1) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)\)

\(=27x^6+9x^4+3x^2-9x^4-3x^2-1\)

\(=27x^6-1\) (hằng đẳng thức dạng a³ - b³)

2) \(\left(x^2-4\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\) 

\(=\left(x-2\right)\left(x+2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)

\(=\left[\left(x-2\right)\left(x^2+2x+4\right)\right].\left[\left(x+2\right)\left(x^2-2x+4\right)\right]\)

\(=\left(x^3-8\right)\left(x^3+8\right)\)

\(=x^6-64\)

a) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)=\left(3x^2-1\right)\left[\left(3x^2\right)^2+3x^2.1+1^2\right]=\left(3x^2\right)^3-1^3=3x^6-1\)

b) \(\left(x^2-4\right).\left(x^2+2x+4\right).\left(x^2-2x+4\right)=\left(x^2-2^2\right).\left(x+2\right)^2.\left(x-2\right)^2=\left(x+2\right).\left(x-2\right).\left(x+2\right)^2.\left(x-2\right)^2=\left(x+2\right)^3.\left(x-2\right)^3\)

a) \(\left(x^2+4x+3\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x-2\right)\left(x-3\right)=0\)

=> \(\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) hoặc \(\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\) hoặc \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Vậy tập nghiệm PT \(S=\left\{-3;-1;2;3\right\}\)

b) \(\left(x^2-7x+12\right)\left(x^2+8x+7\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x+1\right)\left(x+7\right)=0\)

=> \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\) hoặc \(\orbr{\begin{cases}x+1=0\\x+7=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=4\end{cases}}\) hoặc \(\orbr{\begin{cases}x=-1\\x=-7\end{cases}}\)

Vậy tập nghiệm PT \(S=\left\{-7;-1;3;4\right\}\)

a, \(\left(x^2+4x+3\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1;-3\\x=3;2\end{cases}}\)

b, \(\left(x^2-7x+12\right)\left(x^2+8x+7\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x+1\right)\left(x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=4;3\\x=-1;-7\end{cases}}\)

Bạn tham khảo tại link này nha, mình giải rất chi tiết cả 3 câu a; b; c rồi đó nhaaaaaa !!!!!

Link nè: https://olm.vn/hoi-dap/detail/261219264881.html

Bài làm:

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

\(=-\left(2x+1\right)^2-1\le-1< 0\left(\forall x\right)\)

=> đpcm

b) \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

a) Ta có: \(-4x^2-4x-2=-\left(4x^2+4x+1\right)-1\)

                                           \(=-\left(2x+1\right)^2-1\)

    Vì \(-\left(2x+1\right)^2\le0\forall x\)\(\Rightarrow\)\(-\left(2x+1\right)^2-1\le-1\forall x\)

              \(\Rightarrow\)\(-\left(2x+1\right)^2-1< 0\forall x\)

              \(\Rightarrow\)\(-4x^2-4x-2< 0\forall x\)( ĐPCM )

b) Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

        \(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

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

    Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{cases}}\)\(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1\ge1\forall x,y,z\)

          \(\Rightarrow\)\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2+1>0\forall x,y,z\)( ĐPCM )

đkxđ: \(a\ne\pm3\)

\(P=\left(\frac{a}{a+3}+\frac{3-a}{a+3}+\frac{a^2+3a+9}{a^2-9}\right)\div\frac{3}{a+3}\)

\(P=\left[\frac{3}{a+3}+\frac{a^2+3a+9}{\left(a-3\right)\left(a+3\right)}\right].\frac{a+3}{3}\)

\(P=\frac{3\left(a-3\right)+a^2+3a+9}{\left(a-3\right)\left(a+3\right)}.\frac{a+3}{3}\)

\(P=\frac{a^2+6a}{3\left(a-3\right)}\)

Đề nghị xem lại đề

Bài làm:

Ta có: \(x^2-22x+127=\left(x^2+22x+121\right)+6=\left(x+11\right)^2\ge6\left(\forall x\right)\)

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(\left(\sqrt{x-2}+\sqrt{20-x}\right)^2\le\left(1^2+1^2\right)\left[\left(\sqrt{x-2}\right)^2+\left(\sqrt{20-x}\right)^2\right]\)

\(=2\left(x-2+20-x\right)=2.18=36\)

\(\Rightarrow\sqrt{x-2}+\sqrt{20-x}\le\sqrt{36}=6\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-11\right)^2\\x-2=20-x\end{cases}}\Rightarrow x=11\)

đkxđ: \(2\le x\le22\) 

Bài đâu

bạn vào câu hỏi của tôi sửa đề bài đi nhé 

cảm ơn

Ta có E = 2x² + 6x - 5

= 2(x² + 3x + 2,25) - 9,5

= 2(x + 1,5)² - 9,5 \(\ge\)-9,5

Dấu bằng xảy ra <=> x + 1,5 = 0 => x = -1,5

Vậy MIN E = -9,5 <=> x = -1,5

a) x = [((n + 1)(n + 4)].[(n + 2)(n + 3)] + 1

= (n² + 5n + 4)(n² + 5n + 6) + 1 

= (n² + 5n + 5 - 1)(n² + 5n + 5 + 1) + 1

= (n² + 5n + 5)² - 1² + 1 = (n² + 5n + 5)² (đpcm)

b) y = [n(n + 9)].[(n + 3)(n + 6)] + 81 

= (n² + 9n).(n² + 9n + 18) + 81

= (n² + 9n + 9 - 9)(n² + 9n + 9 + 9) + 81

= (n² + 9n + 9)² - 9² + 81 = (n² + 9n + 9)² (đpcm)

a) \(x=\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)+1\)    

\(=\left(n+1\right)\left(n+4\right)\left(n+2\right)\left(n+3\right)+1\)  

\(=\left(n^2+5n+4\right)\left(n^2+5n+6\right)+1\)   ( 1 ) 

Đặt \(t=n^2+5n\)     

\(\left(1\right)\Leftrightarrow=\left(t+4\right)\left(t+6\right)+1\)   

\(=t^2+10+24+1\)    

\(=t^2+10t+25\)          

\(=\left(t+5\right)^2\)      

Vậy x là số chính phương 

b)  \(y=n\left(n+3\right)\left(n+6\right)\left(n+9\right)+81\)          

\(=n\left(n+9\right)\left(n+3\right)\left(n+6\right)+81\)    

\(=\left(n^2+9n\right)\left(n^2+9n+18\right)+81\)    ( 1 ) 

Đặt \(a=n^2+9n\)   

\(\Leftrightarrow\left(1\right)=a\left(a+18\right)+81\)       

\(=a^2+18a+81\)         

\(=\left(a+9\right)^2\)               

Vậy y là số chính phương 

A B C D

Ta có: Vì AB // CD

=> \(\widehat{D}=180^0-\widehat{A}=180^0-30^0=150^0\)

Vì \(\widehat{B}+\widehat{C}=180^0\Leftrightarrow2\widehat{C}+\widehat{C}=180^0\Leftrightarrow3\widehat{C}=180^0\)

\(\Rightarrow\widehat{C}=60^0\Rightarrow\widehat{B}=120^0\)