help meeeee
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8 tháng 11 2021
a) \(\left(3x-1\right)^6=\left(3x-1\right)^4\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-1=1\\3x-1=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
b) Áp dụng t/c dtsbn:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
\(\Rightarrow M=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)
Help meeeee
DT
Đỗ Thanh Hải
CTVVIP
24 tháng 9 2021
1 went
2 have never been
3 haven't seen
4 saw
5 have they been
6 got
7 have already done
25 tháng 9 2021
1. I went to the new Mexican restaurant yesterday.
2. I have never been there. Did you like it?
3. I have not seen Ann for a very long time.
4. Really? I saw her yesterday at the gym.
5. How long have they been married?
6. Only a year. They got married last June.
7. No, you don't have to. I have already done it.
RG
3
30 tháng 4 2017
Ta có :
\(\dfrac{x}{5}=\dfrac{2}{3}\)
\(\Rightarrow x.3=5.2\)
\(x.3=10\)
\(x=10:3\)
\(x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\) là giá trị cần tìm
30 tháng 4 2017
Ta có :\(\dfrac{x}{5}=\dfrac{2}{3}\)
\(\Rightarrow x.3=2.5\)
\(x.3=10\)
\(x=10:3\)
\(\Rightarrow x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\)
Ta có :
\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+3x^2+3x+1+5x^2+14x+9}\) \(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+14x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x^2+5x+9x+9}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)^3+5x\left(x+1\right)+9\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+2\right)\left(x-2\right)}{\left(x+1\right)^3+\left(x+1\right)\left(5x+9\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+7x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x^2+2x+5x+10}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}\)
\(=\dfrac{x-2}{x+5}\)
\(\Rightarrow a=-2;b=5\)
Vậy a + b = -2 + 5 = 3
a+b=3