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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
a) Ta có: \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
b) Ta có: \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=\left(6x^2+x-2\right)\left(3-x\right)\)
\(=18x^2-6x^3+3x-x^2-6+2x\)
\(=-6x^3+17x^2+5x-6\)
c) Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)\)
\(=x^3+3x^2-5x+3x^2+9x-15\)
\(=x^3+6x^2+4x-15\)
d) Ta có: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+1\)
e) Ta có: \(\left(2x^3-3x-1\right)\left(5x+2\right)\)
\(=10x^4+4x^3-15x^2-6x-5x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) Ta có: \(\left(x^2-2x+3\right)\left(x-4\right)\)
\(=x^3-4x^2-2x^2+8x+3x-12\)
\(=x^3-6x^2+11x-12\)
g) Ta có: \(\left(4x-1\right)\left(3x+1\right)-5x\left(x-3\right)-\left(x-4\right)\left(x-3\right)\)
\(=12x^2+4x-3x-1-5x^2+15x-\left(x^2-7x+12\right)\)
\(=7x^2+16x-1-x^2+7x-12\)
\(=6x^2+23x-23\)
h) Ta có: \(\left(5x-2\right)\left(x+1\right)-3x\left(x^2-x-3\right)-2x\left(x-5\right)\left(x-4\right)\)
\(=5x^2+5x-2x-2-3x^3+3x^2+9x-2x\left(x^2-9x+20\right)\)
\(=-3x^3+8x^2+12x-2-2x^3+18x^2-40x\)
\(=-5x^3+26x^2-28x-2\)
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
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
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
a, (x4-2x3+2x-1):(x2-1) = \(\frac{\left(x^4-1\right)-\left(2x^3-2x\right)}{x^2-1}\)
= \(\frac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\) =\(\frac{\left(x^2-1\right)\left(x^2+1-2x\right)}{x^2-1}\)
= \(x^2+1-2x\)= \(\left(x-1\right)^2\)
b, (8x3-6x2-5x+3):((4x+3)