Giải giúp mình bài 2 với ạ
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LM
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20 tháng 12 2022
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HK
3
31 tháng 10 2021
Bạn ơi, làm như vậy thì quá ngắn rồi ạ, với lại bạn làm thiếu mất đề bài của mình rồi
PN
0
BC
2
MP
2 tháng 5 2023
2.
a. \(\dfrac{-4}{9}\) . \(\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\) = \(\dfrac{-4}{9}.\left(\dfrac{7}{15}+\dfrac{8}{15}\right)\) = \(\dfrac{-4}{9}\) . 1 = \(\dfrac{-4}{9}\)
b. \(\dfrac{5}{-4}.\dfrac{16}{25}+\dfrac{-5}{4}.\dfrac{9}{25}\) = \(\dfrac{-5}{4}.\left(\dfrac{16}{25}+\dfrac{6}{25}\right)\) = \(\dfrac{-5}{4}.1\) = \(\dfrac{-5}{4}\)
c. \(4\dfrac{11}{23}-\dfrac{9}{14}+2\dfrac{12}{23}-\dfrac{5}{4}\) = \(\left(4\dfrac{11}{23}+2\dfrac{12}{23}\right)\) \(-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{68}{23}-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{745}{322}\) - \(\dfrac{5}{4}=\dfrac{685}{644}\)
d. \(2\dfrac{13}{27}-\dfrac{7}{15}+3\dfrac{14}{27}-\dfrac{8}{15}\) = \(\left(2\dfrac{13}{27}+3\dfrac{14}{27}\right)\) - \(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) = \(\dfrac{68}{27}\) - \(\dfrac{-1}{15}\) =
e. \(11\dfrac{1}{4}-\left(2\dfrac{7}{5}+5\dfrac{1}{4}\right)\) = \(11\dfrac{1}{4}\) - \(\dfrac{81}{20}\) = \(\dfrac{-13}{10}\)
g. \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\) = \(\dfrac{7}{9}.\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\) = \(\dfrac{7}{9}.1+\dfrac{12}{19}\) = \(\dfrac{7}{19}+\dfrac{12}{19}\) = \(1\)
30 tháng 9 2021
\(3,\\ a,P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\left(x>0;x\ne1;x\ne4\right)\\ P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+4}\\ P=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\\ b,P=\dfrac{1}{4}\Leftrightarrow\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow4\sqrt{x}-8=3\sqrt{x}\\ \Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\)
\(c,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\\ \Leftrightarrow P=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{\left(\sqrt{3}-1\right)\left(3\sqrt{3}-3\right)}{18}\\ P=\dfrac{12-6\sqrt{3}}{18}=\dfrac{2-\sqrt{3}}{3}\)
\(d,P\in Z\Leftrightarrow3P\in Z\Leftrightarrow\dfrac{3\sqrt{x}-6}{3\sqrt{x}}\in Z\Leftrightarrow1-\dfrac{6}{3\sqrt{x}}\in Z\\ \Leftrightarrow6⋮3\sqrt{x}\Leftrightarrow3\sqrt{x}\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;6\right\}\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x\in\left\{1;4;9;36\right\}\)
30 tháng 9 2021
\(4,\\ A=\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}\\ A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-1\right)^2}\\ A=\left|x+1\right|+\left|x-1\right|\\ A=\left|x+1\right|+\left|1-x\right|\ge\left|x+1+1-x\right|=\left|2\right|=2\)
Dấu \("="\Leftrightarrow x=1\)