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\(a,\left(a+2\right)^2+\left|3b-4\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left(a+2\right)^2=0\\\left|3b-4\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}a+2=0\\3b-4=0\end{cases}}\Rightarrow\hept{\begin{cases}a=-2\\b=\frac{4}{3}\end{cases}}\)
\(b,\left(3a^2+27\right)^2+\left|b-1\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left(3a^2+27\right)^2=0\\\left|b-1\right|=0\end{cases}\Rightarrow\hept{\begin{cases}3a^2+27=0\\b-1=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a^2=27\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a^2=9\\b=1\end{cases}\Rightarrow}\hept{\begin{cases}\orbr{\begin{cases}a=3\\a=-3\end{cases}}\\b=1\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}3a^2=27\\b=1\end{cases}\Rightarrow\hept{\begin{cases}a^2=9\\b=1\end{cases}\Rightarrow}\hept{\begin{cases}\orbr{\begin{cases}a=3\\a=-3\end{cases}}\\b=1\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)
b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=1\)
Bài 2:
a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)
b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)
c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)
d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)
Bài 3:
a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)
\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)
\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)
\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)
\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)
\(=9.\dfrac{2}{3}.-1\)
\(=6.-1=-6\)
b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)
\(#Wendy.Dang\)
Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))
![](https://rs.olm.vn/images/avt/0.png?1311)
b.\(ĐK:x;y\in Z^+;x;y\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)
\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)
\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)
\(\Leftrightarrow x=\dfrac{5y}{y-5}\)
\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )
Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
TH1:
\(y-5=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm ) ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )
Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:
\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)
Câu a mik ko bt nên bạn tham khảo nhé:
https://hoc24.vn/cau-hoi/cho-a-b-c-0-va-day-ti-so-dfrac2bc-aadfrac2c-babdfrac2ab-cctinh-p-dfracleft3a-2brightleft3b-2crightleft.177725456910
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)
a)(a+2)2\(\left|3b-4\right|\)=0
vì (a+2)2 \(\ge\) =0 \(\forall\) a
\(\left|3b-4\right|\)\(\ge0\forall b\)
=>(a+2)2=0 và \(\left[3b-4\right]\) =0
=> a-2=0 và 3b-4=0
=>a=2 và b=\(\dfrac{4}{3}\)