5:

a: \(-120x^5y^4=20x^5y^2\cdot\left(-6y^2\right)\)

b: \(60x^6y^2=20x^5y^2\cdot3x\)

c: \(-5x^{15}y^3=20x^5y^2\cdot\left(-\dfrac{1}{4}x^{10}y\right)\)

d: \(2x^{12}y^{10}=20x^5y^2\cdot\left(\dfrac{1}{10}x^7y^8\right)\)

\(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

\(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(b+1\right)}=\dfrac{b}{a}\)

d) Để phân thức \(\dfrac{4x^3+4x^2}{x^2-1}\) có nghĩa thì: \(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

Khi đó: \(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)

e) Để phân thức \(\dfrac{b^2+b}{a+ab}\) có nghĩa thì: \(a+ab\ne0\Leftrightarrow a\ne-ab\)

Khi đó: \(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(1+b\right)}=\dfrac{b}{a}\)

\(x^3-y^3-z^3=3xyz\)

=>\(\left(x-y\right)^3-z^3+3xy\left(x-y\right)-3xyz=0\)

=>\(\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]+3xy\left(x-y-z\right)=0\)

=>\(\left(x-y-z\right)\left[x^2-2xy+y^2+xz-zy+z^2+3xy\right]=0\)

=>\(\left(x-y-z\right)\left(x^2+y^2+z^2+xy+xz-yz\right)=0\)

=>\(\left(x-y-z\right)\left(2x^2+2y^2+2z^2+2xy+2xz-2yz\right)=0\)

=>\(\left(x-y-z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2+2xz+z^2\right)\right]=0\)

=>\(\left(x-y-z\right)\left[\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2\right]=0\)

=>\(\left[{}\begin{matrix}x-y-z=0\\\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=y+z\\y=z=-x\end{matrix}\right.\)

\(H=\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1-\dfrac{z}{x}\right)\)

\(=\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}\cdot\dfrac{x-z}{x}\)

TH1: x=y+z

=>\(H=\dfrac{y-x}{y}\cdot\dfrac{x}{z}\cdot\dfrac{x-z}{x}\)

\(=\dfrac{y}{x}\cdot\dfrac{z}{x}\cdot\dfrac{y-x}{y}=\dfrac{y}{x}\cdot\dfrac{z}{x}\cdot\dfrac{-z}{y}=-1\)

TH2: y=z=-x

=>y+x+z=0(vô lý vì x,y,z đều dương)

Vậy: H=-1

a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

Đặt: \(x^2+x+1=y\), khi đó biểu thức trở thành:

\(y\left(y+1\right)-12\)

\(=y^2+y-12\)

\(=y^2-3y+4y-12\)

\(=y\left(y-3\right)+4\left(y-3\right)\)

\(=\left(y-3\right)\left(y+4\right)\)

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

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

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

\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x^2+x+5\right)\)

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

b) \(\left(x^2+2x\right)^2+9x^2+18x+20\)

\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)

Đặt: \(x^2+2x=a\), khi đó biểu thức trở thành:

\(a^2+9a+20\)

\(=a^2+4a+5a+20\)

\(=a\left(a+4\right)+5\left(a+4\right)\)

\(=\left(a+4\right)\left(a+5\right)\)

\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

Đặt: \(x^2+10x+20=y\), khi đó biểu thức trở thành:

\(\left(y-4\right)\left(y+4\right)+16\)

\(=y^2-16+16\)

\(=y^2\)

\(=\left(x^2+10x+20\right)^2\)

$\text{#}Toru$

a: Xét ΔKAB và ΔKCD có

\(\widehat{KAB}=\widehat{KCD}\)(hai góc so le trong, AB//CD)

\(\widehat{AKB}=\widehat{CKD}\)(hai góc đối đỉnh)

Do đó: ΔKAB đồng dạng với ΔKCD

=>\(\dfrac{KA}{KC}=\dfrac{KB}{KD}\)

=>\(KA\cdot KD=KB\cdot KC\)

b: Ta có: \(\dfrac{KA}{KC}=\dfrac{KB}{KD}\)

=>\(\dfrac{KC}{KA}=\dfrac{KD}{KB}\)

=>\(\dfrac{KC}{KA}+1=\dfrac{KD}{KB}+1\)

=>\(\dfrac{KC+KA}{KA}=\dfrac{KD+KB}{KB}\)

=>\(\dfrac{AC}{KA}=\dfrac{BD}{KB}\)

=>\(\dfrac{AK}{AC}=\dfrac{BK}{BD}\left(1\right)\)

Xét ΔADC có IK//DC

nên \(\dfrac{AK}{AC}=\dfrac{IK}{DC}\left(2\right)\)

Xét ΔBDC có KQ//DC

nên \(\dfrac{KQ}{DC}=\dfrac{BK}{BD}\left(3\right)\)

Từ (1),(2),(3) suy ra IK=KQ

B

đáp án là B nhé bạn mik chắc luôn nhoa !

\(\Leftrightarrow n^5+n^2-n^2+1⋮n^3+1\)

\(\Leftrightarrow-n^3+n⋮n^3+1\)

\(\Leftrightarrow n=1\)

d: ĐKXĐ: \(3x< >k\Omega\)

=>\(x< >\dfrac{k\Omega}{3}\)

\(cot^23x-cot3x-2=0\)

=>\(cot^23x-2cot3x+cot3x-2=0\)

=>\(\left(cot3x-2\right)\left(cot3x+1\right)=0\)

=>\(\left[{}\begin{matrix}cot3x-2=0\\cot3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cot3x=2\\cot3x=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3x=arccot\left(2\right)+k\Omega\\3x=-\dfrac{\Omega}{4}+k\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\cdot arccot\left(2\right)+\dfrac{k\Omega}{3}\\x=-\dfrac{\Omega}{12}+\dfrac{k\Omega}{3}\end{matrix}\right.\)

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