a, \(\left(7.x\right).\left(2.x-3\right)=0\)

\(\Rightarrow7.x=0\)hoặc \(2.x-3=0\)

 \(\Rightarrow x=0\)     \(\Rightarrow x=\frac{3}{2}\)

vậy \(x\in\left\{0;\frac{3}{2}\right\}\)ghi hoặc ở giữa 0 và \(\frac{3}{2}\)nha

kkk cho mình nhé

b, \(\frac{1}{3}.x+\frac{2}{5}.\left(x+1\right)=0\)

\(\frac{1}{3}.x+\frac{2}{5}.x+\frac{2}{5}=0\)

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

\(x.\frac{11}{15}=\frac{-2}{5}\)

\(x=\frac{-2}{5}:\frac{11}{15}\)

\(x=\frac{-6}{11}\)

ta có x/2=y/5 và x.x.y=160

đặt x/2=y/5=k

=>y=5k    x=2k

vì x.x.y=160

=>2k.2k.5k=160

=>(2.2.5).k³ =160

=>20.k³=160

=>k³=8=2³ =>k=2               

+)x=2.2=4

+)y=2.5=10

cho mình bạn nhé .chúc bạn học tốt

Cảm ơn bạn

Gọi số học sinh mỗi khối 6, 7, 8 lần lượt là \(a,b,c\)(học sinh) \(a,b,c\inℕ^∗\).

Vì số học sinh mỗi khối 6, 7, 8 tỉ lệ nghịch với \(8,9,12\)nên \(8a=9b=12c\Leftrightarrow\frac{a}{9}=\frac{b}{8}=\frac{c}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{a}{9}=\frac{b}{8}=\frac{c}{6}=\frac{a-c}{9-6}=\frac{120}{3}=40\)

\(\Leftrightarrow\hept{\begin{cases}a=40.9=360\\b=40.8=320\\c=40.6=240\end{cases}}\).

4/5 . 7 + 4/7 . 9 + ...+ 4/59 . 61 

= 2 . ( 2/5 . 7 + 2/7 . 9 + ...+ 2/59 . 61 ) 

= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 + ...+ 1/59 - 1/61 ) 

= 2 . ( 1/5 - 1/61 ) 

= 2 . 56/305 

=  112/305 

\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)

\(=\) \(2.\left(\frac{2}{5}.7+\frac{2}{7}.9+...+\frac{2}{59.61}\right)\)

\(=\) \(2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=\) \(2.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=\) \(2.\frac{56}{305}\)

\(=\) \(\frac{112}{305}\)

Từ 3x = 4y - 2z = 7z - 4y <=> 3x = 8y = 9z.

=> Ta có:+\(\frac{x}{8}\)\(=\frac{y}{3}\)

+\(\frac{y}{9}\)\(=\frac{z}{8}\)\(=>\frac{x}{24}\)\(=\frac{y}{9}\)\(=\frac{z}{8}\)

\(\frac{x+y-2z}{24+9-2x8}\)\(=\frac{10}{17}\)

\(=>x=\frac{240}{17}\)

\(y=\frac{90}{17}\)

\(z=\frac{80}{17}\)

=>...................

vì 7y = 5z => y/5 = z/7

=> y/5 = z/7 = x/3

=> x/3 = y/5 = z/7 = x - y + z / 3 -5 + 7=45 /5 = 9

=> x/3 = 9 => x =27

=> y/5 = 9 => x = 45

=> z/7 = 9 => z =63

\(2^2\)\(A=2^2\)\(\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-....+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(4A\)\(=1-\frac{1}{2^2}\)\(+\frac{1}{2^4}\)\(-\frac{1}{2^6}\)\(+.....-\frac{1}{2^{4n-2}}\)\(+\frac{1}{2^{4n}}\)\(-....-\frac{1}{2002}\)

\(4A+A=\)\(..........\)

\(5A=1-\frac{1}{2^{2004}}\)

\(A=\left(1-\frac{1}{2^{2004}}\right)\)\(:5\)

\(A=\frac{1}{5}\)\(-\frac{1}{5}\)\(x\frac{1}{2^{2004}}\)\(< \frac{1}{5}\)\(\left(=0,2\right)\)

\(=>A< 0,2\)

Ta có :

\(S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)

\(\Rightarrow2^2S=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)

\(\Rightarrow4S+S=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\right)\)

\(+\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow5S=1-\frac{1}{2^{2004}}< 1\)\(\Rightarrow S=\frac{1-\frac{1}{2^{2004}}}{5}< \frac{1}{5}=0,2\)

bạn tham khảo nha