![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(-x+80=-220-5x\)
\(\Rightarrow-x+5x=-220-80\)
\(\Rightarrow4x=-300\)
\(\Rightarrow x=-\dfrac{300}{4}\)
\(\Rightarrow x=-75\)
b) \(98+\left(x-12\right)+68=-80-x\)
\(\Rightarrow166+\left(x-12\right)=-80-x\)
\(\Rightarrow166+x-12=-80-x\)
\(\Rightarrow154+x=-80-x\)
\(\Rightarrow154+80=-x-x\)
\(\Rightarrow-2x=234\)
\(\Rightarrow x=-\dfrac{234}{2}\)
\(\Rightarrow x=-117\)
c) \(122+x-78=-55-4x-1\)
\(\Rightarrow44+x=-56-4x\)
\(\Rightarrow x+4x=-56-44\)
\(\Rightarrow5x=-100\)
\(\Rightarrow x=-\dfrac{100}{5}\)
\(\Rightarrow x=-20\)
d) \(663+9x=-x-37\)
\(\Rightarrow9x+x=-37-663\)
\(\Rightarrow10x=-700\)
\(\Rightarrow x=-\dfrac{700}{10}\)
\(\Rightarrow x=-70\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-4\right)\left(x-2\right)^3=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(x-2\right)^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\x-2-0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
\(\Rightarrow x=2\)
b) \(3^{x-1}:81=3^3\)
\(\Rightarrow3^{x-1}:3^4=3^3\)
\(\Rightarrow3^{x-1-4}=3^3\)
\(\Rightarrow3^{x-5}=3^3\)
\(\Rightarrow x-5=3\)
\(\Rightarrow x=8\)
c) \(x^{13}=x\)
\(\Rightarrow x^{13}-x=0\)
\(\Rightarrow x\left(x^{12}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
d) \(\left(x-9\right)^4=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^2=x-9\)
\(\Rightarrow\left(x-9\right)^2-\left(x-9\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-9-1\right)=0\)
\(\Rightarrow\left(x-9\right)\left(x-10\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
1, Áp dụng t/c dtsbn:
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{z-x}{3-6}=\dfrac{-21}{-3}=7\\ \Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=21\end{matrix}\right.\)
2, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
Bài 4:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}=\dfrac{130}{\dfrac{13}{12}}=120\)
Do đó: x=60; y=40; z=30
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 63: A
Câu 64: B
Câu 65: C
Câu 71: D
Câu 72: A
Câu 73: C
Câu 74: B
Câu 75: C
Câu 76: B
Câu 77: A
Câu 78: C
Câu 79: B
Câu 80: C
![](https://rs.olm.vn/images/avt/0.png?1311)
h) \(\left(a-b\right)-\left(c-d\right)-\left(a+d\right)-\left(b+c\right)\)
\(=a-b-c+d-a-d-b-c\)
\(=\left(a-a\right)-\left(b+b\right)-\left(c+c\right)+\left(d+d\right)\)
\(=-2b-2c\)
i) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)
\(=a-c+d-c+a-d\)
\(=\left(a+a\right)-\left(c+c\right)+\left(d+d\right)\)
\(=2a-2c\)
k) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)
\(=-a-b+c-d+a-b-c-d\)
\(=\left(-a+a\right)-\left(b+b\right)+\left(c-c\right)-\left(d+d\right)\)
\(=-2b-2d\)
n) \(-a\left(-b-c+d-e\right)-\left(-a\right)-\left(-b+c-d\right)\)
\(=ab+ac-ad+ae+a+b-c+d\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)^7=\left(x-2\right)^5\)
\(\Leftrightarrow\left(x-2\right)^7-\left(x-2\right)^5=0\)
\(\Leftrightarrow\left(x-2\right)^5.\left[\left(x-2\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^5=0\\\left(x-2\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3;x=1\end{cases}}}\)
Vậy \(x\in\left\{1;-1;2\right\}\)
:V sửa lại hộ mình cái kết luận nhé
\(x\in\left\{-1;3;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\left(1-\frac{1}{100}\right)\)
=\(\left(\frac{100}{100}-\frac{1}{100}\right)\)
=\(\frac{99}{100}\)
Chúc bạn học tốt
a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)
\(=a-c+d-c+a-d\)
\(=2a-2c\)
b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)
\(=-a-b+c-d+a-b-c-d\)
\(=-2b-2d\)
c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)
\(=ab-ac-ad-ab-ac+ad\)
\(=-2ac\)
d) \(\left(a-b\right)+\left(c-d\right)+\left(a+c\right)-\left(b+d\right)\)
\(=a-b+c-d+a+c-b-d\)
\(=2a-2b+2c-2d\)