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Sửa đề: \(\dfrac{3}{x^2+6x+9}-\dfrac{3}{x^2-6x+9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
\(=\dfrac{3x^2-18x+27-3x^2-18x-27+x^2+30x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{x^2-6x-27}{\left(x+3\right)^2\cdot\left(x-3\right)^2}=\dfrac{\left(x-9\right)\left(x+3\right)}{\left(x+3\right)^2\cdot\left(x-3\right)^2}\)
\(=\dfrac{\left(x-9\right)}{\left(x^2-9\right)\left(x-3\right)}\)
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Akai Haruma Nguyễn Huy Tú Nguyễn Huy ThắngHồng Phúc NguyễnPhạm Hoàng Giang......và nhiều bạn nữa giúp mik vs
\(\dfrac{3}{x^2+6x+9}+\dfrac{2}{6x-x^2-9}+\dfrac{x^2+30x-27}{x^4-18x^2+81}\)
\(=\dfrac{3}{\left(x+3\right)^2}+\dfrac{-2}{\left(x-3\right)^2}+\dfrac{x^2+30x-27}{x^4-9x^2-9x^2+81}\)
\(=\dfrac{3}{\left(x+3\right)^2}-\dfrac{2}{\left(x-3\right)^2}+\dfrac{x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{3\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}-\dfrac{2\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\dfrac{x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{3x^2-18x+27-2x^2-12x-18+x^2+30x-27}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2x^2-18}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2\left(x^2-9\right)}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2\left(x+3\right)^2}\)
\(=\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x^2-9}\)
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B = \(\frac{8xy-6x^2}{3y\left(3x-4y\right)}=\frac{2x\left(4y-3x\right)}{-3y\left(4y-3x\right)}=-\frac{2x}{3y}\)
C = \(\frac{2x^3-18x}{x^4-81}=\frac{2x\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\frac{2x}{x^2+9}\)
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xin lỗi mình viết nhầm cho gửi lại câu hỏi!
CHO \(A=\left(\left(\frac{x+7}{x+9}+\frac{x+7}{x^2+81-18x}+\frac{x+5}{x^2-81}\right)\left(\frac{x-9}{x+3}\right)^2\right)^{ }:\left(\frac{x+7}{x+3}\right)\)
a) Rút gọn A
b) Tìm số nguyên x để A nguyên
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Cho mình sửa,vừa nãy gõ thiếu chữ x:V
\(B=\dfrac{2x^2-18x}{x^4-81}=\dfrac{2\left(x^2-9x\right)}{x^4-81}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\dfrac{2\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x^2+9\right)}=\dfrac{2}{x^2+9}\)
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a) \(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+8x+16=\left(x+4\right)^2\)
c) \(x^2+6x+9=\left(x+3\right)^2\)
d) \(4x^2+4x+1=\left(2x+1\right)^2\)
e) \(36+x^2-12x=x^2-12x+36=\left(x-6\right)^2\)
f) \(4x^2+12x+9=\left(2x+3\right)^2\)
g) \(x^4+81+18x^2=x^4+18x^2+81=\left(x^2+9\right)^2\)
h) \(9x^2+30xy+25y^2=\left(3x+5y\right)^2\)
a, \(x^2\) + 2\(x\) + 1 = (\(x\) + 1)2
b, \(x^2\) + 8\(x\) + 16 = (\(x\) + 4)2
c, \(x^2\) + 6\(x\) + 9 = (\(x\) + 3)2
d, 4\(x^2\) + 4\(x\) + 1 = (2\(x\) + 1)2
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\(a,x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow x^2+5x-x^2-x+6=0\Leftrightarrow4x=-6\\ \Leftrightarrow x=-\dfrac{3}{2}\)
\(b,2x^3-18x=0\\ \Leftrightarrow2x\left(x^2-9\right)=0\\ \Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
a: Ta có: \(x\left(x+5\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+5x-x^2-3x+2x+6=0\)
\(\Leftrightarrow7x=-6\)
hay \(x=-\dfrac{6}{7}\)
b: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
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Bài 1 : \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
Bài 2 : \(10123^2-123^2=\left(10123-123\right)\left(10123+123\right)\)
\(=10000.10246=102460000\)
Bài 3 : \(x^2-18x=81\Leftrightarrow x^2-18x-81=0\)
\(\Leftrightarrow\left(x-9\right)^2-162=0\Leftrightarrow x=\frac{18\pm8\sqrt{2}}{2}=9\pm9\sqrt{2}\)
\(x^2-18x+81=0\)
\(\Leftrightarrow x^2-2.9x+9^2=0\)
\(\Leftrightarrow\left(x-9\right)^2=0\Leftrightarrow x=9\)
Trả lời:
\(x^2-18x+81=0\)
\(\Leftrightarrow x^2-2.x.9+9^2=0\)
\(\Leftrightarrow\left(x-9\right)^2=0\)
\(\Leftrightarrow x-9=0\)
\(\Leftrightarrow x=9\)
Vậy x = 9 là nghiệm của pt.