Lời giải:

$(160\times 5-25\times 4):7-3\times x=36\times 25:9-60$

$(800-100):7-3\times x=(36:9)\times 25-60$

$700:7-3\times x=4\times 25-60$

$100-3\times x=100-60$

$100-3\times x=40$

$3\times x=100-40=60$

$x=60:3=20$

(160*5-25*4):7-3xX=40

97xX=40

X=97:40

X=2.245

mong bạn đánh giá kq của mình ^^

 ( 160 \(\times\) 5 - 25 \(\times\) 4 ) \(\div\) 7 - 3 \(\times\) \(x\) = 36 \(\times\) 25 \(\div\) 9 - 60

 ( 800 -100) \(\div\) 7 - 3 \(\times\) \(x\) = 900 \(\div\) 9 - 60

 700 \(\div\) 7 - 3 \(\times\) \(x\) = 100 -60

100 - 3 \(\times\) \(x\) = 40

3 \(\times\) \(x\) = 100 - 40

3 \(\times\) \(x\) = 60

       \(x\) = 60 \(\div\) 3

       \(x\) = 20

a.

12054: ( 15 + 67 )

= 12054 : 82

= 147

29150-136x201

=29150-27336

=1814

b)  9700:100+36x12

=97+432

=529

(160x5-25x4):4

=(800-100):4

= 700:4

=175

k mink nhé

\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{41.43}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{43}\)

\(=\frac{1}{3}-\frac{1}{43}\)

\(=\frac{40}{129}\)

p/s: chúc bạn học tốt

\(B=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)

   \(=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)

    \(=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)

    \(=9.\left(\frac{1}{3}-\frac{1}{675}\right)\)

    \(=9.\frac{224}{675}\)

    \(=\frac{224}{75}\)

THIẾU

\(=\frac{224}{75}< \frac{225}{75}=3\)

Vậy \(B< 3\)

Ta có: \(\frac{1}{\left(2n+1\right)\left(2n+3\right)}-\frac{1}{\left(2n+3\right)\left(2n+5\right)}\)

\(=\frac{2n+5-\left(2n+1\right)}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}\)

\(=\frac{4}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}\)

=>\(\frac{1}{\left(2n+1\right)\left(2n+3\right)\left(2n+5\right)}=\frac14\left(\frac{1}{\left(2n+1\right)\left(2n+3\right)}-\frac{1}{\left(2n+3\right)\left(2n+5\right)}\right)\)

Do đó, ta có:

\(\frac{1}{1\cdot3\cdot5}=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}\right)\)

\(\frac{1}{3\cdot5\cdot7}=\frac14\left(\frac{1}{3\cdot5}-\frac{1}{5\cdot7}\right)\)

...

\(\frac{1}{25\cdot27\cdot29}=\frac14\left(\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right)\)

Do đó: \(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\cdots+\frac{1}{25\cdot27\cdot29}=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+\cdots+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\right)\)

\(=\frac14\left(\frac{1}{1\cdot3}-\frac{1}{27\cdot29}\right)\)

=>\(36\left(\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\cdots+\frac{1}{25\cdot27\cdot29}\right)=36\cdot\frac14\left(\frac{1}{1\cdot3}-\frac{1}{27\cdot29}\right)\)

=>\(\frac{36}{1\cdot3\cdot5}+\frac{36}{3\cdot5\cdot7}+\cdots+\frac{36}{25\cdot27\cdot29}=9\left(\frac13-\frac{1}{27\cdot29}\right)=3-\frac{1}{3\cdot29}=\frac{3\cdot3\cdot29-1}{3\cdot29}=\frac{260}{87}\)

1: =-2/9(15/17+2/17)=-2/9

2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)

=-2-7/30=-67/30

3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

1: =-2/9(15/17+2/17)=-2/9

2: 

=-2-7/30=-67/30

3: 

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

a; 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)