Tìm a sao cho 1/9<12/a<3/2
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\(\frac{8}{3}\cdot\frac{2}{5}\cdot\frac{3}{8}\cdot10\cdot\frac{19}{92}=\left(\frac{8}{3}\cdot\frac{3}{8}\right)\cdot\left(\frac{2}{5}\cdot10\right)\cdot\frac{19}{92}=1\cdot4\cdot\frac{19}{92}=4\cdot\frac{19}{92}=\frac{19}{23}\)
\(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}\)
\(=(\frac{8}{3}.\frac{3}{8}).(\frac{2}{5}.10).\frac{19}{92}\)
\(=4.\frac{19}{92}\)
\(=\frac{19}{23}\)
Hok tôt !!!!!!!!!!!!!!!
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Xét \(\Delta ABM\) và \(\Delta DCM\) có
AM = DM (gt)
\(\widehat{AMB}=\widehat{DMC}\) (đối đỉnh_
BM = CM (gt)
=> \(\Delta ABM\) = \(\Delta DCM\) (c.g.c)
=> AB = DC ( 2 cạnh t/ứ)
Xét \(\Delta ACM\) và \(\Delta DBM\) có
AM = DM (gt)
\(\widehat{AMC}=\widehat{BMD}\) (đối đỉnh)
CM = BM (gt)
=> \(\Delta ACM\) = \(\Delta DBM\) (c.g.c)
=> AC = DB ( 2 cạnh t/ứ)
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Trong tam giác ABC, có:
( Mình không viết đc dấu góc nên chỉ viết tên góc)
A + B + C = 180o ( Định lí)
⟹ A = 180o - B - C = 180o - 70o - 30o = 80o
⟹ A = ACD = 80o (so le trong)
⟹ AB//CD (đpcm)
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Bài 2 :
a, \(P\left(x\right)=2x^5+2-6x^2-3x^3+4x^2-2x+x^3+4x^5=6x^5-2x^3-2x^2+2\)
b, sắp xếp rồi, trên ý
c, Bậc : 5
Bài 3 : \(Q\left(x\right)=3x-5=0\Leftrightarrow x=\frac{5}{3}\)
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a)
\(x=\left(\frac{3}{7}\right)^7:\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
b)
\(-\frac{1}{27}\cdot x=\frac{1}{81}\)
\(x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(x=-\frac{1}{3}\)
c)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
d)
\(\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{3}\\x+\frac{1}{2}=\frac{-2}{3}\end{cases}}\)
\(\orbr{\begin{cases}x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\\x=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\end{cases}}\)
Sửa lại câu a : ( nhìn sai số )
\(\frac{3^5}{5^5}\cdot x=\frac{3^7}{7^7}\)
\(x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(x=\frac{3^7}{7^7}\cdot\frac{5^5}{3^5}\)
\(x=\frac{5^5\cdot3^2}{7^7}\)
\(x=\frac{28125}{823453}\)
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