Câu 1 :
Gọi số học sinh khối 6 là a ( \(200\le a\le400\) ; \(a\in\) N )
Theo bài ra ta có :
+) \(\left(a-5\right)⋮12\)
+) \(\left(a-5\right)⋮15\)
+) \(\left(a-5\right)⋮18\)
\(\Rightarrow\left(a-5\right)\in BC\left(12;15;18\right)\)
Ta có : \(12=2^2.3\)
\(15=3.5\)
\(18=3^2.2\)
\(\Rightarrow BCNN\left(12;15;18\right)=3^2.2^2.5=180\)
\(\Rightarrow B\left(180\right)=\left\{0;180;360;720;...\right\}\)
Vì \(200\le a\le400\)
\(\Rightarrow a=360\)
Vậy số học sinh khối 6 là 360 em
Câu 2 :
a) \(\left(2x+1\right)^2=16\)
\(\Rightarrow\left(2x+1\right)^2=4^2\)
\(\Rightarrow\begin{cases}2x+1=4\\2x+1=-4\end{cases}\)
Trường hợp 1
\(\Rightarrow2x+1=4\)
\(\Rightarrow2x=4-1\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=3:2\)
\(\Rightarrow x=1,5\)
Trường hợp 2 :
\(\Rightarrow2x+1=-4\)
\(\Rightarrow2x=\left(-4\right)-1\)
\(\Rightarrow2x=-5\)
\(\Rightarrow x=-5:2\)
\(\Rightarrow-2,5\)
Vậy \(x=\begin{cases}-2,5\\1,5\end{cases}\)
b) \(x-\frac{12}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{12}{4}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
Câu 3 :
\(\frac{7}{8}.\frac{64}{49}-\frac{64}{49}:\left(-\frac{3}{7}+\frac{5}{13}+-\frac{4}{13}\right)\)
\(=\frac{7}{8}.\frac{64}{49}-\frac{64}{49}:\left(-\frac{3}{7}+\frac{1}{13}\right)\)
\(=\frac{7}{8}.\frac{64}{49}-\frac{64}{49}:-\frac{32}{91}\)
\(=\frac{7}{8}.\frac{64}{49}-\frac{64}{49}.-\frac{91}{32}\)
\(=\frac{64}{49}.\left[\frac{7}{8}-\left(-\frac{91}{32}\right)\right]\)
\(=\frac{64}{91}.\frac{119}{32}\)
\(=\frac{34}{13}\)