sửa đề phải là tìm x,y bt nha và x,y nguyên

a) ta có \(\left\vert x-24\right\vert\ge0\)

\(\left\vert y+8\right\vert\ge0\)

=> TH1: \(\left\vert x-24\right\vert=0\) và \(\left\vert y+8\right\vert=1\)

=> x=24

TH1a: y+8 =1

=>y=-7

TH1b: y+8=-1

=> y=-9

TH2: \(\left\vert x-24\right\vert=1\) và \(\left\vert y+8\right\vert=0\)

TH2a: x-24=1

=> x=25

TH2b: x-24=-1

=>x=23

=>y=-8

b) ta có số mũ chẵn luôn lớn hơn 0

=> \(\left(x-2\right)^{10}\ge0\)

\(\left\vert y-2\right\vert\ge0\)

=> \(\left(x-2\right)^{10}=0\) và \(\left\vert y-2\right\vert=0\)

=> x=2 và y=2

c) ta có số số ở đây là: (30-0):1+1=31( số)

<=> \(31x+\left(1+2+3+..+30\right)=1240\)

mà trong ngoặc có số số là: (30-1):1+1=30( số)

áp dụng công thức tính dãy số liên tiếp ta có:

trong ngoặc=\(\frac{30\left(30+1\right)}{2}=465\)

thay vào biểu thức ta vừa suy ra ở trên ta có:

\(31x+465=1240\)

=> \(31x=775\)

=>\(x=25\)

\(A=\left(1-\frac{1}{2018}\right)\left(1-\frac{2}{2018}\right)\left(1-\frac{3}{2018}\right)...\left(1-\frac{2020}{2018}\right).\)

   \(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot\left(1-\frac{2018}{2018}\right)\cdot...\cdot\frac{-2}{2018}\)

   \(=\frac{2017}{2018}\cdot\frac{2016}{2018}\cdot\frac{2015}{2018}\cdot...\cdot0\cdot...\cdot\frac{-2}{2018}\)

   \(=0\)

Bài 1:

ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{100^2}\)

\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)

\(\Rightarrow B< \frac{1}{4}\)

Bài 2:

ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Học tốt nhé bn !!

\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)

= \(\left(\frac{1}{20}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)

= \(0\cdot\left(\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)

Đặt \(\frac{2017}{2018}-\frac{2018}{2019}=A\)

Ta có : 
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)

\(=\left(\frac{5}{20}-\frac{4}{20}-\frac{1}{20}\right).A\)

\(=\left(\frac{1}{20}-\frac{1}{20}\right).A\)

\(=0.A\)

\(=0\)

Vậy ...

Chúc bạn học tốt !!! 

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow x+2=41\)

\(\Leftrightarrow x=41-2\)

\(\Leftrightarrow x=39\)

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C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)

c=\(\frac{1}{1}-\frac{1}{10}\)

c=\(\frac{9}{10}\)

còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!