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mong bạn giúp

Giả sử: \(9^n+63=x^2\)

+) Xét n=2k+1 (lẻ):

\(9^{2k+1}+63=9^{2k}.9+63\equiv\left(-1\right)^{2k}.9+3\equiv2\)(mod 5) -> vô lí vì scp không đòng dư với 2 mod 5 -> n=2k

+) Xét n=2k:

\(9^{2k}+63=x^2\Leftrightarrow x^2-9^{2k}=63\Leftrightarrow\left(x-9^k\right)\left(x+9^k\right)=63\)
Đến đây bạn lập bảng là ra nhé!

Ta có:\(2n^4+3n^2+1=\left(n^2\right)^2+2n^21^2+1^2+\left(n^4+n^2\right)=\left(n^2+1\right)^2+n^2\left(n^2+1\right)\)

\(=\left(n^2+1\right)\left(2n^2+1\right)\)

Vì \(\left(n^2+1\right)\left(2n^2+1\right)\)mà \(2n^2+1\ge n^2+1\)

\(\Rightarrow2n^2+1⋮n^2+1\)

\(\Rightarrow2n^2+2-1=2\left(n^2+1\right)-1⋮n^2+!\)

\(\Rightarrow-1⋮n^2+1\)

Mà \(n^2+1>0\)

\(\Rightarrow n^2+1=1\Rightarrow n=0\)

\(2x^3+5x^2-36=0\)

\(\Leftrightarrow2x^3+9x^2-4x^2+18x-18x-36=0\)

\(\Leftrightarrow\left(2x^3+9x^2+18x\right)-\left(4x^2+18x+36\right)=0\)

\(\Leftrightarrow x\left(2x^2+9x+18\right)-2\left(2x^2+9x+18\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2+9x+18\right)=0\)

\(TH1:x-2=0\Leftrightarrow x=2\)

\(TH2:2x^2+9x+18=0\)

Ta có: \(\Delta=9^2-4.2.18=-63< 0\)

Vậy TH2 ko có nghiệm

Vậy x = 2

thì làm sao???Hỏi xong rồi tự trả lời thì có ích gì

(✿◠‿◠)(๛ČℌUƔÊŇ♥Ť❍Ą́Ňツ)

Ê nhóc đừng có nghĩ lung tung 

x⁴y⁴ + 64

= x⁴y⁴ + 16x²y² + 64 - 16x²y²

= (x²y² + 8)² - (4xy)²

= (x²y² - 4xy + 8)(x²y² + 4xy + 8)

a) Ta có:

\(\frac{1}{2\left(m+1\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{3m+2}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}\)

\(+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{3m+3}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{3\left(m+1\right)}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{3}{2\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{3\left(8m+5\right)}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{24m+15}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{24m+16}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{8\left(3m+2\right)}{2\left(3m+2\right)\left(8m+5\right)}\)

\(=\frac{8}{2\left(8m+5\right)}=\frac{4}{8m+5}\left(đpcm\right)\)

b) Ta có: \(\frac{1}{m+1}+\frac{1}{3m+2}+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)

\(=\frac{3m+2}{\left(m+1\right)\left(3m+2\right)}+\frac{m+1}{\left(m+1\right)\left(3m+2\right)}\)

\(+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)

\(=\frac{4m+4}{\left(m+1\right)\left(3m+2\right)}\)

\(=\frac{4\left(m+1\right)}{\left(m+1\right)\left(3m+2\right)}\)

\(=\frac{4}{3m+2}\left(đpcm\right)\)