X 0,5 X 25% X 1/ 8 X 2017= 2020
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\(\Rightarrow x:\left(0,2+0,5+0,25+0,125\right)=5,7\)
\(\Rightarrow x:1,075=5,7\)
\(\Rightarrow x=5,7.1,075=6,1275\)
x:(0,2+0,5) + (0,25+0,125)=5,7
x:0,7 + 0,375 = 5,7
x:1,075 = 5,7
x=5,7 X 1,075
x=6,1275
\(Q=\frac{2017}{x-8\sqrt{x}+2018}=\frac{2017}{\left(\sqrt{x}-4\right)^2+2002}\)
ta có \(\left(\sqrt{x}-4\right)^2\ge0\)
\(Q\le\frac{2017}{2002}\)
dấu "=" xảy ra khi \(x=16\)
\(MAX:Q=\frac{2017}{2002}\)
Bài 4 : Tính nhanh :
a, 15. 64 + 25. 100 + 36. 15 + 60. 100
= (15 . 64 + 36. 15) + (25. 100 + 60. 100)
= 15.(64 + 36) + 100.(25 + 60)
= 15. 100 + 100. 85
= 100.(15 + 85)
= 100. 100
= 10000
b, 472 + 482 - 25 + 94. 48
= 472 + 2.47. 48 + 482 - 25
= (47 + 48)2 - 52
= (47 + 48 - 5)(47 + 48 + 5)
= (48 + 22)(48 + 52)
= 90. 100
= 9000
c, 93 - 92. ( -1) - 9. 11 + ( -1). 11
= 93 + 92 + 11(- 9 - 1)
= 92.(9 + 1) + 11. (-10)
= 81. 10 - 110
= 810 - 110
= 700
d,2016. 2018 - 20172
= (2017 - 1)(2017 + 1) - 20172
= 20172 - 1 - 20172
= -1
#Học tốt!
\(\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(0,5-1\dfrac{3}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\left(\dfrac{1}{2}-\dfrac{8}{5}\right)\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{-3}{8}\times\dfrac{11}{10}\)
\(\Rightarrow\left(x:2,2\right)\times\dfrac{1}{6}=\dfrac{33}{80}\)
\(\Rightarrow x:2,2=\dfrac{33}{80}:\dfrac{1}{6}\)
\(\Rightarrow x:2,2=\dfrac{99}{40}\)
\(\Rightarrow x=\dfrac{99}{40}\times2,2\)
\(\Rightarrow x=\dfrac{1089}{200}\)
=>(x:2,2)*1/6=-3/8(1/2-8/5)=33/80
=>x:2,2=99/40
=>x=1089/200
Ta có: x + x : 0,5 + x : 25% + x : 1/8 = 3
\(\Rightarrow\)\(x+x\times2+x\times4+x\times8=3\)
\(\Rightarrow\)x \(\times\)< 1 + 2 + 4 + 8 > \(=\) 3
\(\Rightarrow x\times15=3\)
\(\Rightarrow x=3\div15\)
\(\Rightarrow x=\frac{3}{15}\)
\(\Rightarrow x=\frac{1}{5}\)
Nhớ k chị nhé.
b) \(x^2+6x+9=144\)
\(\Leftrightarrow\left(x+3\right)^2=12^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
b, Ta có : \(x^2+6x+9=144\)
=> \(\left(x+3\right)^2=12^2\)
=> \(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{9,-15\right\}\)
c, Ta có : \(\frac{2-x}{2016}-1=\frac{1-x}{2017}-\frac{x}{2018}\)
=> \(\frac{2-x}{2016}-1=\frac{1-x}{2017}+\frac{-x}{2018}\)
=> \(\frac{2-x}{2016}+1=\frac{1-x}{2017}+1+\frac{-x}{2018}+1\)
=> \(\frac{2-x}{2016}+\frac{2016}{2016}=\frac{1-x}{2017}+\frac{2017}{2017}+\frac{-x}{2018}+\frac{2018}{2018}\)
=> \(\frac{2018-x}{2016}=\frac{2018-x}{2017}+\frac{2018-x}{2018}\)
=> \(\frac{2018-x}{2016}-\frac{2018-x}{2017}-\frac{2018-x}{2018}=0\)
=> \(\left(2018-x\right)\left(\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
=> \(2018-x=0\)
=> \(x=2018\)
Vậy phương trình có tập nghiệm là \(S=\left\{2018\right\}\)