mng ơi giúp e với ạ!! e cần gấp :((
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Bài 5:
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-20}{1}=-20\)
Do đó: x=-40; y=-60; z=-80
1.I wish Tracy knew what i feel about her
2.I wish I studied math as well as peter
3.my mother wishes I would do well in the final exam next week
4.Rachel wishes she had straight hair
5.I wish they stopped making so much noise upstairs
9. It's time for us to go home. (It's time for you to go bed :>)
Khuya rồi ngủ đi bạn, sáng mai có ng giải cho nè!
1 The teacher made the students do their homework
2 I wouldn't have done it if it hadn't been good for my health
3 They have called off tomorrow's meeting
4 You can hardly pass your exam if you don't study
5 It was such good beer that I had to order another one
6 The exam was easier than we had thought before
7 I wish you had passed the exam
8 I was on my way to pick my son up from school
9 It is time we went home
10 The class have been put off until next Tuesday
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: \(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
\(e,=\dfrac{1}{x-1}-\dfrac{2x}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x^2-2x+1}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}=\dfrac{x-1}{x^2+1}\\ f,=\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\\ =\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{2\left(3x-1\right)^2}{2\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
\(g,=\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\\ =\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\\ h,=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
a: \(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
b: \(N=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
1. \(M=\dfrac{5}{x-1}-\dfrac{8}{x^2-1}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(M=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)
\(M=\dfrac{1}{x-1}.\)
2. \(N=\dfrac{5}{x-1}+\dfrac{8}{1-x^2}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(N=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}\)
\(N=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}.\)
3. \(Q=\dfrac{1}{2x-1}-\dfrac{4}{4x^2-1}-\dfrac{2}{2x+1}\left(x\ne\pm\dfrac{1}{2}\right).\)
\(Q=\dfrac{2x+1-4-2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(Q=\dfrac{-2x-1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-1}{2x-1}.\)
4. \(F=\dfrac{x+3}{x-2}+\dfrac{x+2}{3-x}+\dfrac{x+2}{x^2-5x+6}\left(x\ne2,x\ne3\right).\)
\(F=\dfrac{x+3}{x-2}-\dfrac{x+2}{x-3}+\dfrac{x+2}{\left(x-3\right)\left(x-2\right)}\)
\(F=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x-2\right)+x+2}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{1}{x-2}.\)