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\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
<=>\(\left(\frac{2}{3}\right)^{\frac{n}{2}}=\left(\frac{2}{3}\right)^5\)
<=>\(\frac{n}{2}=5\)
<=>n=10
\(\left(\frac{4}{9}\right)^n=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow\left(\frac{2}{3}\right)^{2n}=\left(\frac{2}{3}\right)^5\)
\(\Rightarrow2n=5\Rightarrow n=\frac{5}{2}\)
Vậy n = 5/2
1)
\(=-18+9+16=7\)
2)
\(144+n^2=225\)
\(n^2=225-144\)
\(n^2=81=\left(-9\right)^2=9^2\)
\(n=9\)hoặc \(n=-9\)
3) Tính tổng:
(-25)+(-24)+(-23)+..+(-3)+(-2)+(-1).+0+1+2+3+...+23+24+25+...+40
=0+(1-1)+(2-2)+(3-3)+...+(25-25)+...+26+27+...+40
=26+27+28+...+40 ( Số số hạng: (40-26):1+1=15 ( số hạng))
= ( 40 + 26) x15:2=495