(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
Những câu hỏi liên quan
(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
ính
(6^9x2^10x2^10): (2^19x27^3+15x4^9x9^4)
Help ikkk, tui like cho
(69.210.210) : (219.273+15.49.94)
=((2.3)9.(2.2)10): (219.(33)3+15.(22)9.(32)4)
= (29.39.(22)10) : (219.39+15.218.38)
= (29.39.220) : (218.38.(2.3+15))
= (29.39.22.218) :(218.38.21)
=(211.39.218) : (218.39.7)
=211:7
=\(\frac{2048}{7}\)
Đúng 0
Bình luận (0)
Bài 1:
a) A=2^19x27^3+15x4^9x9/6^9x2^10+1^210
Các bạn giải giúp mik nhé, nếu sai đề thì bảo mik, mik cảm ơn^^
Tính :
\(\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
^Mk cần câu TL rõ ràng nhé
=\(\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
=\(\frac{2^{19}.3^9+5.2^{18}.3.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+2^{18}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)
=\(\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)
Đúng 0
Bình luận (0)
Tính giá trị của các biểu thức sau:
a) \(A=\frac{5x2^{13}x4^{11}-16^9}{\left(3x2^{17}\right)^2}\)
b) \(B=\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
a)\(A=\frac{5.2^{13}.2^{22}-2^{36}}{\left(3.2^{17}\right)^2}\)
\(A=\frac{5.2^{35}-2^{36}}{3^2.2^{34}}\)
\(A=\frac{2^{35}\left(5-2\right)}{3^2.2^{34}}\)
\(A=\frac{2.3}{3^2}=\frac{2}{3}\)
Đúng 0
Bình luận (0)
b) \(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(2+3\right)}\)
\(B=\frac{7}{2.5}=\frac{7}{10}\)
Đúng 0
Bình luận (0)
a) (\(\frac{-1}{2}\))3- (\(\frac{3}{4}\))3x(-2)2
b) 2x(-1)5+(\(\frac{3}{4}\))2-\(\frac{3}{8}\)
c) \(\frac{2^{19}x27^3+15x4^9x9^4}{6^9x2^{10}+12^{10}}\)
(69x210+1210):(219x273+15x49x94) = ?
Giúp me zới!!!
Bài 1: Tìm giá trị nhỏ nhất:
a)A=x2-2xy+5y2+4y+51
b)B=121/-4xy2-12x+2
c)C=9/-2x2+4x-7
d)10x2+4y2-4xy+8x-4y+20
e)E=9x2+2y2+6xy-6x-8y+10
a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)
\(=x^2-2xy+y^2+4y^2+4y+1+50\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)
Đúng 1
Bình luận (0)
a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)
\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)
c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)
\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)
d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)
\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)
\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)
Đúng 1
Bình luận (0)
4 tính gia tri bieu thuc
M=219x273+15x49x94
69x210+1210