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Câu a, b, c thì đơn giản òi. Câu d phải chú ý điểm rơi:v
d) Ta có: \(D=\left(x-\frac{1}{2}\right)^4+\frac{1}{2}\left(3x^2-3x+\frac{15}{8}\right)\)
\(=\left(x-\frac{1}{2}\right)^4+\frac{3}{2}\left(x-\frac{1}{2}\right)^2+\frac{9}{16}\ge\frac{9}{16}\)
Đẳng thức xảy ra khi x = 1/2
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\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)
Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)
\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)
\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)
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Ta có : Q = x2 - 2xy -12x +y2 +12y + 36 + 5y2 -10y + 5 + 1976
= [ x2 -2x(y + 6 ) + ( y + 6 )2 ] + 5 (y2 -2y +1 ) +1976
= ( x- y - 6 )2 + 5 (y-1)2 + 1976
Vì ( x - y - 6)2 \(\ge\)0 với mọi x ; y ;5 .(y-1)2 \(\ge\)0 với mọi x ; y và 1976 > 0
Nên biểu thức Q luôn nhận giá trị dương với mọi x ;y
Q=x2+6y2−2xy−12x+2y+2017
Q=(x2-2xy+y2)-(12x-12y)+36+(5y2-10y+5)+1976
=(x-y)2-12(x-y)+36+5(y2-2y+1)+1976
=[(x-y)2-12(x-y)+36]+5(y-1)2+1976
=(x-y-6)2+5(y-1)2+1976
do (x-y-6)2 ≥ 0 ∀ x,y
(y-1)2 ≥ 0 ∀ y
=> (x-y-6)2+5(y-1)2+1976 ≥ 1976
=> Q≥ 1976
=> MinA=1976 khi
y-1=0
=>y=1
x-y-6=0
=>x-1-6=0
=>x-7=0
=>x=7
Vậy GTNN của Q =1976 khi x=7 và y=1
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\(A=x^2+2y^2-2xy+4x-6y+6\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)+\left(y^2-6y+9\right)-7\)
\(=\left(x-y\right)^2+\left(x+2\right)^2+\left(y-3\right)^2-7\)
Đề hình như có gì đó không đúng
Ta có: \(A=x^2+2y^2-2xy+4x-6y+6=\left(x^2-2xy+y^2\right)\) \(+4\left(x-y\right)+4+y^2-2y+1+1=\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]\)\(+\left(y-1\right)^2+1=\left(x-y+2\right)^2+\left(y-1\right)^2+1\)
Ta có: \(\left(x-y+2\right)^2\ge0\forall x,y\); \(\left(y-1\right)^2\ge0\forall y\)nên \(\left(x-y+2\right)^2+\left(y-1\right)^2+1>0\forall x,y\)
Vậy \(A=x^2+2y^2-2xy+4x-6y+6>0\forall x,y\)(đpcm)
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A=\(\left(x-y\right)^2-2.6.\left(x-y\right)+36+5y^2+10y+5+4\)
=\(\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
Dấu bằng xảy ra khi y=1 và x=5
2B=\(2x^2+2y^2-2xy-2x+2y+2\)
=\(\left(x-y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)
=>B\(\ge\)0
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\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)
\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)
\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)
ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)
Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)
T i c k cho mình 1 cái nha mới bị trừ 50 đ
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a) x(x² + x) + x(x + 1)
= x²(x + 1) + x(x + 1)
= (x + 1)(x² + x)
= x(x + 1)² ⋮ (x + 1)
b) xy² - yx² + xy
= xy(y - x + 1) ⋮ xy