\(\frac{69}{157}-\left\{2+\left[3+\left(4+5^{-1}\right)^{-1}\right]^{-1}\right\}^{-1}\)

\(=\frac{69}{157}-\left\{2+\left[3+\left(4+\frac{1}{5}\right)^{-1}\right]^{-1}\right\}^{-1}\)

\(=\frac{69}{157}-\left\{2+\left[3+\left(\frac{21}{5}\right)^{-1}\right]^{-1}\right\}^{-1}\)

\(=\frac{69}{157}-\left[2+\left(3+\frac{5}{21}\right)^{-1}\right]^{-1}\)

\(=\frac{69}{157}-\left[2+\left(\frac{68}{21}\right)^{-1}\right]^{-1}\)

\(=\frac{69}{157}-\left(2+\frac{21}{68}\right)^{-1}\)

\(=\frac{69}{157}-\left(\frac{157}{68}\right)^{-1}\)

\(=\frac{69}{157}-\frac{68}{157}\)

\(=\frac{1}{157}\)

\(a.\left(5x+1\right)^2=\left(5x\right)^2+2\cdot5x1+1^2=25x^2+10x+1=x\cdot\left(25x+1\right)+1\)

\(d.\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2+2\cdot\left(ab+ac+bc\right)\)

\(=a^2+b^2+c^2+2.\left(2a+2b+2c\right)=a^2+b^2+c^2+2^2\cdot\left(a+b+c\right)\)

\(c.\left(1-5x\right)^2=1^2-2\cdot1\cdot5x+\left(5x\right)^2=1-10x+25x^2=1-\left(10x-25x^2\right)=1-\left[x\cdot\left(10-25x\right)\right]\)

\(d.\left(x^2-\frac{1}{2}\right)^3=\left(x^2\right)^3-3\left(x^2\right)^2\frac{1}{2}+3x^2\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3=x^6-\frac{3}{2}x^4+\frac{3}{4}x^2+\frac{1}{8}\)

\(e.\left(x+3\right)^3=x^3+3x^23+3x3^2+3^3=x^3+3^2x^2+3^3x+3^3=x^2\left(x+3^2\right)+3^3\cdot\left(x+1\right)\)

trời tự làm đi hỏi suốt ko giỏi đc lên đâu

??? nhưng đang bị rối não vì nhiều bài quá

help me

a)(x+3)-(x-11)+2011>0

   <=>x+3-x+11+2011>0

   <=>(x-x)+(3+11+2011)>0

   <=>0+2025>0

   <=>2025>0

đề bài hình như thiệu bạn ạ

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

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\)

\(\Rightarrow\)\(\frac{a_1}{a_2}=\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\)

\(\Rightarrow\)\(\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\right)^n\) \(\left(1\right)\)

Lại có : 

\(\left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}.....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}.....\frac{a_n}{a_{n+1}}=\frac{a_1.a_2.a_3.....a_n}{a_2.a_3.a_4.....a_{n+1}}=\frac{a_1}{a_{n+1}}\) \(\left(2\right)\)

Từ (1) và (2) suy ra đpcm : \(\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)

Chúc bạn học tốt ~ 

\(6^{15}.24^8.3=\left(2.3\right)^{15}.\left(2^3.3\right)^8.3=2^{15}.3^{15}.2^{24}.3^8.3==2^{39}.3^{24}\)

\(72^{12}=\left(2^3.3^2\right)^{12}=2^{36}.3^{24}\)

Vì \(\left(2^{39}.3^{24}\right)⋮\left(2^{36}.3^{24}\right)\Rightarrow\left(6^{15}.24^8.3\right)⋮72^{12}\)