Ta có:

+) \(a+b=5\)

\(\Rightarrow\left(a+b\right)^2=25\)

\(\Rightarrow a^2+b^2+2ab=25\)

\(\Rightarrow a^2+b^2=13\)

+) \(a+b=5\)

\(\Rightarrow\left(a+b\right)^3=125\)

\(\Rightarrow a^3+3ab\left(a+b\right)+b^3=125\)

\(\Rightarrow a^3+b^3=35\)

+) \(\left(a^2+b^2\right)\left(a^3+b^3\right)=13.35\)

\(\Rightarrow a^5+a^2b^3+b^2a^3+b^5=455\)

\(\Rightarrow a^5+b^5+a^2b^2\left(a+b\right)=455\)

\(\Rightarrow a^5+b^5=275\)

Vậy...

a, sai đề

b, \(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Rightarrow\dfrac{1}{42}+\dfrac{1}{56}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\) ( nhân cả 2 vế với \(\dfrac{1}{2}\) )

\(\Rightarrow\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\Rightarrow x+1=18\Rightarrow x=17\)

Vậy x = 17

\(S=\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{50^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\dfrac{1}{3}-\dfrac{1}{50}=\dfrac{47}{150}\) (1)\(S=\dfrac{1}{4^2}+\dfrac{1}{5^2}...+\dfrac{1}{50^2}>\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{51}\)

\(=\dfrac{1}{4}-\dfrac{1}{51}=\dfrac{47}{204}\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{47}{150}>S>\dfrac{47}{204}\)

\(\Rightarrow S\notin Z\)

\(\Rightarrowđpcm\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)

\(\Rightarrow\dfrac{y}{x+z+1}=\dfrac{1}{2}\)

\(\Rightarrow2y=x+z+1\)

\(\Rightarrow3y=x+y+z+1\)

\(\Rightarrow3y=\dfrac{1}{2}+1\)

\(\Rightarrow y=\dfrac{1}{2}\)

Vậy...

\(B=3+3^2+...+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=\left(3+3^3+...+3^{99}\right).4⋮4\)

\(\Rightarrowđpcm\)

Thôi viets cho mik kết bài cũng dduocj 

Hay nhé rồi mik tick cho

\(x^2-7xy+12y^2\)

\(=x^2-4xy-3xy+12y^2\)

\(=x\left(x-4y\right)-3y\left(x-4y\right)\)

\(=\left(x-3y\right)\left(x-4y\right)\)

a, \(y^3+y^2-9y-9=y^2\left(y+1\right)-9\left(y+1\right)\)

\(=\left(y^2-9\right)\left(y+1\right)=\left(y-3\right)\left(y+3\right)\left(y+1\right)\)

b, \(y^2+3y+2=y^2+2y+y+2=y\left(y+2\right)+y+2\)

\(=\left(y+1\right)\left(y+2\right)\)

\(P=\dfrac{x^2-1}{x^2+1}=1-\dfrac{2}{x^2+1}\)

Để P nhỏ nhất thì \(\dfrac{2}{x^2+1}\) lớn nhất

Ta có: \(x^2+1\ge1\)

\(\Rightarrow\dfrac{2}{x^2+1}\le2\)

\(\Rightarrow P=1-\dfrac{2}{x^2+1}\le1-2=-1\)

Dấu " = " khi \(x^2=0\Rightarrow x=0\)

Vậy \(MIN_P=-1\) khi x = 0

cách giải đây
1 tấn đường=1000kg
Buổi sáng bán được số đường là
1000x2:5=400
Buổi chiều bán được số đường là
400x1,5=600
Đáp số 600 kg đường
Tick tớ nhá