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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
28 tháng 1
\(N=\left(\dfrac{3x}{2x-2}-\dfrac{3x}{2x+2}-\dfrac{3x-2}{x^2-1}\right)\cdot\dfrac{4x^2-4}{3}\left(x\ne\pm1\right)\)
\(=\left[\dfrac{3x}{2\left(x-1\right)}-\dfrac{3x}{2\left(x+1\right)}-\dfrac{3x-2}{\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4x^2-4}{3}\)
\(=\left[\dfrac{3x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{3x\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{2\cdot\left(3x-2\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{3}\)
\(=\dfrac{3x^2+3x-3x^2+3x-6x+4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\)
\(=\dfrac{4\cdot2}{3}\)
\(=\dfrac{8}{3}\)
Vậy giá trị của \(N\) không phụ thuộc vào biến \(x\).
H
14 tháng 2
\(N=\left(\dfrac{3x}{2x-2}-\dfrac{3x}{2x+2}-\dfrac{3x-2}{x^2-1}\right)\cdot\dfrac{4x^2-4}{3}\\ =\left(\dfrac{3x}{2\left(x-1\right)}-\dfrac{3x}{2\left(x+1\right)}-\dfrac{3x-2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{4x^2-4}{3}\\ =\dfrac{3x\left(x+1\right)-3x\left(x-1\right)-2\left(3x-2\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4x^2-4}{3}\\ =\dfrac{\left(3x^2+3x\right)-\left(3x^2-3x\right)-\left(6x-4\right)}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4x^2-4}{3}\)
\(=\dfrac{3x^2+3x-3x^2+3x-6x+4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\\ =\dfrac{4}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{3}\\ =\dfrac{4\cdot2}{3}\)
`=8/3`
`->N` không phụ thuộc vào biến `x`
`color{blue}{ @ A_ri}`
4 tháng 12 2023
a: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10\)
\(=\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x^2+2x-x-2\right)\)
\(=\left(x^2+x+5\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
b: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)\)
\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
c: \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x\right)^2+24\left(x^2+10x\right)+16\left(x^2+10x\right)+384+16\)
\(=\left(x^2+10x\right)^2+40\left(x^2+10x\right)+400\)
\(=\left(x^2+10x+20\right)^2\)
3 tháng 7 2022
Gọi số chi tiết máy tổ 1 và tổ 2 sản xuất được lần lượt là a,b
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+b=900\\\dfrac{11}{10}a+b=1010\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1100\\b=-200\left(loại\right)\end{matrix}\right.\)
22 tháng 1
ĐKXĐ: \(x\ne\pm y\)
VT=\(\left(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}\right)\cdot\dfrac{2x}{x+y}+\dfrac{y}{y-x}\)
\(=\left(\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{2\left(x+y\right)}\right)\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)
\(=\dfrac{4xy+\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{2x}{x+y}-\dfrac{y}{x-y}\)
\(=\dfrac{\left(x+y\right)^2\cdot x}{\left(x-y\right)\cdot\left(x+y\right)^2}-\dfrac{y}{x-y}\)
\(=\dfrac{x}{x-y}-\dfrac{y}{x-y}=\dfrac{x-y}{x-y}=1\)=VP
H9
HT.Phong (9A5)
CTVHS
18 tháng 11 2023
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) \(\left(x+5\right)^2-16x^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x+5-4x\right)\left(x+5+4x\right)=0\)
\(\Leftrightarrow\left(5-3x\right)\left(5x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-3x=0\\5x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=5\\5x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
18 tháng 11 2023
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
a: \(x\left(x-1\right)-4x+4=0\)
=>\(x\left(x-1\right)-\left(4x-4\right)=0\)
=>\(x\left(x-1\right)-4\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-4\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
b: \(x^2\left(x-6\right)+24-4x=0\)
=>\(x^2\left(x-6\right)-\left(4x-24\right)=0\)
=>\(x^2\left(x-6\right)-4\left(x-6\right)=0\)
=>\(\left(x-6\right)\left(x^2-4\right)=0\)
=>\(\left(x-6\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-6=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=2\\x=-2\end{matrix}\right.\)