Ta có: \(A=\frac{1}{101^2}+\frac{1}{102^2}+......\frac{1}{105^2};\frac{1}{2^2.3.5^2.7}\)

\(A>\frac{1}{\left(101.101\right)}+\frac{1}{\left(101.102\right)}+\frac{1}{\left(102.103\right)}+......\frac{1}{\left(104.105\right)}\)

Ta thấy mỗi mẫu đều < thì => sẽ lớn hơn

\(A>\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+........\)

\(A>\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{\left(2^2.3.5^2.7\right)}=B\)

=> gọi vế \(\frac{1}{\left(2^2.2.5^2.7\right)}\) là B

=> A>B

\(\text{Ta có :}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100.101}+\frac{1}{101.102}+.....+\frac{1}{105.106}\)

                \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+....+\frac{1}{105}-\frac{1}{106}\)\

               \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{105}\)

              \(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2100}\)

             \(\text{Mà :}\)\(\frac{1}{2100}=\frac{1}{2^2.3.5^2.7}\)

             \(\text{Nên:}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2^2.3.5^2.7}\)

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)

\(A=5\left(1+5\right)+...+5^{11}\left(1+5\right)\)

\(=6\cdot\left(5+...+5^{11}\right)⋮30\)

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)

khó quá

c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)

d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\) 

l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)

Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)

\(\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)

\(< \frac{1}{100.101}+\frac{1}{101.102}+\frac{1}{102.103}+\frac{1}{103.104}+\frac{1}{104.105}\)

\(< \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)

\(< \frac{1}{100}-\frac{1}{105}=\frac{1}{2100}\)

\(< \frac{1}{2^2.3.5^2.7}\)