250 : x + 3.5 = 5^2

=> 250:x = 25 - 15 = 10

=> x = 250 : 10 = 25

\(B=\frac{2^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2^2}{195.197}=2\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{195.197}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{195}-\frac{1}{197}\right)=2\left(1-\frac{1}{197}\right)=2.\frac{196}{197}=\frac{392}{197}\)

\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)

\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)

\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)

=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)

a

câu a

nha bạn !

\(\dfrac{4}{2}=\dfrac{2}{1}=\dfrac{16}{8}\)

32

\(\frac{3}{x}+\frac{y}{2}=\frac{1}{16}\)

=>\(\frac{6+xy}{2x}=\frac{1}{16}\)

=>16(xy+6)=2x

=>8(xy+6)=x

=>8xy-x=-48

=>x(8y-1)=-48

=>(x;8y-1)∈{(1;-48);(-48;1);(-1;48);(48;-1);(2;-24);(-24;2);(-2;24);(24;-2);(3;-16);(-16;3);(-3;16);(16;-3);(4;-12);(-12;4);(-4;12);(12;-4);(6;-8);(-8;6);(-6;8);(8;-6)}

=>(x;8y)∈{(1;-47);(-48;2);(-1;49);(48;0);(2;-23);(-24;3);(-2;25);(24;-1);(3;-15);(-16;4);(-3;17);(16;-2);(4;-11);(-12;5);(-4;13);(12;-3);(6;-7);(-8;7);(-6;9);(8;-5)}

=>(x;y)∈{\(\left(1;-\frac{47}{8}\right);\left(-48;\frac14\right);\left(-1;\frac{49}{8}\right);\left(48;0\right);\left(2;-\frac{23}{8}\right);\left(-24;\frac38\right);\left(-2;\frac{25}{8}\right);\left(24;-\frac18\right);\left(3;-\frac{15}{8}\right);\left(-16;\frac12\right)\) ; \(\left(4;-\frac{11}{8}\right);\left(-12;\frac58\right);\left(-4;\frac{13}{8}\right);\left(12;-\frac38\right);\left(6;-\frac78\right);\left(-8;\frac78\right)\) ; \(\left(-6;\frac98\right);\left(8;-\frac58\right)\) }

a.

\(A=B\)

\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Leftrightarrow8\left(x+2\right)=0\)

\(\Leftrightarrow x=-2\left(ktm\right)\)

Vậy không có giá trị x thỏa mãn A=B

b.

\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)

\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{-16}< 0\)

\(\Leftrightarrow\dfrac{8x}{16}>0\)

\(\Leftrightarrow\dfrac{x}{2}>0\)

\(\Leftrightarrow x>0\)