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Bài 1:
a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-5+20x-4x^2-12x-9\)
\(=-30\)
b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
\(=-175\)
d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)
\(=-11x^2-32x+3-48+32x+11x^2-44\)
=-89
$\textbf{1)}$
$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$
$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$
$=\left[(4x-1)-(3x-7)\right]^2$
$=(x+6)^2.$
Tại $x=44$: $(44+6)^2=50^2=2500.$
$\textbf{2)}$
$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$
$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$
$=\left[(2x-5)-(3x-4)\right]^2$
$=(x+1)^2.$
Tại $x=24$: $(24+1)^2=625.$
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
a) \(\left(6x^3+3x^2+4x+2\right):\left(3x^2+2\right)\)
\(=\left[3x^2\left(2x+1\right)+2\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=\left[\left(3x^2+2\right)\left(2x+1\right)\right]⋮\left(3x^2+2\right)\)
\(=2x+1\)
b) \(\left(2x^3-22x^2-5x^2+60x+55x-150\right):\left(x-5\right)\)
\(=\left[\left(2x^3-22x^2+60x\right)-\left(5x^2-55x+150\right)\right]:\left(x-5\right)\)
\(=\left[2x\left(x^2-11x+30\right)-5\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-11x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x^2-5x-6x+30\right)\right]:\left(x-5\right)\)
\(=\left[\left(2x-5\right)\left(x-5\right)\left(x-6\right)\right]:\left(x-5\right)\)
\(=\left(2x-5\right)\left(x-6\right)\)
\(=2x^2-17x+30\)
d) \(\left(x^5+4x^3+3x^2-5x+15\right):\left(x^3-x+3\right)\)
\(=\left(x^5+5x^3+3x^2-x^3-5x+15\right):\left(x^3-x+3\right)\)
\(=\left[\left(x^5-x^3+3x^2\right)+\left(5x^3-5x^2+15\right)\right]:\left(x^3-x+3\right)\)
\(=\left[x^2\left(x^3-x+3\right)+5\left(x^3-x^2+3\right)\right]:\left(x^3-x+3\right)\)
\(=\left[\left(x^2+5\right)\left(x^3-x+3\right)\right]:\left(x^3-x+3\right)\)
\(=x^2+5\)