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\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{2^3.3^3+2.2^2.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\=\dfrac{2^3.3^3+2^3.3^2+2^3}{37}\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2^3.\left(3^3+3^2+1\right)}{37}=\dfrac{2^3.37}{37}=2^3=8\)
\(\dfrac{6^3+2.6^2+2^3}{37}=\dfrac{216+72+8}{37}=\dfrac{296}{37}=8\)
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gọi số bi của 3 bạn Tâm, Bình , An lần lượt là : x, y, z\
Ta có :\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)= \(\dfrac{z}{7}\) và x + y + z = 45
áp dụng tính chất dãy tỉ số bằng nhau:
\(\Rightarrow\)\(\dfrac{x}{3}\)=\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)= \(\dfrac{x+y+z}{3+5+7}\)= 3
\(\Rightarrow\)x = 3.3 =9
y = 3.5 = 15
z = 3.7 = 21
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27/82 và 26/75
Ta có:
27/82 = 2025/6150
26/75 = 2132/6150
Vì 2025/6150<2132/6150 nên 27/82<26/75.
Vậy: 27/82<26/75.
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\(\left(\dfrac{-1}{3}\right)^{-1}=-3\)
Vì \(\left(\dfrac{-1}{3}\right)^{-1}=\left(\dfrac{1}{-3}\right)^{-1}=\left(-3^{-1}\right)^{-1}=-3^{-1\times\left(-1\right)}=-3^1=-3\)
=> \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)
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\(\dfrac{12^3.18^2}{24^2}=\dfrac{12^3.6^2.3^2}{6^2.4^2}=\dfrac{4^3.3^3.3^2}{4^2}=4.3^3.3^2=4.3^5=972\)
\(\dfrac{12^3.18^2}{24^2}=\dfrac{1728.324}{576}=\dfrac{559872}{576}=972\)
Vậy giá trị cần tìm là 972
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Cộng trừ hai số hữu tỉ
B6:
a, \(\frac{-1}{21}+\frac{-1}{28}\) = \(\frac{-1}{12}\)
\(b,\frac{-8}{18}-\frac{15}{27}\) = -1
\(c,\frac{-5}{12}+0,75=\frac{1}{3}\)
\(d,3,5-\left(\frac{-2}{7}\right)=\frac{53}{14}\)
\(B8:\)
\(a,\frac{3}{7}+\left(\frac{-5}{2}\right)+\left(\frac{-3}{5}\right)\)
= \(\frac{3}{7}-\frac{5}{2}-\frac{3}{5}\)
\(=\frac{-187}{70}\)
\(b,\left(\frac{-4}{3}\right)+\left(\frac{-2}{5}\right)+\left(\frac{-3}{2}\right)\)
\(=\frac{-4}{3}-\frac{2}{5}-\frac{3}{2}\)
\(=\frac{-97}{30}\)
\(c,\frac{4}{5}-\left(\frac{2}{7}\right)-\frac{7}{10}\)
\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)
\(=\left(\frac{4}{5}-\frac{7}{1}\right)+\frac{2}{7}\)
\(=\frac{1}{10}+\frac{2}{7}\)
\(=\frac{27}{70}\)
\(d,\frac{2}{3}-\left[\left(\frac{-7}{4}\right)-\left(\frac{1}{2}+\frac{3}{8}\right)\right]\)
\(=\frac{2}{3}-\left(\frac{-7}{4}-\frac{7}{8}\right)\)
\(=\frac{2}{3}-\left(\frac{-21}{8}\right)\)
\(=\frac{79}{24}\)
Làm từng bài một nha
Chẳng biết đúng hay sai nữa
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\(5,\\ 5\cdot625=25\cdot125\\ \Leftrightarrow\dfrac{5}{25}=\dfrac{125}{625};\dfrac{5}{125}=\dfrac{25}{625};\dfrac{625}{25}=\dfrac{125}{5};\dfrac{625}{125}=\dfrac{25}{5}\)
\(6,\)
\(a,\) Áp dụng t/c dtsbn: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-21}{7}=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\cdot2=-6\\y=-3\cdot5=-15\end{matrix}\right.\)
\(b,7x=3y\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{7}\)
Áp dụng t/c dtsbn: \(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-4\cdot3=-12\\y=-4\cdot7=-28\end{matrix}\right.\)
\(7,\\ a:b:c:d=2:3:4:5\Leftrightarrow\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{d}{5}\)
Áp dụng t/c dtsbn: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{d}{5}=\dfrac{a+b+c+d}{2+3+4+5}=\dfrac{-42}{14}=-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-3\cdot2=-6\\b=-3\cdot3=-9\\c=-3\cdot4=-12\\d=-3\cdot5=-15\end{matrix}\right.\)
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\(5,\\ \left\{{}\begin{matrix}AB\perp AD\\CD\perp AD\end{matrix}\right.\Rightarrow AB//CD\\ \Rightarrow\widehat{ABC}+\widehat{BCD}=180^0\left(2.góc.TCP\right)\\ \Rightarrow60^0+\widehat{BCD}=180^0\\ \Rightarrow\widehat{BCD}=120^0\)
\(6,\\ a,\left\{{}\begin{matrix}a\perp AB\\a//b\end{matrix}\right.\Rightarrow b\perp AB\\ b,a//b\Rightarrow\widehat{C}+\widehat{D}=180^0\left(2.góc.TCP\right)\\ \Rightarrow\widehat{D}=180^0-120^0=60^0\\ c,\widehat{ICD}=\dfrac{1}{2}\widehat{ACD}=\dfrac{1}{2}\cdot120^0=60^0\left(t/c.phân.giác\right)\\ \Delta CID.có:\widehat{CID}+\widehat{IDC}+\widehat{ICD}=180^0\\ \Rightarrow60^0+60^0+\widehat{CID}=180^0\\ \Rightarrow\widehat{CID}=60^0\)
a) Xét 2 tam giác AHB và AHC có:
AB = AC(GT); góc AHB = AHC(Do AH vuông góc BC); AH cạnh chung
=) 2 tam giác AHB và AHC bằng nhau
Phần b và c tối mik làm giúp cho còn bây h mik hok trực tuyến