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.

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) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)

\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)

\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)

\(=1+1=2\)

Bài 1:

1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3

= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3

= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3 

= \(\frac{11}{10}\)- 3

= \(\frac{-19}{10}\)

2) \(\frac{5}{6}\):  \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)

= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)

= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))

= \(\frac{5}{6}\).( -30)

= -25

mách mình mấy câu kia với

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

nhớ làm giúp mình nhá mai mình phải đi học r :<

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !