Ta có: 8. 25 = 200

a) (-8). 25 = -200

b) 8. (-25) = -200

c) (-8). (-25) = 200

Ta có: 8 . 25 = 200

=> a) (- 8) . 25 = - 200.

     b) 8 . (- 25) = - 200.

     c) (- 8) . (- 25) = 200.

Ta có: 8. 25 = 200

a) (-8). 25 = -200

b) 8. (-25) = -200

c) (-8). (-25) = 200

Ta có: 225.8 = 1800

8.(-225) = -1800

Ta có: 225.8 = 1800

(-225).8 = -1800

Ta có: 225.8 = 1800

(-8).225 = -1800

 a. \(160 - \left( {{2^3}{{.5}^2} - 6.25} \right)\)

\(\begin{array}{l} = 160 - \left( {8.25 - 6.25} \right)\\ = 160 - 25.\left( {8 - 6} \right)\\ = 160 - 25.2\\ = 160 - 50\\ = 110\end{array}\)

Ta có: 110 = 2.5.11

b. \(37.3 + 225:{15^2}\)

\(\begin{array}{l} = 37.3 + 225:225\\ = 37.3 + 1\\ = 111 + 1\\ = 112\end{array}\)

Ta có: \(112 = 2^4.7\)

c. \(5871:103 - 64:{2^5}\)

\(\begin{array}{l} = 5871:103 - 64:32\\ = 57 - 2 = 55\end{array}\)

Ta có: 55 = 5. 11

d. \(\left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8} \right){.5^2} - 850:2\)

\(\begin{array}{l} = \left[ {\left( {1 + 8} \right) + \left( {2 + 7} \right) + \left( {3 + 6} \right) + \left( {4 + 5} \right)} \right]{.5^2} - 850:2\\ = \left( {9 + 9 + 9 + 9} \right){.5^2} - 850:2\\ = {9.4.5^2} - 850:2\\ = {36.5^2} - 425\\ = {36.5^2} - {5^2}.17\\ = {5^2}.\left( {36 - 17} \right)\\ = {5^2}.19=475\end{array}\)

Ta có: \(475 = 5^2.19\)

a: \(160-\left(2^3\cdot5^2-6\cdot25\right)\)

\(=160-\left(8\cdot25-150\right)\)

\(=160-200+150=10=2\cdot5\)

b: \(=111+225:225=112=2^4\cdot7\)

c: \(=57-64:32=57-2=55=5\cdot11\)

d: \(=\left(9\cdot\dfrac{8}{2}\right)\cdot25-425=36\cdot25-425=25=5^2\)

\(213.3 = 639\)

a) \(\left( { - 213} \right).3 =  - \left( {213.3} \right) =  - 639\)

b) \(\left( { - 3} \right).213 =  - \left( {3.213} \right) =  - 639\)

c) \(\left( { - 3} \right).\left( { - 213} \right) = 3.213 = 639\)

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

a) \( - \left( {4 + 7} \right) =  - 11\)

\(\begin{array}{l}\left( { - 4 - 7} \right) = \left( { - 4} \right) + \left( { - 7} \right)\\ =  - \left( {4 + 7} \right) =  - 11\\ \Rightarrow \left( { - 4 - 7} \right) =  - \left( {4 + 7} \right)\end{array}\)

b)

\(\begin{array}{l} - \left( {12 - 25} \right) =  - \left[ {12 + \left( { - 25} \right)} \right]\\ =  - \left[ { - \left( {25 - 12} \right)} \right] =  - \left( { - 13} \right) = 13\end{array}\)

\(\begin{array}{l}\left( { - 12 + 25} \right) = 25 - 12 = 13\\ \Rightarrow  - \left( {12 - 25} \right) = \left( { - 12 + 25} \right)\end{array}\)

c)

\(\begin{array}{l} - \left( { - 8 + 7} \right) =  - \left[ { - \left( {8 - 7} \right)} \right] =  - \left( { - 1} \right) = 1\\\left( {8 - 7} \right) = 1\\ \Rightarrow  - \left( { - 8 + 7} \right) = \left( {8 - 7} \right)\end{array}\)

d)

\(\begin{array}{l} + \left( { - 15 - 4} \right) =  + \left[ {\left( { - 15} \right) + \left( { - 4} \right)} \right]\\ =  + \left[ { - \left( {15 + 4} \right)} \right] =  + \left( { - 19} \right) =  - 19\\\left( { - 15 - 4} \right) = \left( { - 15} \right) + \left( { - 4} \right)\\ =  - \left( {15 + 4} \right) =  - 19\\ \Rightarrow  + \left( { - 15 - 4} \right) = \left( { - 15 - 4} \right)\end{array}\)

e)

\(\begin{array}{l} + \left( {23 - 12} \right) =  + 11 = 11\\\left( {23 - 12} \right) = 11\\ \Rightarrow  + \left( {23 - 12} \right) = \left( {23 - 12} \right)\end{array}\)

Viết hết tất cả dưới dạng tự luận luôn à bạn?

Câu 2: C

C1. Mik ko bik vì bấm máy tính thì nó ra kết quả quá lớn.

C2. \(\dfrac{15}{19}.\dfrac{2}{3}-\dfrac{7}{19}.\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\left(\dfrac{15}{19}-\dfrac{7}{19}\right)+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\dfrac{8}{19}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{16}{57}+\dfrac{136}{57}\\ =\dfrac{152}{57}\\ =\dfrac{8}{3}\left(C\right)\)

C3. \(x=4\left(B\right)\)

C4. Gọi tổng số tiền ba đợt ủng hộ lần lượt là a, b, c (a, b, c ϵ N*).

Vì số tiền ủng hộ ba đợt lần lượt tỉ lệ với 7; 8; 9

Nên \(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}\)

Vì tổng số tiền ủng hộ đợt hai và đợt ba nhiều hơn số tiền ủng hộ đợt một 80 tỉ

Nên \(\left(b+c\right)-a=80\)

Theo tính chất của dãy tỉ số bằng nhau

Ta có:

\(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}=\dfrac{\left(b+c\right)-a}{\left(8+9\right)-7}=\dfrac{80}{10}=8\)

Do đó:

\(\dfrac{a}{7}=8=>a=8.7=>a=56\\ \dfrac{b}{8}=8=>b=8.8=>b=64\\ \dfrac{c}{9}=8=>c=8.9=>c=72\)

Vậy số tiền ủng hộ ba đợt lần lượt là: 56; 64; 72.

Số tiền ủng hộ đợt hai là 64 tỉ (B).