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Bài 4:
a) Ta có: \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^7\right)-\left(x^6-x^4\right)-\left(x^5-x^3\right)+\left(x^2-1\right)\)
\(=x^7\left(x^2-1\right)-x^4\left(x^2-1\right)-x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^7-x^4-x^3+1\right)\)
\(=\left(x^2-1\right)\cdot\left[x^4\left(x^3-1\right)-\left(x^3-1\right)\right]\)
\(=\left(x^2-1\right)\cdot\left(x^3-1\right)\cdot\left(x^4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)^2\cdot\left(x^2+1\right)\cdot\left(x^2+x+1\right)\)
a, Ta có : \(x^5-x^4-x^3-x^2-x-2\)
\(=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)
\(=x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)\)
Bài 8:
a) Ta có: \(2^9-1=\left(2^3-1\right)\cdot\left(2^6+2^3+1\right)\)
\(=7\cdot\left(64+8+1\right)=7\cdot73⋮73\)(đpcm)
b) Ta có: \(5^6-10^4=5^4\cdot5^2-5^4\cdot2^4=5^4\left(5^2-2^4\right)\)
\(=5^4\left(25-16\right)=5^4\cdot9⋮9\)(đpcm)
c) Ta có: \(\left(n+3\right)^2-\left(n-1\right)^2\)
\(=\left(n+3-n+1\right)\left(n+3+n-1\right)\)
\(=4\cdot\left(2n+2\right)=4\cdot2\cdot\left(n+1\right)=8\left(n+1\right)⋮8\)(đpcm)
d) Ta có: \(\left(n+6\right)^2-\left(n-6\right)^2\)
\(=\left(n+6-n+6\right)\left(n+6+n-6\right)\)
\(=12\cdot2n=24n⋮24\)(đpcm)
Lời giải:
a)
$8^3:(-8)^{-5}=8^3.(-8)^5=8^3.(-8^5)=-8^3.8^5=-8^{3+5}=-8^{13}$
b)
$x^3y^4:(x^3y)=x^{3-3}.y^{4-1}=x^0.y^3=y^3$
c)
$5x^2y^4:(10x^2y)=(5:10).(x^2:x^2)(y^4:y)=\frac{1}{2}.1.y^3=\frac{1}{2}y^3$
d)
$\frac{3}{4}(xy)^3:(\frac{-1}{2}x^2y^2)$
$=(\frac{3}{4}: \frac{-1}{2})(x^3:x^2).(y^3:y^2)$
$=\frac{-3}{2}xy$
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
Sửa lại ạ!
a) \(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
b) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-4-2x\right)\left(-4+12x\right)\)
c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x+7\right)\left(8x+11\right)\)
P/s: Ko chắc!
Lời giải:
a) $8^3:(-8)^{-5}=8^3:(-8^{-5})=-(8^3:8^{-5})$
$=-8^{3-(-5)}=-8^{13}$
b) $(\frac{-5}{16})^{12}:(\frac{5}{-16})^4$
$=(\frac{-5}{16})^{12}:(\frac{-5}{16})^4$
$=(\frac{-5}{16})^{12-4}=(\frac{-5}{16})^8=(\frac{5}{16})^8$
c) $(\frac{5}{3})^6:(\frac{5}{3})^4=(\frac{5}{3})^{6-4}=(\frac{5}{3})^2$
d)
$(\frac{9}{7})^9:(\frac{-9}{-7})^3=(\frac{9}{7})^9:(\frac{9}{7})^3$
$=(\frac{9}{7})^{9-3}=(\frac{9}{7})^4$