cho a ≠ \(\dfrac{-7}{3}\); b ≠\(\dfrac{7}{2}\) và 2a-b=7. Tính giá trị của biểu thức: \(\dfrac{5a-b}{3a+7}-\dfrac{3b-2a}{2b-7}\)
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BC
1
12 tháng 9 2021
Ta có \(\left\{{}\begin{matrix}a+b=\dfrac{-2+\sqrt{3}}{3}+\dfrac{-2-\sqrt{3}}{3}=-\dfrac{4}{3}\\ab=\dfrac{\left(-2+\sqrt{3}\right)\left(-2-\sqrt{3}\right)}{9}=\dfrac{1}{9}\end{matrix}\right.\)
\(\left(a+b\right)^2=a^2+b^2+2ab=16\\ \Leftrightarrow a^2+b^2=\dfrac{16}{9}-2\cdot\dfrac{1}{9}=\dfrac{14}{9}\left(1\right)\\ \left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3+\dfrac{1}{3}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{64}{27}\\ \Leftrightarrow a^3+b^3=-\dfrac{64}{27}+\dfrac{4}{9}=-\dfrac{52}{27}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\left(a^2+b^2\right)\left(a^3+b^3\right)=a^5+b^5+a^2b^2\left(a+b\right)=\dfrac{14}{9}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5+\dfrac{1}{81}\cdot\left(-\dfrac{4}{3}\right)=-\dfrac{728}{243}\\ \Leftrightarrow a^5+b^5=-\dfrac{728}{243}+\dfrac{4}{243}=-\dfrac{724}{243}\left(3\right)\)
\(\left(1\right)\left(3\right)\Rightarrow\left(a^2+b^2\right)\left(a^5+b^5\right)=a^7+b^7+a^2b^2\left(a^3+b^3\right)=\dfrac{14}{9}\cdot\left(-\dfrac{724}{243}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7+\dfrac{1}{81}\cdot\left(-\dfrac{52}{27}\right)=-\dfrac{10136}{2187}\\ \Leftrightarrow a^7+b^7=-\dfrac{10136}{2187}-\dfrac{52}{2187}=-\dfrac{10188}{2187}=\dfrac{1132}{243}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 4 2021
\(\pi< a< \dfrac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\)
\(sin\left(\dfrac{7\pi}{2}+a\right)=sin\left(4\pi-\dfrac{\pi}{2}+a\right)=sin\left(-\dfrac{\pi}{2}+a\right)=-sin\left(\dfrac{\pi}{2}-a\right)=-cosa>0\)
Đáp án A
20 tháng 4 2021
a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)
\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)
\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)
\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)
\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
TC
6 tháng 5 2022
a)\(\dfrac{a}{b}=5-\dfrac{3}{5}=\dfrac{25}{5}-\dfrac{3}{5}=\dfrac{22}{5}\)
b)\(\dfrac{a}{b}=\dfrac{5}{6}+\dfrac{4}{7}=\dfrac{35}{42}+\dfrac{24}{42}=\dfrac{59}{42}\)
c)\(\dfrac{a}{b}=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{9}{10}\)
H9
HT.Phong (9A5)
CTVHS
10 tháng 6 2023
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
1 tháng 7 2021
\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)
H
1 tháng 7 2021
1. Tính
\(a,5\times\dfrac{7}{3}=\dfrac{35}{3}\)
\(b,\dfrac{13}{4}:7=\dfrac{13}{4}\times\dfrac{1}{7}=\dfrac{13}{28}\)
2. Tính
\(a,\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}\)
\(=\dfrac{15}{35}+\dfrac{14}{35}+\dfrac{3}{4}\)
\(=\dfrac{29}{35}+\dfrac{3}{4}\)
\(=\dfrac{116}{140}+\dfrac{105}{140}\)
\(=\dfrac{221}{140}\)
\(b,\dfrac{9}{7}-\dfrac{5}{11}\times\dfrac{11}{7}\)
\(=\dfrac{9}{7}-\dfrac{55}{77}\)
\(=\dfrac{99}{77}-\dfrac{55}{77}\)
\(=\dfrac{44}{77}=\dfrac{4}{7}\)
\(c,\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{4}{7}\)
\(=\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}\right)\)
\(=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(=\dfrac{27}{35}\)
\(d,\dfrac{7}{9}\times\dfrac{2}{5}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}:\dfrac{3}{11}\)
\(=\dfrac{14}{45}\times\dfrac{11}{3}\)
\(=\dfrac{154}{135}\)
\(e,\dfrac{9}{7}+\dfrac{2}{3}-\dfrac{1}{4}\)
\(=\dfrac{27}{21}+\dfrac{14}{21}-\dfrac{1}{4}\)
\(=\dfrac{41}{21}-\dfrac{1}{4}\)
\(=\dfrac{164}{84}-\dfrac{21}{84}\)
\(=\dfrac{143}{84}\)
\(g,\dfrac{4}{9}:\dfrac{3}{5}\times\dfrac{2}{11}\)
\(=\dfrac{4}{9}\times\dfrac{5}{3}\times\dfrac{2}{11}\)
\(=\dfrac{20}{27}\times\dfrac{2}{11}\)
\(=\dfrac{40}{297}\)
\(h,\dfrac{7}{2}-\dfrac{3}{10}:\dfrac{2}{5}\)
\(=\left(\dfrac{7}{2}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\left(\dfrac{35}{10}-\dfrac{3}{10}\right):\dfrac{2}{5}\)
\(=\dfrac{32}{10}:\dfrac{2}{5}\)
\(=\dfrac{16}{5}\times\dfrac{5}{2}\)
\(=\dfrac{80}{10}=8\)
6 tháng 12 2021
Sửa đề: \(x+\dfrac{1}{x}=a\)
\(A=x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=a^3-3a\\ B=x^6+\dfrac{1}{x^6}=\left(x^3+\dfrac{1}{x^3}\right)^2-2=\left(a^3-3a\right)^2-2=a^6-6a^4+9a^2-2\\ C=x^7+\dfrac{1}{x^7}=\left(x^3+\dfrac{1}{x^3}\right)\left(x^4+\dfrac{1}{x^4}\right)-\left(x+\dfrac{1}{x}\right)\)
Mà \(x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=\left[\left(x+\dfrac{1}{x}\right)^2-2\right]^2-2=\left(a^2-2\right)^2-2=a^4-4a^2+2\)
\(\Leftrightarrow C=\left(a^3-3a\right)\left(a^4-4a^2+2\right)-a=...\)
\(\dfrac{5a-b}{3a+7}\)-\(\dfrac{3b-2a}{2b-7}\)
=\(\dfrac{5a-b}{3a+2a-b}\)-\(\dfrac{3b-2a}{2b-\left(2a-b\right)}\)
=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{2b-2a+b}\) (vì 2a-b=7)
=\(\dfrac{5a-b}{5a-b}\)-\(\dfrac{3b-2a}{3b-2a}\)
=1-1
=0