Rình mãi ms được 1 câu!
Bài 3:
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
Đặt \(A=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(A=\left[\left(x+1\right).\left(x+7\right)\right].\left[\left(x+3\right).\left(x+5\right)\right]+15\)
\(A=\left(x^2+7x+x+7\right).\left(x^2+5x+3x+15\right)+15\)
\(A=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)
Đặt \(t=x^2+8x+7\Rightarrow t+8=x^2+8x+15\)
\(\Rightarrow A=t.\left(t+8\right)+15\)
\(A=t^2+8t+15=t^2+3t+5t+15\)
\(A=\left(t^2+3t\right)+\left(5t+15\right)=t.\left(t+3\right)+5.\left(t+3\right)\)
\(A=\left(t+3\right).\left(t+5\right)\)
Vì \(t=x^2+8x+7\) nên
\(A=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)
\(A=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)
\(A=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)
\(A=\left(x^2+8x+10\right).\left[\left(x^2+2x\right)+\left(6x+12\right)\right]\)
\(A=\left(x^2+8x+10\right).\left[x.\left(x+2\right)+6.\left(x+2\right)\right]\)
\(A=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)
Chúc bạn học tốt!!!
Lâu lâu lm 1 câu cho vui!
Câu 2:
Ta có: \(x^2+4y^2+z^2-2x-6z-8y+15\)
= \(\left(x^2-2x+1\right)+\left(4y^2-8y+4\right)+\left(z^2-6x+9\right)+1\)
= \(\left(x-1\right)^2+\left(2y-2\right)^2+\left(z-3\right)^2+1\) \(\ge\) 1 > 0
=> đpcm
(x-1)2\(\ge0\);4(y-1)2\(\ge0\);(z-3)2\(\ge0\) nên (x-1)2+4(y-1)2+(z-3)2\(\ge0\)
=>(x-1)2+4(y-1)2+(z-3)2+1\(\ge1\)>0(đpcm)
Câu 1:
Ta có: \(5x^2+5y^2+8xy-2x+2y+2=0\)
<=> \(\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
<=> \(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\) (1)
Vì \(\left\{{}\begin{matrix}\left(2x+2y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\) (2)
Từ (1); (2) => \(\left\{{}\begin{matrix}2x+2y=0\\x-1=0\\y+1=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\)
Vậy x = 1; y = -1
Bài 2:
Ta có:\(x^2+4y^2+z^2-2x-6z-8y+15\)
\(=\left(x^2-2x\right)+\left(4y^2-8y\right)+\left(z^2-6z\right)+15\)
\(=\left(x^2-x-x+1\right)+\left(4y^2-4y-4y+4\right)+\left(z^2-3z-3z+9\right)+1\)
\(=\left[\left(x^2-x\right)-\left(x-1\right)\right]+\left[\left(4y^2-4y\right)-\left(4y-4\right)\right]+\left[\left(z^2-3z\right)-\left(3z-9\right)\right]+1\)
\(=\left[x.\left(x-1\right)-\left(x-1\right)\right]+\left[4y.\left(y-1\right)-4.\left(y-1\right)\right]+\left[z.\left(z-3\right)-3.\left(z-3\right)\right]+1\)
\(=\left(x-1\right)^2+\left(y-1\right).\left(4y-4\right)+\left(z-3\right)^2+1\)
\(=\left(x-1\right)^2+\left(y-1\right).4\left(y-1\right)+\left(z-3\right)^2+1\)
\(=\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\)
Với mọi giá trị của \(x;y;z\in R\) ta có:
\(\left(x-1\right)^2\ge0;4\left(y-1\right)^2\ge0;\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2\ge0\)
\(\Rightarrow\left(x-1\right)^2+4\left(y-1\right)^2+\left(z-3\right)^2+1\ge1>0\) với mọi giá trị của \(x;y;z\in R\)(đpcm)
Chúc bạn học tốt!!!
Bài 3:
(x + 1)(x + 3)(x + 5)(x + 7) + 15
= (x + 1)(x + 7)(x + 3)(x + 5) + 15
= ( x2 + 8x + 7)(x2 + 8x + 15) + 15 (1)
Đặt : x2 + 8x + 11 = t
(1) <=> ( t - 4)( t + 4) + 15
= t2 - 16 + 15
= t2 - 1
= (t + 1)( t - 1)
= ( x2 + 8x + 12)( x2 + 8x + 10)
= (x2 + 2x + 6x + 12 )(x2 + 8x + 10)
= ( x + 2)( x + 6)( x2 + 8x +10)