Ai giúp mik vs mik đang cần gấp lắm ạ
Giúp mik vs mik đang cần gấp lắm ạ
Giúp mik vs ạ, mik đang cần gấp lắm!
Bài 34:
a: =>x+28=0
=>x=-28
b: =>27-x=0 hoặc x+9=0
=>x=27 hoặc x=-9
c: =>x(x-43)=0
=>x=0 hoặc x=43
Giải bài tập giúp mik vs ạ 🥺... mik đang cần gấp lắm ạ...
Câu 3:
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(2x\left(x+7\right)-3x\left(x+1\right)\)
\(=2x^2+14x-3x^2-3x\)
\(=-x^2+11x\)
Câu 2:
a: Ta có: \(\left(-8x^5+12x^3-16x^2\right):4x^2\)
\(=-8x^5:4x^2+12x^3:4x^2-16x^2:4x^2\)
\(=-2x^3+3x-4\)
b: Ta có: \(\left(12x^3y^3-18x^2y+9xy^2\right):6xy\)
\(=12x^3y^3:6xy-18x^2y:6xy+9xy^2:6xy\)
\(=2x^2y^2-3x+\dfrac{3}{2}y\)
c: Ta có: \(\dfrac{x^3-11x^2+27x-9}{x-3}\)
\(=\dfrac{x^3-3x^2-8x^2+24x+3x-9}{x-3}\)
\(=x^2-8x+3\)
d: Ta có: \(\dfrac{6x^4-13x^3+7x^2-x-5}{3x+1}\)
\(=\dfrac{6x^4+2x^3-15x^3-5x^2+12x^2+4x-5x-\dfrac{5}{3}-\dfrac{10}{3}}{3x+1}\)
\(=2x^3-5x^2+4x-\dfrac{5}{3}-\dfrac{\dfrac{10}{3}}{3x+1}\)
ai giúp mik vs!!!
mik đang cần gấp lắm
giúp đi mik k cho
Ai giúp mik vs mik cần gấp lắm ạ
1. D => bỏ
2. C
3. C
4. A => the
5. A
6. C => next
7. C => taught
8. C => the most
9. D => from
10. B => would
II
1. I wish I would become a singer.
2. She used to walk to school when she was a child.
3. If I were you, I would learn Chinese
4. The man who I met yesterday was my uncle
5. The trees are watered by me everyday
Các bạn giúp mik nhanh vs ạ! Huhu mik đang cần gấp lắm
\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)
\(=\dfrac{x^3-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)
\(=\dfrac{x^3-x^2-2x+2x-4}{x^2-4}\)
\(=\dfrac{x^3-x^2-4}{x^2-4}\)
a,\(C=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)
\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{\left(x^2-4\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=\dfrac{\left(x-2\right)\left(x+2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow C=x-1\)
b, C=0\(\Rightarrow x-1=0\Rightarrow x=1\)
c, Để C nhận giá trị dương thì \(x-1\ge0\Rightarrow x\ge1\)
Ai giúp mik với mik cảm ơn mik đang cần gấp lắm ạ
c: \(=\dfrac{-27\cdot100}{-30}=\dfrac{2700}{30}=90\)
Ai bt giải giúp mik vs ạ, mik đg cần gấp lắm
20. eat
21. is playing
22. washes
23. rings
24. bring
eat
is playing
washes
rings
bring
Chúc em học tốt
20 eat
21 is playing
22 washes
23 rings
24 bring
|x+3|≥4
các bn giúp mik vs ạ mik đang cần gấp lắm. Cảm ơn nhìu nha
\(\Rightarrow x+3\ge4\\ \Rightarrow x\ge1\)
\(\left|x+3\right|\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3\ge4\\x+3\le4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le1\end{matrix}\right.\)