bài 1: tính
a) 3/4+(-5/2)+(-3/5)
b) \(\sqrt{\left(7\right)^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}\)
c)\(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}\)
K
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KV
26 tháng 12 2023
a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1\)
\(=-\dfrac{3}{2}\)
b) \(8.\sqrt{9}-\sqrt{64}\)
\(=8.3-8\)
\(=24-8\)
\(=16\)
c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)
\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)
\(=-1+\dfrac{5}{2}\)
\(=\dfrac{3}{2}\)
25 tháng 10 2022
a: \(=2\cdot\dfrac{5}{4}-3\cdot\dfrac{7}{6}+4\cdot\dfrac{9}{8}=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{9}{2}=\dfrac{7}{2}\)
b: \(=18-16\cdot\dfrac{1}{2}+\dfrac{1}{16}\cdot\dfrac{3}{4}\)
=10+3/64
=643/64
c: \(=\dfrac{2}{3}\cdot\dfrac{9}{4}-\dfrac{3}{4}\cdot\dfrac{8}{3}+\dfrac{7}{5}\cdot\dfrac{5}{14}=\dfrac{3}{2}-2+\dfrac{1}{2}=2-2=0\)
20 tháng 12 2022
a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
25 tháng 10 2022
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
24 tháng 8 2021
a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
8 tháng 6 2021
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
2 tháng 12 2023
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
Bài 1:
a) \(\sqrt{1,44\cdot1,21-1,44\cdot0,4}\)
\(=\sqrt{1,44\cdot\left(1,21-0,4\right)}\)
\(=\sqrt{1,44\cdot0,81}\)
\(=\sqrt{1,44}\cdot\sqrt{0,81}\)
\(=1,2\cdot0,9\)
\(=1,08\)
b) \(\dfrac{\sqrt{5}-2}{\sqrt{5}+2}+\sqrt{80}\)
\(=\dfrac{\left(\sqrt{5}-2\right)^2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}+4\sqrt{5}\)
\(=\dfrac{5-4\sqrt{5}+4}{1}+4\sqrt{5}\)
\(=9-4\sqrt{5}+4\sqrt{5}\)
\(=9\)
c) \(\sqrt[3]{16}+\sqrt[3]{2}\left(\sqrt[3]{4}-\sqrt[3]{2}\right)\)
\(=\sqrt[3]{2^3\cdot2}+\sqrt[3]{2\cdot4}-\sqrt[3]{2\cdot2}\)
\(=2\sqrt[3]{2}+\sqrt[3]{8}-\sqrt[3]{4}\)
\(=2\sqrt[3]{2}+2-\sqrt[3]{4}\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
Bài 2: Ta có:
\(VT=\dfrac{1}{\sqrt{a}-\sqrt{b}}:\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}:\dfrac{\sqrt{ab}\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\cdot\dfrac{1}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{1}{a-b}=VP\left(dpcm\right)\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)