Giải:

a)A=1/56+1/72+1/90+1/110+1/132+1/156

   A=1/7.8+1/8.9+1/9.10+1/10.11+1/11.12+1/12.13

   A=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13

   A=1/7-1/13

  A=6/91

b)B=4/21+4/77+4/165+4/285+4/437+4/621

   B=4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27

   B=1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27

   B=1/3-1/27

   B=8/27

c) C=1/21+1/77+1/165+1/285+1/437+1/621

    C=1/3.7+1/7.11+1/11.15+1/15.19+1/19.23+1/23.27

    C=1/4.(4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27)

    C=1/4.(1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27)

    C=1/4.(1/3-1/27)

    C=1/4.8/27

    C=2/27

d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/21.26+1/26.31

    D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31)

    D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31)

    D=1/5.(1/1-1/31)

    D=1/5.30/31

    D=6/31

Nếu câu d cậu viết thiếu thì làm như vầy nhé!

Chúc bạn học tốt!

Nếu như câu d ko chép sai thì làm thế này nha:

d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/26.31

    D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21)+1/806

    D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)+1/806

    D=1/5.(1/1-1/21)+1/806

    D=1/5.20/21+1/806

    D=4/21+1/806

    D=3245/16926

Chúc bạn học tốt!

\(C=\dfrac{1}{21}+\dfrac{1}{77}+\dfrac{1}{165}+\dfrac{1}{285}+\dfrac{1}{437}+\dfrac{1}{621}\) 

\(C=\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\dfrac{1}{15.19}+\dfrac{1}{19.23}+\dfrac{1}{23.27}\) 

\(C=\dfrac{4}{4}.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+\dfrac{1}{15.19}+\dfrac{1}{19.23}+\dfrac{1}{23.27}\right)\)  \(C=\dfrac{1}{4}.\left(\dfrac{4}{3.7}+\dfrac{4}{7.11}+\dfrac{4}{11.15}+\dfrac{4}{15.19}+\dfrac{4}{19.23}+\dfrac{4}{23.27}\right)\) \(C=\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{27}\right)\) \(C=\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{27}\right)\) 

\(C=\dfrac{1}{4}.\left(\dfrac{9}{27}-\dfrac{1}{27}\right)\) 

\(C=\dfrac{1}{4}.\dfrac{8}{27}=\dfrac{2}{27}\)  

Mà \(\dfrac{2}{27}< \dfrac{4}{27}\) hay \(C< \dfrac{4}{27}\) 

\(\Rightarrow C< \dfrac{4}{27}\) ( điều phải chứng tỏ ) 

A = \(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+\frac{1}{285}+\frac{1}{437}\)

A = \(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+\frac{1}{15.19}+\frac{1}{19.23}\)

A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)+\frac{1}{4}.\left(\frac{1}{7}-\frac{1}{11}\right)+\frac{1}{4}.\left(\frac{1}{11}-\frac{1}{15}\right)+\frac{1}{4}.\left(\frac{1}{15}-\frac{1}{19}\right)+\frac{1}{4}.\left(\frac{1}{19}-\frac{1}{23}\right)\)

A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}\right)\)

A = \(\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{23}\right)\)

A = \(\frac{1}{4}.\frac{20}{69}\)

A = \(\frac{5}{69}\)

A=\(\frac{5}{69}\)

1.

a) \(125+80+375+220\)

\(=\left(125+375\right)+\left(80+220\right)\)

\(=500+300\)

\(=800\)

b) \(25.11.8.4.125\)

\(=\left(25.4\right).\left(8.125\right).11\)

\(=100.1000.11\)

\(=1100000\)

c) \(18.74+18.27-18\)

\(=18.\left(74+27-1\right)\)

\(=18.100\)

\(=1800\)

d) \(75.23+75.77-500\)

\(=75.\left(23+77\right)-500\)

\(=75.100-500\)

\(=7500-500\)

\(=7000\)

e) \(150:\left[25.\left(18-4^2\right)\right]\)

\(=150:\left[25.\left(18-16\right)\right]\)

\(=150:\left[25.2\right]\)

\(=150:50\)

\(=3\)

f) \(125.23.2.8.50\)

\(=\left(125.8\right).\left(2.50\right).23\)

\(=1000.100.23\)

\(=2300000\)

g) \(235+88+165+12\)

\(=\left(235+165\right)+\left(88+12\right)\)

\(=400+100\)

\(=500\)

h) \(71.32+71.68-1100\)

\(=71.\left(32+68\right)-1100\)

\(=71.100-1100\)

\(=7100-1100\)

\(=6000\)

i) \(6^7:6^6+4^3.2-24^0\)

\(=6+64.2-1\)

\(=6+128-1\)

\(=133\)

j) \(546-6.\left[158:\left(30+7^2\right)\right]\)

\(=546-6.\left[158:\left(30+49\right)\right]\)

\(=546-6.\left[158:79\right]\)

\(=546-6.2\)

\(=546-12\)

\(=534\)

k) \(268+147+132+253\)

\(=\left(268+132\right)+\left(147+253\right)\)

\(=400+400\)

\(=800\)

2. 

a) \(7x=707\)

\(x=707:7\)

\(x=101\)

b) \(6x+5=47\)

\(6x=47-5\)

\(6x=42\)

\(x=42:6\)

\(x=7\)

c) \(35:\left(x+1\right)=7\)

\(x+1=35:7\)

\(x+1=5\)

\(x=5-1\)

\(x=4\)

d) \(12+\left(29-3x\right)=35\)

\(29-3x=35-12\)

\(29-3x=23\)

\(3x=29-23\)

\(3x=6\)

\(x=6:3\)

\(x=2\)

e) \(2.4^x+10^1=138\)

\(2.4^x=138-10\)

\(2.4^x=128\)

\(4^x=128:2\)

\(4^x=64=4^3\)

\(\Rightarrow x=3\)

f) \(7x=140\)

\(x=140:7\)

\(x=20\)

g) \(7\left(15-x\right)+14=84\)

\(7\left(15-x\right)=84-14\)

\(7\left(15-x\right)=70\)

\(15-x=70:7\)

\(15-x=10\)

\(x=15-10\)

\(x=5\)

h) \(3^x.5-15=390\)

\(3^x.5=390+15\)

\(3^x.5=405\)

\(3^x=405:5\)

\(3^x=81=3^4\)

\(\Rightarrow x=4\)

3.

Ta có:

\(\Rightarrow n+11⋮n+1\)

\(\Rightarrow n+1+10⋮n+1\)

\(\Rightarrow10⋮n+1\) hay \(n+1\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

\(\Rightarrow\) \(n\in\left\{-11;-6;-3;-2;0;1;4;9\right\}\)

Vì \(n\in N\) nên \(n\in\left\{0;1;4;9\right\}\)

 Ta có 1n2+4n=14(1n−1n+4)1n2+4n=14(1n−1n+4) Khi đó pt tương đương: 14(13−17+17−111+...+1n−1n+4)=5667314(13−17+17−111+...+1n−1n+4)=56673 ⟺13−1n+4=224673=>n=2015

Sai rồi

\(B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{1}{5}B=\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)

\(B-\frac{1}{5}B=\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}-\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\)\(\frac{1}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{21}+\frac{4}{77}+\frac{4}{165}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\)

\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{7}+\frac{4}{7}-\frac{4}{11}+\frac{4}{11}-\frac{4}{15}+...+\frac{4}{4n-1}-\frac{4}{4n+3}\)

\(\frac{4}{5}B=\frac{4}{3}-\frac{4}{4n-3}\)

\(\frac{4}{5}B=\frac{16n-24}{12n-9}\)

\(B=\frac{\frac{16n-24}{12n-9}}{\frac{4}{5}}\)

\(B=\frac{20n-30}{12n-9}\)

B = \(\frac{5}{21}+\frac{5}{77}+\frac{5}{165}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{3.7}+\frac{5}{7.11}+\frac{5}{11.15}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}.\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}+\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}.\left(\frac{1}{3}-\frac{1}{4n+3}\right)=\frac{5}{12}-\frac{5}{4\left(4n+3\right)}=\frac{5}{12}-\frac{5}{16n+12}\)

sửa lại

\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)

\(\frac{4}{5}B=\frac{1}{3}-\frac{1}{4n+3}\)

\(\frac{4}{5}B=\frac{4n}{12n+9}\)

\(B=\frac{\frac{4n}{12n+9}}{\frac{4}{5}}\)

\(B=\frac{5n}{12n+9}\)

\(\frac{8}{19}\)

giải cả bài ra

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