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VD
16 tháng 6 2023
= (x+1/2)(x-1/2 x + 1/4)
=(x+1/2)(1/2x + 1/4)
=x(1/2x + 1/4) + 1/2(1/2x+1/4)
=1/2 x^2 + 1/4 x + 1/4 x + 1/8
= x^2/2 + 1/2 x + 1/8
H9
HT.Phong (9A5)
CTVHS
16 tháng 6 2023
\(\left(x-\dfrac{1}{2}\right)\left(x-\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(=\left(x+\dfrac{1}{2}\right)\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{2}x^2+\dfrac{1}{4}x+\dfrac{1}{4}x+\dfrac{1}{8}\)
\(=\dfrac{1}{2}x^2+\dfrac{1}{2}x+\dfrac{1}{8}\)
K
1
8 tháng 12 2021
\(=\dfrac{3x^2+5x+14+x^2-1-4x^2+4x-4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{9x+9}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{9}{x^2-x+1}\)
15 tháng 12 2021
\(a,=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{x-y}{xy\left(y-x\right)}=\dfrac{-1}{xy}\\ b,=\dfrac{x+3-x-4}{x-2}=\dfrac{-1}{x-2}\)
10 tháng 1 2022
1) \(A=\left[x^4-\left(x-1\right)^2\right]:\left(x^2+x-1\right)-x^2+x=\left[\left(x^2-x+1\right)\left(x^2+x-1\right)\right]:\left(x^2+x-1\right)-x^2+x=x^2-x+1-x^2+x=1\)
2) \(B=\dfrac{\left(x+1\right)\left(x+2\right)+4\left(x-2\right)+2-7x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4}{x^2-4}=1\)
H
22 tháng 7 2023
\(a,\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}-\dfrac{x-4}{1-x}\\ =\dfrac{x+2}{x-1}-\dfrac{x-3}{x-1}+\dfrac{x-4}{x-1}\\ =\dfrac{x+2-x+3+x-4}{x-1}\\ =\dfrac{x+1}{x-1}\)
\(b,\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{x^2-25}\\ =\dfrac{1}{x+5}-\dfrac{1}{x-5}+\dfrac{2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{x-5-x-5+2x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x-10}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2}{x+5}\)
\(c,x+\dfrac{2y^2}{x+y}-y\\ =\dfrac{x\left(x+y\right)+2y^2-y\left(x+y\right)}{x+y}\\ =\dfrac{x^2+xy+2y^2-xy-y^2}{x+y}\\ =\dfrac{x^2+y^2}{x+y}\)
5 tháng 1 2022
\(\dfrac{1}{x^2+x}=\dfrac{x-1}{x\left(x-1\right)\left(x+1\right)};\dfrac{x^2-4}{x^2-1}=\dfrac{x\left(x^2-4\right)}{x\left(x-1\right)\left(x+1\right)}\\ \dfrac{1}{y-1}-\dfrac{1}{y}=\dfrac{y-y+1}{y\left(y-1\right)}=\dfrac{1}{y\left(y-1\right)}\)
TN
23 tháng 7 2023
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
25 tháng 12 2021
Nhận thấy \(x^3-x=x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)\)
\(\dfrac{3}{x}-\dfrac{x}{x-1}-\dfrac{x^2}{x+1}-\dfrac{x^2-3}{x^3-x}\\ =\dfrac{3x^2-3-x^3-x^2-x^4+x^3-x^2+3}{x\left(x-1\right)\left(x+1\right)}\\ =\dfrac{-x^4+x^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{-x^2\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)\left(x+1\right)}=-x\)
\(\dfrac{x+1}{x^2-4}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+1}{x+2}=\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x+1}=x-2\)
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