45- { [(120-60):15+5]:3} + (-55)

= 45 - { [60:15+5]:3} + (-55)

= 45 - {9:3} +(-55)

= 45 - 3 + (-55) = -13

b) \(6^2-2^3.5^0.6=36-8.1.6\)

\(36-48=-12\)

TICK MIK NHA

(0.6)^5/(0.2)^6 
= [(0.2)^5. 3^5]/(0.2)^6 
= 3^5/0.2 
=243/0.2 
= 1215 

6^3+3.6^2+3^3/-13 
= 2^3.3^3+3.2^2.3^2+3^3/-13 
=3^3(2^3+2^2+1)/-13 
=3^3.13/-13 
=-27

\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{\left(3\cdot2\right)^3+3\cdot\left(3\cdot2\right)^2+3^3}{-13}=\frac{3^3\cdot2^3+3\cdot3^2\cdot2^2+3^3}{-13}=\frac{3^3\cdot\left(2^3+2^2+1\right)}{-13}=\frac{27\cdot13}{-13}=-27\)

Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)

\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)

mình không hiểu

c) \(3^2+2^4-\left(6^8:6^6-6^2\right)< 5^x< 125\)

\(=9+16-\left(6^{8-6}-36\right)< 5^x< 5^3\)

\(=25-\left(6^2-36\right)< 5^x< 5^3\)

\(=25-\left(36-36\right)< 5^x< 5^3\)

\(=25-0< 5^x< 5^3\)

\(=25< 5^x< 5^3\)

\(=5^2< 5^x< 5^3\)

Vì \(5^2=25\) và \(5^3=125\) nên \(x\) không thể thỏa mãn đề bài

⇒ \(x\) không thỏa mãn đề bài

\(B=136.25+75.136-6^2.10^2\)

\(=136.\left(75+25\right)-36.100\)

\(=136.100-36.100\)

\(=\left(136-36\right).100\)

\(=100.100=10000\)

\(C=2^3.5^3-\left\{7^2.2^3.5^2.\left[4^3:8+11^2:121-2\left(37-5.7\right)\right]\right\}\)

\(=10^3-\left\{49.8.25.\left[8+1+2.2\right]\right\}\)

\(=1000-\left(49.200.13\right)\)

\(=1000-127400\)

\(=-126400\)

Bài này đơn giản thôi bạn

B= 136.(25+75)-36.100

B= 136. 100-36.100

B= 100.(136-36)

B= 100.100

B= 10000

Bài 1.

\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)

\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)

\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)

\(=1200:\left[450:\left(450-300\right)\right]\)

\(=1200:\left(450:150\right)\)

\(=1200:3\)

\(=400\)

\(---\)

\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)

\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)

\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)

\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)

\(=675-20\left[90-\left(164-100\right)\right]\)

\(=675-20\left(90-64\right)\)

\(=675-20\cdot26\)

\(=675-520\)

\(=155\)

\(---\)

\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)

\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)

\(=\left(1\cdot2022-1986\right)\cdot5-169\)

\(=\left(2022-1986\right)\cdot5-169\)

\(=36\cdot5-169\)

\(=180-169\)

\(=11\)

Bài 2.

\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)

Vậy \(x=2\).

\(---\)

\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)

Vậy \(x=\dfrac{52}{5}\).

\(Toru\)

b)

\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

c)

\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)

\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)

\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)

\(=\frac{406}{183}\)

d)

\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)

\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)

=57/37

Cô giáo à bạn bị sao thế. Ng` ta có bài kg biết thì phải hỏi , đằng này bạn lại còn đi chửi ng` khác nữa. Đây chắc chắn là giáo viên giả mạo rồi.

A = 1 /1.2 + 1/ 2.3 + 1 /3.4 + . . . + 1/ 49.50 + 1/ 50.51

 A = 2 − 1/ 1.2 + 3 − 2 /2.3 + 4 − 3 /3.4 + . . . + 50 − 49 /49.50 + 51 − 50/ 50.51

A = 1 − 1/ 2 + 1/ 2 − 1 /3 + 1 /3 − 1/ 4 + . . . + 1 /50 − 1 /51

A=1-1/51

A=50/51

Cảm ơn bn