\(\sqrt{5^2-4^2}-\sqrt{0,25}+\sqrt{\frac{9}{4}}\)

\(=\sqrt{3^2}-\sqrt{\left(0,5\right)^2}+\sqrt{\left(\frac{3}{2}\right)^2}\)

\(=3-\left(0,5\right)+\frac{3}{2}\)

\(=4\)

\(=\sqrt{25-16}-0,5+\frac{3}{2}\)

\(=\sqrt{9}-0,5+1,5\)

\(=3-0,5+1,5\)

\(=4\)

a,b,c tỉ lệ với 3,2,5

\(\Rightarrow\frac{a}{3}=\frac{b}{2}=\frac{c}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{a}{3}=\frac{b}{2}=\frac{c}{5}=\frac{a-b+c}{3-2+5}=\frac{12}{6}=2\)

\(\Rightarrow a=2.3=6;b=2.2=4;c=2.5=10\)

Vậy ...

trả lời giúp mình cần gấp

\(\sqrt{x}=12\)

\(\Rightarrow12^2=x\)hay x = 144

\(\frac{3x-4}{2}=\frac{1}{3}\)

\(\Rightarrow\left(3x-4\right).3=2\)

\(\Rightarrow9x-12=2\)

\(\Rightarrow9x=14\)

\(\Rightarrow x=\frac{19}{4}\)

à, cho mình sửa \(\frac{19}{4}\)thành \(\frac{14}{9}\)nhé

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(A=1-\frac{1}{20}\)

\(A=\frac{19}{20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

Ta có :

\(\frac{8^{14}}{4^{12}}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}=\frac{2^{42}}{2^{24}}=2^{18}\)