\(\left[-\sqrt{2,25}+4\sqrt{\left(-2,15\right)^2}-\left(3\sqrt{\dfrac{7}{6}}\right)^2\right]\sqrt{1\dfrac{9}{16}}\)

\(=\left[-1,5+4\sqrt{2,15^2}-9\cdot\dfrac{7}{6}\right]\sqrt{\dfrac{25}{16}}\)

\(=\left[4\cdot\dfrac{43}{20}-10,5-1,5\right]\cdot\dfrac{5}{4}\)

\(=\left[\dfrac{43}{5}-12\right]\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{5}\cdot\dfrac{5}{4}-12\cdot\dfrac{5}{4}\)

\(=\dfrac{43}{4}-15=\dfrac{-17}{4}\)

\(5^x-5^{x-1}-5^{x-2}=2375\\ \Leftrightarrow5^x-\dfrac{5^x}{5}-\dfrac{5^x}{25}=2375\\ \Leftrightarrow5^x\left(1-\dfrac{1}{5}-\dfrac{1}{25}\right)=2375\\ \Leftrightarrow5^x.\dfrac{19}{25}=2375\\ \Leftrightarrow5^x=3125\\ \Leftrightarrow5^x=5^5\\ \Leftrightarrow x=5\)

\(5^x-5^{x-1}-5^{x-2}=2375\)

\(\Rightarrow5^{x-2}\left(5^2-5^1-1\right)=2375\)

\(\Rightarrow5^{x-2}.19=2375\)

\(\Rightarrow5^{x-2}=125=5^3\Rightarrow x-2=3\Rightarrow x=5\)

17/21 < 1 < 51/31

=) 17/21 < 51/31

\(\dfrac{17}{21}< \dfrac{17+10}{21+10}=\dfrac{27}{31}< \dfrac{51}{31}\)

\(\Rightarrow\dfrac{17}{21}< \dfrac{51}{31}\)

g) \(\sqrt[]{64}+2\sqrt[]{\left(-3\right)^2}-7\sqrt[]{1,69}+3\sqrt[]{\dfrac{25}{16}}\)

\(=8+2.3-7.1,3+3.\dfrac{5}{4}\)

\(=14-9,1+\dfrac{15}{4}\)

\(=5,1+3,75=8,85\)

ko tính đc

số âm ko có căn bậc 2 số học

\(\sqrt[]{2^2+\sqrt[]{4^2}+\sqrt[]{\left(-6\right)^2}+\sqrt[]{\left(-8\right)^2}}\)

\(=\sqrt[]{4+4+6+8}=\sqrt[]{22}\)

CR HCN là:

27/4 - 3/5 = 123/20(m)

P HCN là:

(27/4 + 123/20).2 = 129/5(m)

ĐS: 129/5m

dấu chấm là dấu nhân nha em

P là chu vi nha

chúc em học tốt!

tick cho mình nha

Câu khó thế

um khó mà giúp tui đi mn

\(\dfrac{x^2-3x}{2x^2-3x-9}=\dfrac{x^2+3x}{A}\)

\(\Rightarrow A=\dfrac{\left(x^2+3x\right)\left(2x^2-3x-9\right)}{x^2-3x}\)

\(\Rightarrow A=\dfrac{x\left(x+3\right)\left(2x^2-3x-9\right)}{x\left(x-3\right)}\)

\(\Rightarrow A=\dfrac{\left(x+3\right)\left(2x^2-3x-9\right)}{\left(x-3\right)}\)

mà \(x=-\dfrac{3}{2}\)

\(\Rightarrow A=\dfrac{\left(-\dfrac{3}{2}+3\right)\left(2\left(-\dfrac{3}{2}\right)^2-3\left(-\dfrac{3}{2}\right)-9\right)}{\left(-\dfrac{3}{2}-3\right)}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(2.\dfrac{9}{4}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}=0\)

A = 0