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a,<=> x2-4x+22+y2-8y+42-14
<=> (x2-2x2+22)+(y2-2x4+42)-14
<=> (x-2)2+(y-4)2-14
Vì (x-2)2+(y-4)2>= 0
=> F >= -14 => MIn F = -14 <=> x=2, y=4
b, <=> (x2+52+(2y)2-4xy+10x-20y) +(y2-2y+1)+2
<=> (x+5-2y )2+(y-1)2+2
Vì (x+5-2y) 2+(y-1)2 >= 0
=> G >= 2 => Min =2 <=> y=1, x= -3
\(F=x^2-4x+y^2-8y+6\)
\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)
\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y-4\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)
\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)
Ta có : A = 2x2 + 10x - 15
= 2x2 + 10x - \(\frac{50}{4}-\frac{5}{2}\)
= 2(x2 + 5x - \(\frac{25}{4}\)) - \(\frac{5}{2}\)
= 2(x - \(\frac{5}{2}\) )2 - \(\frac{5}{2}\)
Mà ; 2(x - \(\frac{5}{2}\) )2 \(\ge0\forall x\)
Nên : 2(x - \(\frac{5}{2}\) )2 - \(\frac{5}{2}\) \(\ge-\frac{5}{2}\forall x\)
Vậy Amin = \(-\frac{5}{2}\) , dấu bằng xảy ra khi x = \(\frac{5}{2}\)
a) Đặt \(A=-x^2+9x-12\)
\(-A=x^2-9x+12\)
\(-A=\left(x^2-9x+\frac{81}{4}\right)-\frac{33}{4}\)
\(-A=\left(x-\frac{9}{2}\right)^2-\frac{33}{4}\)
Mà \(\left(x-\frac{9}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-A\ge-\frac{33}{4}\Leftrightarrow A\le\frac{33}{4}\)
Dấu "=" xảy ra khi : \(x-\frac{9}{2}=0\Leftrightarrow x=\frac{9}{2}\)
Vậy \(A_{Max}=\frac{33}{4}\Leftrightarrow x=\frac{9}{2}\)
b) Đặt \(B=2x^2+10x-1\)
\(B=2\left(x^2+5x+\frac{25}{4}\right)-\frac{29}{4}\)
\(B=2\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\)
Mà \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow B\ge-\frac{29}{4}\)
Dấu "=" xảy ra khi : \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{29}{4}\Leftrightarrow x=-\frac{5}{2}\)
c) Đặt \(C=\left(2x+6\right)\left(x-1\right)\)
\(C=2x^2-2x+6x-6\)
\(C=2x^2+4x-6\)
\(C=2\left(x^2+2x+1\right)-8\)
\(C=2\left(x+1\right)^2-8\)
Mà \(\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow C\ge-8\)
Dấu "=" xảy ra khi : \(x+1=0\Leftrightarrow x=-1\)
Vậy \(C_{Min}=-8\Leftrightarrow x=-1\)
d) Đặt \(D=3x-2x^2\)
\(-2D=4x^2-6x\)
\(-2D=\left(4x^2-6x+\frac{9}{4}\right)-\frac{9}{4}\)
\(-2D=\left(2x-\frac{3}{2}\right)^2-\frac{9}{4}\)
Mà \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow-2D\ge-\frac{9}{4}\)
\(\Leftrightarrow D\le\frac{9}{8}\)
Dấu "=" xảy ra khi : \(2x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{4}\)
Vậy \(D_{Max}=\frac{9}{8}\Leftrightarrow x=\frac{3}{4}\)
a) \(A=\left(x+1\right)\left(2x-1\right)\)
\(A=2x^2+x-1\)
\(A=2\left(x^2+\frac{1}{2}x-\frac{1}{2}\right)\)
\(A=2\left[x^2+2\cdot x\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2-\frac{9}{16}\right]\)
\(A=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]\)
\(A=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge\frac{-9}{8}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{4}\)
Vậy Amin = -9/8 khi và chỉ khi x = -1/4
b) \(B=4x^2-4xy+2y^2+1\)
\(B=\left(2x\right)^2-2\cdot2x\cdot y+y^2+y^2+1\)
\(B=\left(2x-y\right)^2+y^2+1\ge1\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x-y=0\\y=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}\Rightarrow}}x=y=0\)
Vậy Bmin = 1 khi và chỉ khi x = y = 0
a, \(x^2+y^2-2x+6y-30\)
\(=x^2-2x+1+y^2+6y+9-40\)
\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)
\(min=-40\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
a)x^2+y^2-2x+6y-30=(x-1)^2+(y+3)^2-40\(\ge\) -40
dấu = xảy ra khi x=1,y=-3
\(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(=9x=9.15=135\)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)\(\Leftrightarrow x^2+x=0\)\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
So sánh với ĐKXĐ ta thấy: \(x=0\)không thoả mãn
Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)
Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x.\left(x+2\right)-\left(x-2\right)}{\left(x-2\right).x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-x+2}{x^2-2x}=\frac{2}{x^2-2x}\)
\(\Rightarrow x^2+x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy \(S=\left\{-1;0\right\}\)
\(\left(x^2-5\right)\left(x+2\right)+5x=2x^2+17\)
\(\Leftrightarrow x^2\left(x+2\right)-5\left(x+2\right)+5x-2x^2-17=0\)
\(\Leftrightarrow x^3+2x^2-5x-10+5x-2x^2-17=0\)
\(\Leftrightarrow x^3-3^3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x^2+3x+9=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\left(tm\right)\\\left(x+\frac{3}{2}\right)^2+\frac{27}{4}>0\left(loai\right)\end{array}\right.\)
M.n giúp mk nhanh nka!
Vì mk gửi câu hỏi bị lỗi nên nhờ Nguyễn Hoàng Gia Bảo đăng giúp
B = 5x - x2
B = -x2 + 5x
-B = x2 - 5x
-4B = 4x2 - 20x
-4B = (2x-5)2 -25
B = -(2x-5)2 / 4 + 6,25
GTLN của B = 6,25 <=> 2x-5 = 0 => x = 5/2
A = 2x2 + 10x - 1
2A = 4x2 + 20x - 2
2A = (2x+5)2 - 27
A = (2x+5)2 / 2 - 13,5
GTNN của A là -13,5 <=> 2x+5 = 0 => x = -5/2