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a) Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3\left(x^2-1\right)\)
\(=4x-3x^2+3\)
\(=-3x^2+4x+3\)
b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)
\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)
\(=5x^2-20-32x^2+48x-16+17\)
\(=-27x^2+48x-19\)
a: Ta có: \(\left(x+1\right)^2+\left(x-1\right)^2-2\left(1+x\right)\left(1-x\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x+1+x-1\right)^2\)
\(=4x^2\)
c: Ta có: \(3\left(x+2\right)^2-\left(3x+1\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)
\(=x^2+6x+32\)
a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)
\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)
\(=-16x+8\)
b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
=27x-55
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
a: Ta có: \(A=\left(x+2\right)\left(x-4\right)+\left(x+1\right)\left(x-6\right)\)
\(=x^2-4x+2x-8+x^2-6x+x-6\)
\(=2x^2-7x-14\)
b: \(B=\left(2a-b\right)\left(4a^2+2ab+b^2\right)=8a^3-b^3\)
c: \(C=\left(2+x\right)\left(2-x\right)\left(x+4\right)\)
\(=\left(4-x^2\right)\left(x+4\right)\)
\(=4x+16-x^3-4x^2\)
a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)
\(=\left(x+\dfrac{1}{x-1}\right):\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(=\dfrac{x^2-x+1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{x^2-x+1}\)
\(=x-1\)
b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)
\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{-1}{y}\)
a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)
\(=\dfrac{2x\left(x-1\right)}{x-1}\)
=2x
b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)
\(=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}\)
\(=\dfrac{x+1}{3x}\)
c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)
\(=\dfrac{x}{3\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1+3}{3\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+4}{3x^2-3}\)
Bài 1:
a.
$=(x^3+2^3)-(x^3-2)=2^3+2=10$
b.
$=(x^2+10x+25)-4x(4x^2+12x+9)-(2x-1)(x^2-9)$
$=x^2+10x+25-16x^3-48x^2-36x-(2x^3-18x-x^2+9)$
$=-18x^3-46x^2-8x+16$
2.
a.
$301^2=(300+1)^2=300^2+2.300+1=90000+600+1$
$=90601$
b.
$198^2=(200-2)^2=4(100-1)^2=4(100^2-2.100+1)$
$=4(10000-200+1)=4.9801=39204$
c.
$93.107=(100-7)(100+7)=100^2-7^2$
$=10000-49=9951$
d.
$127^2+146.127+73^2$
$=127^2+2.73.127+73^2$
$=(127+73)^2=200^2=40000$
a) \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+3\)
\(=4x+3\)
b) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)
\(=5\left(x^2-4\right)-3+4x+17\)
\(=5x^2-20-3+4x+17\)
\(=5x^2-6+4x\)