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Ta có : (x + 1)2 - (x + 2)(x - 2) = 0
<=> (x + 1)2 - (x2 - 22) = 0
<=> x2 + 2x + 1 - x2 + 4 = 0
<=> 2x + 5 = 0
=> 2x = -5
=> x = \(-\frac{5}{2}\)
<=> 4(x^2 + 2x + 1) + 4x^2 - 4x +1 - 8(x^2 - 1) = 11
<=> 4x^2 + 8x + 4 + 4x^2 - 4x +1 - 8x^2 +8 - 11 = 0
<=> 4x + 2 = 0
<=> x = - 1/2
B = x2y2+2x2+24xy+16x+191 = [ (xy)^2 + 24xy + 144] + \(\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.4\sqrt{2}+32\right]\)+15
= (xy+12)^2 +(\(\sqrt{2}x\)+\(4\sqrt{2}\))^2 + 15
( ở đây mik làm tắt) => Min B = 15 khi \(\sqrt{2}x+4\sqrt{2}=0=>x=-4\)và xy+12 = 0 => -4y = -12= > y=3
A= 2x^2+9y^2-6xy-6x-12y+2004
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2004
A = [(x -3y)^2 +4(x -3y) + 4] + (x^2 -10x +25) + 1975
A= (x -3y +2)^2 + (x -5)^2 + 1975
( mik rút mấy cái bước (x-3y+2)^2 = 0, bn làm thì nên thêm vào=> Min A = 1975 vs x= 5 và y = 7/3
D=-x^2+2xy-4y^2+2x+10y-8
D = (-x^2 - y^2 - 1 + 2xy + 2x - 2y) + (-3y^2 + 12y - 12) + 5
D = -(x^2+y^2+1 - 2xy - 2x + 2y) - 3(y^2 - 4y + 4) + 5
D= - (x - y - 1)^2 - 3(y - 2)^2 +5
=> Max D = 5 khi x= 3 và y=2
\(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2xz+z^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x;y;z\)
Điều này xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow x=y=z}\)
Vậy \(x=y=z\)
\(\frac{x^2+2x+1}{x^2+2x+1}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
\(\Leftrightarrow\frac{x^2+2x+2-1}{x^2+2x+2}+\frac{x^2+2x+3-1}{x^2+3x+3}=\frac{7}{6}\)
\(\Leftrightarrow1-\frac{1}{x^2+2x+2}+1-\frac{1}{x^2+2x+3}=\frac{7}{6}\)
Đặt \(y=x^2+2x+1\), ta được:
\(2-\left(\frac{1}{y+1}+\frac{1}{y+2}\right)=\frac{7}{6}\)
\(\Leftrightarrow\frac{1}{y+1}+\frac{1}{y+2}=2-\frac{7}{6}=\frac{5}{6}\)
\(\Leftrightarrow\frac{1}{y+1}+\frac{1}{y+2}-\frac{5}{6}=0\)
\(\Leftrightarrow\frac{6\left(y+2\right)+6\left(y+1\right)-5\left(y+1\right)\left(y+2\right)}{6\left(y+1\right)\left(y+2\right)}=0\)
\(\Leftrightarrow6y+12+6y+6-\left(5y+5\right)\left(y+2\right)=0\)
\(\Leftrightarrow6y+12+6y+6-5y^2-10y-5y-10=0\)
\(\Leftrightarrow-5y^2-3y+8=0\)
\(\Leftrightarrow-5y^2+5y-8y+8=0\)
\(\Leftrightarrow-5y\left(y-1\right)-8\left(y-1\right)=0\)
\(\Leftrightarrow-\left(y-1\right)\left(5y+8\right)=0\)
Th1 \(y-1=0\Leftrightarrow y=1\)
\(\Leftrightarrow x^2+2x+1=1\)
\(\Leftrightarrow\left(x+1\right)^2=1\Leftrightarrow x+1=1;x=1=-1\)
\(\Leftrightarrow x=0\) hoặc \(x=-2\)
Th2 \(5y+8=0\Leftrightarrow5y=-8\Leftrightarrow y=\frac{-8}{5}\)
\(\Leftrightarrow x^2+2x+1=\frac{-8}{5}\)
\(\Leftrightarrow\left(x+1\right)^2=-\frac{8}{5}\)
Vì \(\left(x+1\right)^2\ge0\) mà \(\left(x+1\right)^2=\frac{-8}{5}\) ( vô lý) nên k có giá trị của x
Vậy \(S=\left\{0;-2\right\}\)
\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)
\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8-x^3-2x=4\)
\(\Leftrightarrow-2x+8=4\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Chúc bạn học giỏi
Kết bạn với mình nha
\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-2\right)-\left(2x-1\right)\right]\cdot\left[\left(2x-2\right)+\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-2-2x+1\right)\cdot\left(2x-2+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-2x-2+1\right)\cdot\left(2x+2x-2-1\right)=0\)
\(\Leftrightarrow\left(-1\right)\cdot\left(4x-3\right)=0\)
\(\Leftrightarrow4x-3=0\div\left(-1\right)\)
\(\Leftrightarrow4x-3=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\left[2x-2-\left(2x-1\right)\right]\left[2x-2+\left(2x-1\right)\right]=0\)
\(\left(2x-2-2x+1\right)\left(2x-2+2x-1\right)=0\)
\(-1\left(4x-3\right)=0\)
\(-4x+3=0\)
\(-4x=-3\)
\(x=\frac{3}{4}\)