(\left(-\frac{5}{7}\right).14=\left(-\frac{5}{7}\right).\frac{14}{1}=-10\)
d)\(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
=\(\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
=\(\frac{4}{5}.\frac{225}{16}\)
=\(\frac{900}{80}=\frac{45}{4}\)
Nhớ tick cho mình nha!
Bài 1: Tính
\(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)
\(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
\(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
\(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) b) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
c) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi
Bài 1:Thực hiện phép tính (Tính bằng cách tính hợp lí nếu có thể)
a. \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
b. \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
c. \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
d. \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
Bài 2: Cho \(\Delta\)ABC có AB=AC và \(\widehat{B}=45^0\). Tam giác ABC là tam giác gì? Vì sao?
Bài 1:
a. \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{1}{5}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)\)
\(=1+1\)
\(=2\)
b. \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}\)
\(=12.\frac{4}{9}+\frac{4}{3}\)
\(=\frac{16}{3}+\frac{4}{3}\)
\(=\frac{20}{3}\)
c. \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\)
\(=\frac{-5}{7}.\left(12,5+1,5\right)\)
\(=\frac{-5}{7}.14\)
\(=-10\)
d. \(\frac{4}{5}.\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}.\left(\frac{15}{4}\right)^2\)
\(=\frac{4}{5}.\frac{225}{16}\)
\(=\frac{45}{4}\)
Bài 1:
a) Ta có: \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
b) Ta có: \(12\cdot\left(\frac{-2}{3}\right)^2+\frac{4}{3}\)
\(=12\cdot\frac{4}{9}+\frac{4}{3}\)
\(=\frac{48}{9}+\frac{12}{9}=\frac{60}{9}=\frac{20}{3}\)
c) Ta có: \(12,5\cdot\left(-\frac{5}{7}\right)+1,5\cdot\left(-\frac{5}{7}\right)\)
\(=\frac{-5}{7}\cdot\left(12,5+1,5\right)\)
\(=\frac{-5}{7}\cdot14=-\frac{70}{7}=-10\)
d) Ta có: \(\frac{4}{5}\cdot\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}\left(\frac{14}{4}+\frac{1}{4}\right)^2\)
\(=\frac{4}{5}\cdot\frac{225}{16}=\frac{900}{80}=\frac{45}{4}\)
Bài 2:
Xét ΔABC có AB=AC(gt)
nên ΔABC cân tại A(định nghĩa tam giác cân)
⇒\(\widehat{A}=180^0-2\cdot\widehat{B}\)(số đo của góc ở đỉnh trong ΔABC cân tại A)
hay \(\widehat{A}=180^0-2\cdot45^0=90^0\)
Xét ΔABC cân tại A có \(\widehat{A}=90^0\)(cmt)
nên ΔABC vuông cân tại A(định nghĩa tam giác vuông cân)
Bài 2:
Ta có AB=AC
\(\Rightarrow\Delta\)ABC là tam giác cân
\(\Rightarrow\widehat{A}=180^0-\widehat{B}.2\)
\(=180^0-45^0.2\\ =180^0-90^0\\ =90^0\)
\(\Rightarrow\Delta\)ABC là tam giác cân tại A.
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{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$v
I. Nội qui tham gia "Giúp tôi giải toán"
1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;
2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.
3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.
Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web.
Cho A=\(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\) và B=\(\frac{2}{5}.\frac{4}{7}.\frac{6}{9}.....\frac{44}{47}.\frac{46}{49}\).So sánh A và B
\(\frac{2}{3}+\frac{4}{9}+\frac{2}{7}-\frac{7}{5}+\frac{1}{21}-\frac{2}{45}+4\frac{2015}{2016}\)
tính nhanh
A =\(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
B = \(\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)
\(A=-1+1+\frac{-1}{5}\)
\(A=\frac{-1}{5}\)
\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)
\(B=\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{2}{5}+\frac{-3}{5}+\frac{1}{5}\right)\)
\(B=-1+0\)
\(B=-1\)
