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Tìm x,biết
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)
\(=\dfrac{4.4^5.6.6^5}{3.3^5.2.2^5}\)
\(=\dfrac{4^6.6^6}{3^6.2^6}\)
\(=\dfrac{2^6.2^6.2^6.3^6}{3^6.2^6}\)
\(=2^{12}=2^{3^4}=8^4=8^x\)
Vậy x = 4
45+45+45+4535+35+35.65+65+65+65+65+6525+2545+45+45+4535+35+35.65+65+65+65+65+6525+25
=4.45.6.653.35.2.25=4.45.6.653.35.2.25
=46.6636.26=46.6636.26
=26.26.26.3636.26=26.26.26.3636.26
=212=234=84=8x=212=234=84=8x
Vậy x = 4
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
\(\Rightarrow\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\)
\(\Rightarrow\dfrac{4^5.4.6^5.6}{3^5.3.2^5.2}=2^n\)
\(\Rightarrow\dfrac{\left(2.2\right)^5.2.2.\left(3.2\right)^5.3.2}{3^5.3.2^5.2}=2^n\)
\(\Rightarrow\dfrac{2^5.2^5.2.2.3^5.2^5.3.2}{3^5.3.2^5.2}=2^n\)
Rút gọn vế trái ta có :
\(2^5.2.2.^5=2^n\)
\(\Rightarrow2^{12}=2^n\)
\(\Rightarrow n=12\) ( Thỏa mãn điều kiện \(n\in N\) )
Vậy n =12
Lời giải:
\(\text{VT}=\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)
\(=\frac{4.4^5}{3.3^5}.\frac{6.6^5}{2.2^5}=\frac{4^6.6^6}{3^6.2^6}=\frac{2^{12}.2^6.3^6}{3^6.2^6}=2^{12}\)
Do đó: \(8^{|2x+6|}=2^{12}\Leftrightarrow 2^{3|2x+6|}=2^{12}\)
\(\Leftrightarrow 3|2x+6|=12\Leftrightarrow |2x+6|=4\)
\(\Rightarrow\left[{}\begin{matrix}2x+6=4\\2x+6=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
=>\(\dfrac{4^5\left(1+1+1+1\right)}{3^5\left(1+1+1\right)}.\dfrac{6^5\left(1+1+1+1+1+1\right)}{2^5\left(1+1\right)}=2^n\)
=>\(\dfrac{4^5.4}{3^5.3}.\dfrac{6^5.6}{2^5.2}=2^n\) =>\(\dfrac{4^6}{3^6}.\dfrac{6^6}{2^6}=2^n\)
=>\(\left(\dfrac{4.6}{3.2}\right)^6=2^n\) =>\(4^6=2^n\) =>\(2^{12}=2^n\) =>n=12.
\(1\)/
\(a\)) \(=\left(\dfrac{7}{5}-\dfrac{8}{7}\right)+\dfrac{17}{5}:0,6\)
\(=\dfrac{9}{35}+\dfrac{17}{3}\)
\(=\dfrac{622}{105}\)
\(b\)) \(=\dfrac{11}{6}+\dfrac{-14}{15}\)
\(=\dfrac{9}{10}\)
\(c\)/ \(=\dfrac{7}{4}-\dfrac{2}{3}\)
\(=\dfrac{13}{12}\)
Câu hỏi của Lê Khánh Nhi - Toán lớp 7 - Học toán với OnlineMath sửa n thành x cho sửa cho nó thành lũy thừa luôn
