giúp mình 2 bài dưới với ạ
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Bài 1
1) (-35) + (-9)
= -(35 + 9)
= -44
2) (-17) + (-53)
= -(17 + 53)
= -70
3) (-7) + (-14)
= -(7 + 14)
= -21
4) (-26) + (-6)
= -(26 + 6)
= -32
5) (-75) + (-50)
= -(75 + 50)
= -125
6) (-5) + (-248)
= -(5 + 248)
= -253
7) (-37) + (-23)
= -(37 + 23)
= -60
8) (-25) + (-15)
= -(25 + 15)
= -40
9) (-36) + (-6)
= -(36 + 6)
= -42
Bài 2
1) (-65) + 37
= -(65 - 37)
= -28
2) (-136) + 122
= -(136 - 122)
= -14
3) 19 + (-49)
= -(49 - 19)
= -30
4) 21 + (-78)
= -(78 - 21)
= -57
5) 363 + (-400)
= -(400 - 363)
= -37
6) (-15) + 7
= -(15 - 7)
= -8
7) (-237) + 137
= -(237 - 137)
= -100
8) 78 + (-123)
= -(123 - 78)
= -45
9) (-75) + 50
= -(75 - 50)
= -25
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = \(\sum\limits^{50}_1\left(2x\right)-\sum\limits^{50}_1\left(2x-1\right)\) = 5050
b) B = \(\sum\limits^{2010}_1x^3\) = 4084663313000
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\(2x=3y\\ =>\dfrac{x}{3}=\dfrac{y}{2}\\ 4y=5z\\ =>\dfrac{y}{5}=\dfrac{z}{4}\\ \dfrac{x}{3}=\dfrac{y}{2}\\ =>\dfrac{x}{3.5}=\dfrac{y}{2.5}\\ =>\dfrac{x}{15}=\dfrac{y}{10}\\ \dfrac{y}{5}=\dfrac{z}{4}\\ =>\dfrac{y}{5.2}=\dfrac{z}{4.2}\\ =>\dfrac{y}{10}=\dfrac{z}{8}\\ =>\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{x+y+z}{15+10+8}=\dfrac{11}{33}=\dfrac{1}{3}\\ =>\left\{{}\begin{matrix}x=\dfrac{1}{3}.15=5\\y=\dfrac{1}{3}.10=\dfrac{10}{3}\\z=\dfrac{1}{3}.8=\dfrac{8}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{3^{10}.11+9^5.5}{27^3.2^4}.x=-9\\ =>\dfrac{3^{10}.11+\left(3^2\right)^5.5}{\left(3^3\right)^3.2^4}.x=-9\\ =>\dfrac{3^{10}.\left(11+5\right)}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.16}{3^9.2^4}.x=-9\\ =>\dfrac{3^{10}.2^4}{3^9.2^4}.x=-9\\ =>3^1.x=-9\\ =>x=-9:3\\ =>x=-3\)
bài 10
b) x3-5x-14=x(x2-5x-14)=x(x2+7x-2x-14)=x[(x2-2x)+(7x-14)]=x[x(x-2)+7(x-2)]=x(x-2)(x+7)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-7\end{matrix}\right.\)
Bài 10:
c: Ta có: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)