• (x/1/2)*5/3=7/4-1/2
• (x-1/2)*5/3=5/4
• x-1/2=5/4:5/3
• x-1/2=3/4
• x=3/4+1/2
• x=5/4

(2x - 1)² - 2(4x² - 1) + (2x + 1)²

= (2x - 1)² - 2(2x - 1)(2x + 1) + (2x + 1)²

= (2x - 1 - 2x - 1)²

= (-2)² = 4

\(\left(2x-1\right)^2-2\left(4x^2-1\right)+\left(2x+1\right)^2\)

\(=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\)

\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)

\(=\left(2x-1-2x-1\right)^2=\left(-2\right)^2=4\)

Ta có: \(4x^2+2y^2-4xy+4+\sqrt{\left(x+y+z\right)^2}=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+y^2+4+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|=-4\)

Mà \(VT\ge0\left(\forall x,y,z\right)\) => vô lý

=> PT vô nghiệm

\(4x^2+2y^2-4xy+4+\sqrt{\left(x+y+z\right)^2}=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+y^2+4+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|+4=0\)(1)

Vì \(\left(2x-y\right)^2\ge0\); \(y^2\ge0\); \(\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(2x-y\right)^2+y^2+\left|x+y+z\right|+4\ge4\forall x,y,z\)(2)

Từ (1) và (2) \(\Rightarrow\)Vô lý 

Vậy không tìm được giá trị của x, y, z thỏa mãn đề bài

Ta có: \(2x^2+2y^2+z^2+25-6y-2xy-8x+2z\left(y-x\right)=0\)

\(\Leftrightarrow\left(x^2-8x+16\right)+\left(y^2-6y+9\right)+\left(x^2-2xy+y^2\right)-2\left(x-y\right)z+z^2=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(y-3\right)^2+\left[\left(x-y\right)^2-2\left(x-y\right)z+z^2\right]=0\)

\(\Leftrightarrow\left(x-4\right)^2+\left(y-3\right)^2+\left(x-y-z\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y-3\right)^2=0\\\left(x-y-z\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=3\\z=1\end{cases}}\)

Chỗ (x²-8x+16) 

16 là ở đâu ra vậy bạn

Chỗ (y²-6y+9 ) 

9 là ở đâu ra nx v

Ta có:

\(G=x^2+3y^2+2xy-6y+3\)

\(G=\left(x^2+2xy+y^2\right)+\left(2y^2-6y+\frac{18}{4}\right)-\frac{3}{2}\)

\(G=\left(x+y\right)^2+2\left(y-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+y\right)^2=0\\2\left(y-\frac{3}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

Vậy Min(G) = -3/2 khi \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

G = x² + 3xy² + 2xy - 6y + 3

<=> G = ( x² + 2xy + y² ) + ( y² - 6y + 9 ) - 6

<=> G = ( x + y )² + ( y - 3 )² - 6

Vì ( x + y )²\(\ge\)0 ; ( y - 3 )²\(\ge\)0\(\forall\)x ; y

=> G = ( x + y )² + ( y - 3 )² - 6\(\ge\)- 6

Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\y=3\end{cases}}\)

Vậy minG = - 6 <=> x = - 3 ; y = 3

\(5120=2^{10}.5\)

Ta có: C=\(2^{2000}+2^{2002}=2^{2000}\left(1+2^2\right)=2^{2000}.5⋮2^{10}.5\)

Vậy. C chia hết cho 5120

\(axz^2-ax-ayz^2+ax+ay+az^3\)

= \(axz^2-ayz^2+ay+az^3\)

\(=a\left(xz^2-yz^2+y+z^3\right)\)

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