a) ĐK: \(x\inℝ\).

\(A=\left(\frac{x^2+1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)

\(A=\frac{\left(x^2+1\right)^2-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}.\frac{x^4\left(1+x^2\right)+1-x^4}{1+x^2}\)

\(A=\frac{x^4+2x^2+1-x^4+x^2-1}{x^6+1}.\frac{x^4+x^6+1-x^4}{1+x^2}\)

\(A=\frac{3x^2}{x^6+1}.\frac{x^6+1}{1+x^2}=\frac{3x^2}{1+x^2}\)

b) \(A=\frac{3x^2}{1+x^2}\ge0\)

Dấu \(=\)xảy ra khi \(x=0\). Vậy GTNN của \(A\)là \(0\).

Câu 1: 

a) 2Cu: 2 nguyên tử đồng

    5K: 5 nguyên tử Kali

    2O₂: 2 phân tử khí oxi

    2H₂:2 phân tử khí Hiđrô

b)CTHH:H₂SO₄

   Ý nghĩa: cứ 1 phân tử H2SO4 thì có 2 nguyên tử H,1 nguyên tử S,4 nguyên tử O

Gấp thì được

1) \(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-4\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

2) \(A=x^2-2x+2020=\left(x^2-2x+1\right)+2019=\left(x-1\right)^2+2019\ge2019>0\)

Bài 1:

\(\left(x-2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x-2\right)-\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-x-2\right)=0\)

\(\Leftrightarrow-4\left(x-2\right)=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Bài 2:

Ta có: \(A=x^2-2x+2020=x^2-2x+1+2019=\left(x-1\right)^2+2019\)

Vì \(\left(x-1\right)^2\ge0\forall x\)\(\Rightarrow\left(x-1\right)^2+2019\ge2019\forall x\)

hay \(A>0\)( đpcm )

Ta có:

\(M=-x^2-y^2+8x+4y-21\)

\(M=-\left(x^2-8x+16\right)-\left(y^2-4y+4\right)-1\)

\(M=-\left(x-4\right)^2-\left(y-2\right)^2-1\le-1\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-4\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=4\\y=2\end{cases}}\)

Vậy Max(M) = -1 khi x = 4 và y = 2

\(M=-x^2-y^2+8x+4y-21\)

\(=-x^2+8x-16-y^2+4y-4-1\)

\(=-\left(x^2-8x+16\right)-\left(y^2-4y+4\right)-1\)

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

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow-\left(x-4\right)^2\le0\forall x\)

\(\left(y-2\right)^2\ge0\forall y\)\(\Rightarrow-\left(y-2\right)^2\le0\forall y\)

\(\Rightarrow-\left(x-4\right)^2-\left(y-2\right)^2\le0\forall x,y\)

\(\Rightarrow-\left(x-4\right)^2-\left(y-2\right)^2-1\le-1\forall x,y\)

hay \(A\le-1\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-4=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=2\end{cases}}\)

Vậy \(maxM=-1\)\(\Leftrightarrow\hept{\begin{cases}x=4\\y=2\end{cases}}\)

1) Để \(A=x^4-x^3-3x^2+x+a\) chia hết cho \(x-1\) thì:

Nghiệm của x - 1 cũng là nghiệm của A, khi đó:

Tại x = 1 thì \(A=0\)

\(\Leftrightarrow1-1-3+1+a=0\)

\(\Rightarrow a=2\)

Vậy a = 2 thì A chia hết cho x - 1

2) Ta có: 

\(A=x^2-4x+5=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-2\right)^2=0\)

\(\Rightarrow x=2\)

Vậy Min(A) = 1 khi x = 2

\(x^3-2x^2-4xy^2+x\)

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

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

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

\(n_{H_2}=0,4\left(mol\right)\), \(n_{O_2}=0,3\left(mol\right)\)

Phương trình: \(H_2+\frac{1}{2}O_2\rightarrow H_2O\)

\(H_2\)sẽ phản ứng hết, \(O_2\)còn dư. \(n_{H_2O}=n_{H_2}=0,4\left(mol\right)\Rightarrow m_{H_2O}=0,4.18=7,2\left(g\right)\)

Ta có:

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2\)

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

\(P=\left(2y-2x\right)\left(2y+2x\right)\)

\(P=4y^2-4x^2\)

P = ( x - y )² + ( x + y )² - 2( x + y )( x - y ) - 4x²

= x² - 2xy + y² + x² + 2xy + y² - 2( x² - y² ) - 4x²

= -2x² + 2y² - 2x² + 2y²

= 4y² - 4x²