Tìm x biết :\(\frac{32}{2^x}=2\)
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Ta có:
\(\frac{2}{x+32}=\frac{-1}{3x+5}\)
\(\Leftrightarrow2\left(3x+5\right)=-1\left(x+32\right)\)
\(\Leftrightarrow6x+10=-x-32\)
\(\Leftrightarrow7x=-42\)
\(\Rightarrow x=-6\)
Vậy................
hok tốt
\(\frac{2}{x+32}=\frac{-1}{3x+5}\)
\(\Rightarrow\frac{2}{x+32}=\frac{2}{-6x-10}\)
\(\Rightarrow x+32=-6x-10\)
\(x+6x=-10+32\)
\(\Rightarrow7x=22\Rightarrow x=\frac{22}{7}\)
\(\frac{x}{2}=\frac{y}{5}=\frac{2}{3}\Rightarrow\frac{2x}{4}=\frac{3y}{15}=\frac{2}{3}\)
Áp dụng dãy tỉ số = nhau ta có:
\(\frac{2x}{4}=\frac{3y}{15}=\frac{2}{3}=\frac{2x+3y-2}{4+15-3}=\frac{32}{16}=2\)
Suy ra: \(\frac{x}{2}=2\Rightarrow x=4\)
\(\frac{y}{5}=2\Rightarrow y=10\)
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5
\(32\left(\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+...+\frac{1}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+....+\frac{3}{197.200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{197}-\frac{1}{200}\right)-x=\frac{1}{2}\)
\(\frac{32}{3}\left(\frac{1}{8}-\frac{1}{200}\right)-x=\frac{1}{2}\)
x=0.78
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}=\frac{27}{9}=3\)
\(\Rightarrow x=\frac{3\cdot4+2}{2}=7\)
\(\Rightarrow y=\frac{3\cdot9+6}{3}=11\)
\(\Rightarrow z=3\cdot4+3=15\)
Suy ra \(\frac{x}{3}\)=\(\frac{y}{2}\) suy ra \(\frac{x}{3}\)=\(\frac{2y}{4}\) suy ra \(\frac{x}{21}\)=\(\frac{2y}{28}\)
Từ 5x=7z suy ra \(\frac{x}{7}\)=\(\frac{z}{5}\)suy ra \(\frac{x}{21}\)= \(\frac{z}{15}\)
Suy ra \(\frac{x}{21}\)=\(\frac{2y}{28}\)=\(\frac{z}{15}\)
Áp dụng t/c dãy tỉ số = nhau ta có
\(\frac{x}{21}\)=\(\frac{2y}{28}\)=\(\frac{z}{15}\)=\(\frac{x-2y+z}{21-28+15}\)=\(\frac{32}{8}\)=4
Suy ra \(\frac{x}{21}\)=4 suy ra x=84
Suy ra \(\frac{2y}{28}\)=4 suy ra y=56
Suy ra \(\frac{z}{15}\)=4 suy ra z=60
Hok tốt nhớ t i ck nhé
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97
Ta có 3A= \(^{3^2+3^3+3^4+...+3^{100}}\)
3A-A=2A= (\(3^2+3^3+3^4+...+3^{100}\))-(\(3+3^2+3^3+...+3^{99}\))
2A= \(3^{100}-3\)
theo bài ra ta có
2A+3=\(3^n\)= \(3^{100}-3+3=3^n\)=\(^{3^{100}}\)\(\Rightarrow\)n=100
\(\frac{32}{2^x}=2\)
\(\Rightarrow\frac{2^5}{2^x}=2^1\)
\(\Rightarrow5-x=1\)
\(\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
\(\frac{32}{2^x}=2\)
=> 2x = 32 : 2
=> 2x = 16
=> 2x = 24
=> x = 4
Năm mới vui vẻ nha!
Hk tốt