a\(^2\) + b\(^2\) = (a - b)\(^2\) + 2ab = 1\(^2\) + 2.2 = 1 + 4 = 5

a\(^2\) + b\(^2\) = (a + b)\(^2\) - 2ab = 2\(^2\) + 2.1 = 4 + 2 = 6

A = \(a^2\) + 2\(a^2b\) + 2\(ab^2\) + b\(^2\)

A = (\(a^2+2ab+b^2\)) - 2ab + (2\(a^2b+2ab^2\))

A = (a + b)\(^2\) + 2ab.(a+ b - 1) (1)

Thay a + b = 1 vào biểu thức (1) ta có:

A = 1\(^2\) + 2ab.(1 - 1)

A = 1 + 2.0

A = 1 + 0

A = 1

C =(a - b - c)\(^2\) - a\(^2\) - b\(^2\) - c\(^2\)

C = (a\(^{}\) - b)\(^2\) - 2(a -b)c + c\(^2\) - a\(^2\) - b\(^2\) - \(c^2\)

C = a\(^2\) - 2ab + b\(^2\) - 2ac + 2bc + c\(^2\) - \(a^2\) - \(b^2-c^2\)

C = (a\(^2\) - a\(^2\))+(\(b^2\) - b\(^2\))+(c\(^2\) - \(c^2\))-2ab - 2ac + 2bc

C = 0 + 0 + 0 - 2ab - 2ac + 2bc

C = -2ab - 2ac + 2bc

Bài 1

a: \(2x\left(2x-3\right)-\left(2x-5\right)\left(2x+5\right)\)

\(=4x^2-6x-\left(4x^2-25\right)\)

\(=4x^2-6x-4x^2+25=-6x+25\)

b: \(\left(x^2+16\right)\left(x-4\right)\left(x+4\right)\)

\(=\left(x^2+16\right)\left(x^2-16\right)\)

\(=x^4-256\)

Bài 2:

a: \(12x^2-3\)

\(=3\left(4x^2-1\right)\)

\(=3\left(2x-1\right)\left(2x+1\right)\)

b: \(x^3-x^2=x^2\cdot x-x^2\cdot1=x^2\left(x-1\right)\)

c: \(4x^2y-y^3=y\left(4x^2-y^2\right)=y\left(2x-y\right)\left(2x+y\right)\)

d: \(x^2-y^2-4x+4\)

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

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

e: \(4x^2+9y^2-12xy-4\)

\(=\left(4x^2-12xy+9y^2\right)-4\)

\(=\left(2x-3y\right)^2-2^2=\left(2x-3y-2\right)\left(2x-3y+2\right)\)

\(=x^2-2xy+y^2-\left(y^2-2yx+x^2\right)\)

=0

Ta có: \(4\left(3x-5\right)^2-9\left(9x^2-25\right)=0\)

=>\(4\left(3x-5\right)^2-9\left(3x-5\right)\left(3x+5\right)=0\)

=>(3x-5)[4(3x-5)-9(3x+5)]=0

=>(3x-5)(12x-20-27x-45)=0

=>(3x-5)(-17x-65)=0

=>(3x-5)(17x+65)=0

=>\(\left[\begin{array}{l}3x-5=0\\ 17x+65=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=5\\ 17x=-65\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53\\ x=-\frac{65}{17}\end{array}\right.\)

4(3x - 5)² - 9(9x² - 25) = 0

4(9x² - 30x + 25) - 81x² + 225 = 0

36x² - 120x + 100 - 81x² + 225 = 0

-45x² - 120x + 325 = 0

45x² + 120x - 325 = 0

(15x - 25)(3x + 13) = 0 hoặc sử dụng công thức nghiệm.

x = 5/3 hoặc x = -13/3.

Vậy x ∈ {5/3, -13/3}.

Ta có: \(25x^4-x^2=0\)

=>\(x^2\left(25x^2-1\right)=0\)

=>\(x^2\left(5x-1\right)\left(5x+1\right)=0\)

=>\(\left[\begin{array}{l}x^2=0\\ 5x-1=0\\ 5x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac15\\ x=-\frac15\end{array}\right.\)

25x^4 - x^2 = 0

x^2(25x^2 - 1) = 0

x^2 = 0 hoặc 25x^2 - 1 = 0

x = 0 hoặc x = ±1/5

Vậy x ∈ {0, 1/5, -1/5}

A = -x^2 - 6x + 1

= -(x^2 + 6x + 9) + 10

= -(x + 3)^2 + 10 ≤ 10

GTLN của A là 10 khi x = -3.

Ta có: \(A=-x^2-6x+1\)

\(=-\left(x^2+6x-1\right)\)

\(=-\left(x^2+6x+9-10\right)\)

\(=-\left(x+3\right)^2+10\le10\forall x\)

Dấu '=' xảy ra khi x+3=0

=>x=-3

x=2

Ta có: \(x^3-3x^2+3x-2=0\)

=>\(x^3-3x^2+3x-1-1=0\)

=>\(\left(x-1\right)^3=1\)

=>x-1=1

=>x=2