Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x}{3}=\frac{y}{6}=\frac{z}{8}=\frac{3x}{9}=\frac{2y}{12}=\frac{3x-2y-z}{9-12-8}=\frac{20}{-11}\)

=>x=60/-11; y=120/-11; z=160/-11

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

\(\frac{x}{3}=\frac{y}{6}=\frac{z}{8}=\frac{3x-2y-z}{3\times3-2\times6-8}=\frac{20}{-11}\)

Do đó: \(x=\frac{-60}{11}\), \(y=\frac{-120}{11}\),\(z=\frac{-160}{11}\)

\(\Rightarrow\left(-\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^{10}:\left(\frac{3}{2}\right)^6\Rightarrow\left(-\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^4=\left(-\frac{3}{2}\right)^4\)

\(\Rightarrow y=4\)

\(\left(\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^{10}:\left(\frac{3}{2}\right)^6\Rightarrow\left(\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^4\)

\(\Rightarrow\)y=4

x⁶.y⁶=64  (x,y khác 0)

<=> (x.y)⁶=2⁶ (64=2⁶)

=> x.y=2 => x=2/y

Lại có: \(\frac{x^3+y^3}{6}=\frac{x^3-2y^2}{4}\) <=> \(\frac{x^3+y^3}{3}=\frac{x^3-2y^2}{2}\)

<=> 2x³+2y³=3x³-6y²

<=> 2y³=x³-6y² . Thay x=y/2 vào ta được:

\(2y^3=\frac{y^3}{8}-6y^2\)  <=> 16y³=y³-48y² 

<=> 15y³+48y² =0

<=> y²(15y+48)=0

Do y khác 0 => 15y+48=0 => \(y=-\frac{48}{15}=-\frac{16}{5}\)

x=y/2 => \(x=-\frac{8}{5}\)

Đáp số: \(x=-\frac{8}{5}\);  \(y=-\frac{16}{5}\)

\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)

\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)

\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)

\(\frac{2}{7}:y=\frac{10}{21}.\frac{9}{14}\)

\(\frac{2}{7}:y=\frac{15}{49}\)

\(y=\frac{2}{7}:\frac{15}{49}\)

\(y=\frac{2}{7}.\frac{49}{15}\)

\(y=\frac{14}{15}\)

\(y-\frac{1}{3}=\frac{10}{21}:\frac{15}{28}\)

\(y-\frac{1}{3}=\frac{10}{21}.\frac{28}{15}\)

\(y-\frac{1}{3}=\frac{8}{9}\)

\(y=\frac{8}{9}+\frac{1}{3}\)

\(y=\frac{8}{9}+\frac{3}{9}\)

\(y=\frac{11}{9}\)

a) Ta có :  \(y-\frac{1}{3}=\frac{10}{21}\div\frac{15}{28}\)

\(\Rightarrow\)       \(y-\frac{1}{3}=\frac{8}{9}\)

\(\Rightarrow\)       \(y\)        \(=\frac{8}{9}+\frac{1}{3}\)

\(\Rightarrow\)      \(y\)         \(=\frac{11}{9}\)

Vậy \(y=\frac{11}{9}\)

b) Ta có : \(\frac{2}{7}\div y=\frac{10}{21}\times\frac{9}{14}\) 

\(\Rightarrow\)      \(\frac{2}{7}\div y=\frac{15}{49}\)

\(\Rightarrow\)                  \(y=\frac{2}{7}\div\frac{15}{49}\)      

\(\Rightarrow\)                  \(y=\frac{14}{15}\)

Vậy \(y=\frac{14}{15}\)

                     Cbht !!!

Ta có: 

\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{99^2}{99.100}.\frac{100^2}{100.101}\)

\(=\frac{1}{2}.\frac{4}{6}.\frac{9}{12}....\frac{9801}{9900}.\frac{10000}{10100}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}.\frac{100}{101}=\frac{1.2.3...99.100}{2.3.4...100.101}=\frac{1}{101}\)(Tối giản)

1. n = 15;16;17;18;19;20;21;22;23;24 

2. y = 20

Bạn tự đăng câu hỏi thì bạn phải tự trả lời chứ.

k nha !