Giúp e dzới, e đang cần gấp kinh khủng
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g: \(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
h: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
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a, Ta có phương trình cân bằng nhiệt
\(Q_{toả}=Q_{thu}\\ \Leftrightarrow2.380\left(150-t_{cb}\right)=1,5.4200\left(t_{cb}-30\right)\\ \Rightarrow t_{cb}=42,91^o\)
b, Theo đề bài ta đc
( đoạn này mik chưa hiểu đề lắm nói rõ ra đc ko bạn )
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a) \(x=x_0+v_0t+\dfrac{1}{2}at^2=10+5t+\dfrac{1}{2}\cdot1\cdot t^2=10+5t+0,5t^2\)
b) Sau 20s:
Vận tốc vật: \(v=v_0+at=5+1\cdot20=25\)m/s
Vật cách gốc tọa độ: \(x=10+5\cdot20+0,5\cdot20^2=310\left(m\right)\)
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a, Do MN ⊥ Ax , MN ⊥ Cy => Ax // Cy ( từ vuông góc đến song song)
b, Kẻ Bz // AB // Cy
Do Ax // Bz \(\Rightarrow\widehat{A}=\widehat{B_1}=40^o\)(2 góc so le trong)
Do Bz // Cy \(\Rightarrow\widehat{B_2}=\widehat{C}=50^o\)
Ta thấy: \(\widehat{ABC}=\widehat{B_1}+\widehat{B_2}=40^o+50^o=90^ohayAB\perp BC\left(dpcm\right)\)