Rút gọn các biểu thức sau :
A = 318 : 315
B= \(\dfrac{2^{11}.9^2}{3^5.2^8}\)
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\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)
C=3
D=9
F=1/20
G=5/2
Em ko giải chi tiết vì nó lâu
Mong thông cảm!
a.
=(5^7+5^9)(6^8+6^10)(16-4^2)
=(5^7+5^9)(6^8+6^10)(16-16)
=(5^7+5^9)(6^8+6^10).0
=0
Bài 1:
\(a,ĐK:2+8x\ge0\Leftrightarrow x\ge-\dfrac{1}{4}\\ b,ĐK:-\dfrac{1}{5}x+9\ge0\Leftrightarrow-\dfrac{1}{5}x\ge-9\Leftrightarrow x\le45\\ c,ĐK:11-7x\ge0\Leftrightarrow x\le\dfrac{11}{7}\)
Bài 2:
\(a,=\sqrt{144a^2}-2a=12\left|a\right|-2a=12a-2a=10\\ b,=\sqrt{6}-6\sqrt{6}-\sqrt{6}=-6\sqrt{6}\)
Bài 3:
\(a,\Leftrightarrow\left|2x+3\right|=3\Leftrightarrow\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=4\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
a: Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Để \(A\ge0\) thì \(\sqrt{x}-3>0\)
hay x>9
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
Ta có: M=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
a: \(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Khi x=11+6 căn 2 thì \(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}=\dfrac{9+3\sqrt{2}}{\sqrt{2}}=\dfrac{9\sqrt{2}+6}{2}\)
c: M<1
=>\(\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
=>căn x-3<0
=>0<x<9
`a,` \(M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\) \(\left(x\ne\pm3;x>0\right)\)
\(M=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{2x-6\sqrt{x}}{x-9}+\dfrac{x+3\sqrt{x}+\sqrt{x}+3}{x-9}-\dfrac{3+11\sqrt{x}}{x-9}\)
\(M=\dfrac{3x+9\sqrt{x}}{x-9}\)
\(M=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
`b,`Ta có :
\(M=\dfrac{3\sqrt{11+6\sqrt{2}}}{\sqrt{11+6\sqrt{2}}-3}\)
\(M=\dfrac{3\sqrt{\left(3+\sqrt{2}\right)^2}}{\sqrt{\left(3+\sqrt{2}\right)^2}-3}\)
\(M=\dfrac{3\left(3+\sqrt{2}\right)}{3+\sqrt{2}-3}\)
\(M=\dfrac{9+3\sqrt{2}}{\sqrt{2}}\)
\(M=\dfrac{6+9\sqrt{2}}{2}\)
`c,` Để `M<1` Ta có :
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)
\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\sqrt{x}-3< 0\) ( vì \(2\sqrt{x}+3>0\) )
\(\sqrt{x}< 3\)
\(x< 9\)
Đối chiếu ĐKXĐ ta có : `0<x<9`
A=33=27
\(B=\dfrac{2^{11}.9^2}{3^5.2^8}=\dfrac{2^{11}.3^4}{3^5.2^8}=\dfrac{2^3}{3}=\dfrac{8}{3}\)
A=3^3=9
B=2^3/3^5=8/243