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\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\\ =5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\\ =5\left(1-\dfrac{1}{106}\right)=5\cdot\dfrac{105}{106}=\dfrac{525}{106}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
a, 1+6+11+16+...+46+51
Số số hạng là : (51-1):5+1 = 11 ( số )
Tổng là : (51+1).11:2=286
b, Đặt A = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+\dfrac{5^2}{16.21}+\dfrac{5^2}{21.26}+\dfrac{5^2}{26.31 } \)
\(\dfrac{1}{5}A=\) \(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\)
\(\dfrac{1}{5}A=\) \(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\)
\(\dfrac{1}{5}A=1-\dfrac{1}{31}\)
\(\dfrac{1}{5}A=\dfrac{30}{31}\)
\(A=\dfrac{30}{31}:\dfrac{1}{5}=\dfrac{150}{31}\)
Vậy..
Bài 1:
1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)
\(=-12-18\)
=-30
2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)
\(=36-2020+2019-136-27\)
\(=1-100-27\)
\(=-126\)
3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)
\(=144-97-244+197\)
\(=-100+100=0\)
4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)
\(=-24\cdot13+24\cdot3\)
\(=24\cdot\left(-13+3\right)\)
\(=24\cdot\left(-10\right)=-240\)
5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)
\(=54+55+56+57+58-64-65-66-67-68\)
\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)
\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)
=-50
6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)
\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)
\(=-24\cdot5+16\cdot5\)
\(=5\cdot\left(-24+16\right)\)
\(=-5\cdot8=-40\)
7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)
\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)
\(=47\cdot50-23\cdot50\)
\(=50\cdot\left(47-23\right)\)
\(=50\cdot24=1200\)
8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)
\(=-31\cdot\left(47+52+1\right)\)
\(=-31\cdot100=-3100\)
Bài 2:
1) Ta có: \(-17-\left(2x-5\right)=-6\)
\(\Leftrightarrow-17-2x+5+6=0\)
\(\Leftrightarrow-2x-6=0\)
\(\Leftrightarrow-2x=6\)
hay x=-3
Vậy: x=-3
2) Ta có: \(10-2\left(4-3x\right)=-4\)
\(\Leftrightarrow10-8+6x+4=0\)
\(\Leftrightarrow6x+6=0\)
\(\Leftrightarrow6x=-6\)
hay x=-1
Vậy: x=-1
3) Ta có: \(-12+3\left(-x+7\right)=-18\)
\(\Leftrightarrow-12-3x+21+18=0\)
\(\Leftrightarrow-3x+27=0\)
\(\Leftrightarrow-3x=-27\)
hay x=9
Vậy: x=9
4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)
\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)
\(\Leftrightarrow-2x-3=-3\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: x=0
5) Ta có: x(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3\right\}\)
6) Ta có: (x-2)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-1;3\right\}\)
Bài 1:
1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)
=−12−18=−12−18
=-30
2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27
=36−2020+2019−136−27=36−2020+2019−136−27
=1−100−27=1−100−27
=−126
Tớ chcs cậu học thật giỏi nha !
a) \(...=-\dfrac{12}{46}-\dfrac{15}{21}+\dfrac{3}{7}-\dfrac{7}{23}\)
\(=-\dfrac{12}{46}-\dfrac{7}{23}-\dfrac{15}{21}+\dfrac{3}{7}\)
\(=-\dfrac{6}{23}-\dfrac{7}{23}-\dfrac{5}{7}+\dfrac{3}{7}\)
\(=-\dfrac{13}{23}-\dfrac{2}{7}\)
\(=-\dfrac{13.7}{23.7}-\dfrac{2.23}{23.7}\)
\(=-\dfrac{91}{161}-\dfrac{46}{161}=-\dfrac{137}{161}\)
b) \(...=-21+\dfrac{7}{9}-\dfrac{3}{17}+\dfrac{25}{9}\)
\(=-21-\dfrac{3}{17}+\dfrac{7}{9}+\dfrac{25}{9}\)
\(=-21-\dfrac{3}{17}+\dfrac{32}{9}\)
\(=-\dfrac{3213}{153}-\dfrac{27}{153}+\dfrac{544}{153}=-\dfrac{2696}{153}\)
Đặt \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(\Rightarrow A=\frac{5^2}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+\frac{1}{26}-\frac{1}{31}\right)\)
\(\Rightarrow A=5.\left(1-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)
Bài 1:
1: =15+37+52-37-17=52-2=50
2: =38-42+14-25+27+15=62-42+29=20+29=49
Bài 1: Bỏ ngoặc rồi tính
3) (21-32) - (-12+32)=21-32-(-12)-32=21-32+12-32=-31
4) (12+21) - (23-21+10)=12+21-23+21-10=21
5) (57-725) - (605-53)=57-725-605+53=-1220
6) (55+45+15) - (15-55+45)=55+45+15-15+55-45=55+55=110
Bài 2: Tính các tổng sau một cách hợp lí
1) (-37) + 14 + 26 + 37=(-37+37)+(14+26)=0+40=40
2) (-24) +6 + 10 + 24=(-24+24)+(6+10)=0+16=16
3) 15 + 23 + (-25) + (-23)=(15-25)+(23-23)=-10+0=-10
4) 60 + 33 + (-50) + (-33)=(60-50)+(33-33)=10+0=10
5) (-16) + (-209) + (-14) + 209=(-16-14)+(-209+209)=-30+0=-30
6) (-12) + (-13) + 36 + (-11)=(-11-12-13)+36=-36+36=0
\(1)\) \(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(2)\) Đặt \(A=4+2^2+2^4+...+2^{20}\)
\(4A=2^4+2^4+2^6+...+2^{22}\)
\(4A-A=\left(2^4+2^4+2^6+...+2^{22}\right)+\left(2^2+2^2+2^4+...+2^{20}\right)\)
\(3A=2^4+2^{22}-2^2-2^2\)
\(3A=2^{22}+2^4-2^3\)
\(A=\frac{2^{22}+2^4-2^3}{3}\)
\(3)\) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\) ( bạn ghi đầy đủ ra nhé ở đây mk viết "..." cho nhanh )
\(=\)\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=\)\(5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\)\(5\left(1-\frac{1}{31}\right)\)
\(=\)\(5.\frac{30}{31}\)
\(=\)\(\frac{150}{31}\)
Chúc bạn học tốt ~
Ta có:
\(\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
_ (1+51)+(6+46)+(11+41)+(16+36)+(21+31)+26
= 52+52+52+52+52+26= 52 x 5+26= 286
_ 5 . ( 5 / 1.6+ 5/ 6.11+ 5/ 11.16+.... + 5/ 26.31)
=5. ( 1/1- 1/6+1/6 -1/11+ 1/11-1/16 +....+1/26-1/31)
= 5. ( 1/1 - 1/31)
= 5. 30/31= 150/31
câu a thì nguyễn thị kim phụng giải đúng rồi còn câu b mình nghĩ cậu ấy làm sai mình sẽ làm lại
b)=5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31
=5.(5/1-5/6+5/6-5/11+5/11-5/16+5/16-5/21+5/21-5/26+5/26-5/31)
=5.(5-5/31)
=5.150/31
=750/31