42x-1= 34.42-272
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9 ngày gấp 3 ngày số lần là
9 : 3 = 3 (lần)
Nếu vẫn số người như vậy ở bếp cần số kg gạo là
22 x 3 = 66 (kg)
đáp số 66 kg
![](https://rs.olm.vn/images/avt/0.png?1311)
`#040911`
\(2^{10}\div8^5\\ =2^{10} \div\left(2^3\right)^5\\=2^{10}\div2^{15}=2^{-5} \)
Đề có nhầm gì k bạn?
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta dễ chứng minh \(\widehat{BIC}=90^o+\dfrac{\widehat{A}}{2}\).
Ta thấy \(\widehat{BFK}=\widehat{A}+\widehat{AEF}=\dfrac{\widehat{A}}{2}+\widehat{IAE}+\widehat{AEF}\) \(=90^o+\dfrac{\widehat{A}}{2}\)
Nên \(\widehat{BIC}=\widehat{BFK}\)
Xét 2 tam giác BIC và BFK, ta có:
\(\widehat{FBK}=\widehat{IBC}\) (do BI là tia phân giác của \(\widehat{FBC}\)) và \(\widehat{BIC}=\widehat{BFK}\left(cmt\right)\)
\(\Rightarrow\Delta BIC~\Delta BFK\left(g.g\right)\) (đpcm)
b) Từ \(\Delta BIC~\Delta BFK\Rightarrow\dfrac{BI}{BF}=\dfrac{BC}{BK}\) \(\Rightarrow\dfrac{BI}{BC}=\dfrac{BF}{BK}\)
Xét 2 tam giác BIF và BCK, ta có
\(\dfrac{BI}{BC}=\dfrac{BF}{BK}\) và \(\widehat{IBF}=\widehat{CBK}\)
\(\Rightarrow\Delta BIF~\Delta BCK\left(c.g.c\right)\)
\(\Rightarrow\widehat{BKC}=\widehat{BFI}\)
Mà \(\widehat{BFI}=90^o\) nên \(\widehat{BKC}=90^o\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8x^3+12x^2y+6xy^2+y^3-z^3\)
\(=\left(2x+y\right)^3-z^3\)
\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)
a, 8a3 - 36a2 +54ab2 - 27b3
=(8a3-36a2b +54ab2 - 27b3)
=(2a-3b)2
=(2a-3b)(2a-3b)(2a-3b)
b, 8x3 + 12x2y + 6xy2 + y3 - z 3
=(8x3 + 12x2y + 6xy2 + y3) - z3
=(2x + y)3 - y3
=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2
= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz
=y.(4\(x^3\) + \(\dfrac{1}{2}\)z)
b, (a2 + b2 - 5)2 - 2.(ab + 2)2
= [a2 + b2 - 5 - \(\sqrt{2}\)(ab + 2) ].[ a2 + b2 - 5 + \(\sqrt{2}\)(ab +2)]
a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)
b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)
\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)
\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)
\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)
\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{1}{4}-1\right).\left(\dfrac{1}{9}-1\right)....\left(\dfrac{1}{100}-1\right).\)
\(\Rightarrow A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)\)
mà A có 9 dấu - \(\left(4;9;16;25;36;49;64;81;100\right)\)
\(\Rightarrow0>A=\left(-\dfrac{3}{4}\right).\left(-\dfrac{8}{9}\right)....\left(-\dfrac{99}{100}\right)=-\dfrac{1}{2}\)
Ta lại có \(\left\{{}\begin{matrix}\dfrac{1}{2}=\dfrac{21}{42}\\\dfrac{11}{21}=\dfrac{22}{42}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}< \dfrac{11}{21}\Rightarrow-\dfrac{1}{2}>-\dfrac{11}{21}\)
\(\Rightarrow A>-\dfrac{11}{21}\)
\(A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)...\left(\dfrac{1}{100}-1\right)\)
\(A=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)...\left(-\dfrac{10^2-1}{10^2}\right)\)
\(A=\left[-\dfrac{1\cdot3}{2\cdot2}\right]\left[-\dfrac{2\cdot4}{3\cdot3}\right]...\left[-\dfrac{9\cdot11}{10\cdot10}\right]\)
Dễ thấy A có 9 thừa số, suy ra
\(A=-\dfrac{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}{2\cdot2\cdot3\cdot3\cdot...\cdot10.10}=-\dfrac{1\cdot11}{2\cdot10}=\dfrac{-11}{20}\)
Vì 20 < 21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\), suy ra \(\dfrac{-11}{20}< \dfrac{-11}{21}\)
Vậy \(A< \dfrac{-11}{21}\)
42x-1 = 81. 16 - 272
42x-1 = 1296 - 272
42x-1 = 1024
42x : 4 = 1024
42x = 1024 x 4
42x = 4096
42x = 42.3
=> x = 3