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Những câu hỏi liên quan
26 tháng 8 2023
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: \(2x^{12}y^{10}=20x^5y^2\cdot\left(\dfrac{1}{10}x^7y^8\right)\)
HT
2
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
\(\dfrac{4x^3+4x^2}{x^2-1}=\dfrac{4x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x^2}{x-1}\)
\(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(b+1\right)}=\dfrac{b}{a}\)
T
12 tháng 1
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: \(\dfrac{b^2+b}{a+ab}=\dfrac{b\left(b+1\right)}{a\left(1+b\right)}=\dfrac{b}{a}\)
T
3 tháng 12 2023
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:
\(\left(y-4\right)\left(y+4\right)+16\)
\(=y^2-16+16\)
\(=y^2\)
\(=\left(x^2+10x+20\right)^2\)
$\text{#}Toru$
3 tháng 12 2023
\(x^3-y^3-z^3=3xyz\)
=>\(\left(x-y\right)^3-z^3+3xy\left(x-y\right)-3xyz=0\)
=>\(\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]+3xy\left(x-y-z\right)=0\)
=>\(\left(x-y-z\right)\left[x^2-2xy+y^2+xz-zy+z^2+3xy\right]=0\)
=>\(\left(x-y-z\right)\left(x^2+y^2+z^2+xy+xz-yz\right)=0\)
=>\(\left(x-y-z\right)\left(2x^2+2y^2+2z^2+2xy+2xz-2yz\right)=0\)
=>\(\left(x-y-z\right)\left[\left(x^2+2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2+2xz+z^2\right)\right]=0\)
=>\(\left(x-y-z\right)\left[\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2\right]=0\)
=>\(\left[{}\begin{matrix}x-y-z=0\\\left(x+y\right)^2+\left(y-z\right)^2+\left(x+z\right)^2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=y+z\\y=z=-x\end{matrix}\right.\)
\(H=\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1-\dfrac{z}{x}\right)\)
\(=\dfrac{y-x}{y}\cdot\dfrac{z+y}{z}\cdot\dfrac{x-z}{x}\)
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
8 tháng 12 2023
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
11 tháng 11 2021
\(\Leftrightarrow n^5+n^2-n^2+1⋮n^3+1\)
\(\Leftrightarrow-n^3+n⋮n^3+1\)
\(\Leftrightarrow n=1\)
NT
1
29 tháng 10 2023
d: ĐKXĐ: \(3x< >k\Omega\)
=>\(x< >\dfrac{k\Omega}{3}\)
\(cot^23x-cot3x-2=0\)
=>\(cot^23x-2cot3x+cot3x-2=0\)
=>\(\left(cot3x-2\right)\left(cot3x+1\right)=0\)
=>\(\left[{}\begin{matrix}cot3x-2=0\\cot3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cot3x=2\\cot3x=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=arccot\left(2\right)+k\Omega\\3x=-\dfrac{\Omega}{4}+k\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\cdot arccot\left(2\right)+\dfrac{k\Omega}{3}\\x=-\dfrac{\Omega}{12}+\dfrac{k\Omega}{3}\end{matrix}\right.\)
B
6
Bài 13:
a) \(x^3-49x=0\)
\(\Leftrightarrow x\left(x^2-49\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=7\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
c) \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow x\left(x^2+2\cdot3\cdot x+3^2\right)=0\)
\(\Leftrightarrow x\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
d) \(\left(x+5\right)^2-16x^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x+5-4x\right)\left(x+5+4x\right)=0\)
\(\Leftrightarrow\left(5-3x\right)\left(5x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-3x=0\\5x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=5\\5x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
12:
a: \(4x^2+4x+1=0\)
=>\(\left(2x\right)^2+2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>2x=-1
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-2x=-1\)
=>\(x^2-2x+1=0\)
=>\(x^2-2\cdot x\cdot1+1^2=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
c: \(-x^2+10x=25\)
=>\(-x^2+10x-25=0\)
=>\(x^2-10x+25=0\)
=>\(x^2-2\cdot x\cdot5+5^2=0\)
=>\(\left(x-5\right)^2=0\)
=>x-5=0
=>x=5
d: \(12x-4x^2=9\)
=>\(-4x^2+12x-9=0\)
=>\(4x^2-12x+9=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot3+3^2=0\)
=>\(\left(2x-3\right)^2=0\)
=>2x-3=0
=>2x=3
=>x=3/2