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

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

C1 : Cardano (mk chưa học )

C2 : Mode set up -> 5 -> ax^3 + bx^2 + cx + d = 0 

PT <=> \(x_1=-1,209...;x_2=2,104....\)

dùng lượng giác hóa cũng được các bạn nhé 

a) \(5\left(x-2\right)>3\left(x-4\right)\)

\(\Leftrightarrow5x-10>3x-12\)

\(\Leftrightarrow2x>-2\)

\(\Rightarrow x>-1\)

b) \(7\left(x+3\right)< 9\left(x-1\right)\)

\(\Leftrightarrow7x+21< 9x-9\)

\(\Leftrightarrow2x>30\)

\(\Rightarrow x>15\)

c) Vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\left(\forall x\right)\)

=> \(2x-5>0\Rightarrow2x>5\Rightarrow x>\frac{5}{2}\)

d) \(x^2-2x+5=\left(x-1\right)^2+4>0\left(\forall x\right)\)

\(\Rightarrow3x-8< 0\Rightarrow3x< 8\Rightarrow x< \frac{8}{3}\)

Đặt \(A=x^2+15y^2+xy+8x+y+2020\)

\(\Rightarrow4A=4x^2+60y^2+4xy+32x+4y+8080\)

\(=\left(4x^2+4xy+y^2\right)+59y^2+32x+4y+8080\)

\(=\left(2x+y\right)^2+16.\left(2x+y\right)+64+59y^2+4y-16y+8016\)

\(=\left(2x+y+8\right)^2+59y^2-12y+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y^2-\frac{59}{12}y\right)+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y^2-2\cdot y\cdot\frac{59}{24}+\frac{59^2}{24^2}-\frac{59^2}{24^2}\right)+8016\)

\(=\left(2x+y+8\right)^2+59\cdot\left(y-\frac{59}{24}\right)^2+7659,439236\ge7659,439236\)

\(\Rightarrow A\ge1914,859809\)

Dấu "=" xảy ra \(\Leftrightarrow y=\frac{59}{14};x=-\frac{171}{28}\)

P/s : Bài này hơi xấu .....

Đặt \(A=x^2+15y^2+xy+8x+y+2020\)

Ta có: \(A=x^2+x\left(y+8\right)+15y^2+y+2020=\left(x^2+x\left(y+8\right)+\frac{\left(y+8\right)^2}{4}\right)\)\(+\left(15y^2+y-\frac{\left(y+8\right)^2}{4}\right)+2020=\left(x+\frac{y+8}{2}\right)^2+\frac{59y^2-12y-64}{4}+2020\)\(=\left(x+\frac{y+8}{2}\right)^2+\frac{59\left(y-\frac{6}{59}\right)^2-\frac{3812}{59}}{4}+2020\ge\frac{\frac{-3812}{59}}{4}+2020=\frac{118227}{59}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}y-\frac{6}{59}=0\\x=-\frac{y+8}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-239}{59}\\y=\frac{6}{59}\end{cases}}\)

C = 25x² + 20x + 5/2

C = 25( x² + 4/5x + 4/25 ) - 3/2

C = 25( x + 2/5 )² - 3/2

25( x + 2/5 )² ≥ 0 ∀ x => 25( x + 2/5 )² - 3/2 ≥ -3/2

Đẳng thức xảy ra <=> x + 2/5 = 0 => x = -2/5

=> MinC = -3/2 <=> x = -2/5

\(C=25x^2+20x+\frac{5}{2}=25x^2+20x+4-\frac{3}{2}\)

\(=25\left(x+\frac{2}{5}\right)^2-\frac{3}{2}\)

Vì \(\left(x+\frac{2}{5}\right)^2\ge0\forall x\)\(\Rightarrow25\left(x+\frac{2}{5}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow25\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy Cmin = -3/2 <=> x = -2/5