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\(2^x:4=32\\ \Rightarrow2^x=128\\ \Rightarrow2^x=2^7\\ \Rightarrow x=7\)
\(3^{x-2}:3=243\\ \Rightarrow3^{x-2}=729\\ \Rightarrow3^{x-2}=3^6\\ \Rightarrow x-2=6\\ \Rightarrow x=8\)
\(256:4^{x+1}=4^2\\ \Rightarrow4^{x+1}=4^2\\ \Rightarrow x+1=2\\ \Rightarrow x=1\)
\(4^{2x-1}:4=4^4\\ \Rightarrow4^{2x-1}=4^5\\ \Rightarrow2x-1=5\\ \Rightarrow x=3\)
\(5^{x-1}:5=5^3\\ \Rightarrow5^{x-1}=5^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
\(3^{2x+1}:3=3^4\\ \Rightarrow3^{2x+1}=3^5\\ \Rightarrow2x+1=5\\ \Rightarrow x=3\)
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
a) \(\frac{-2}{3}x+\frac{1}{5}=\frac{1}{10}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{1}{10}-\frac{1}{5}\)
\(\Leftrightarrow\frac{-2}{3}x=\frac{-1}{10}\)
\(\Leftrightarrow x=\frac{-1}{10}\div\frac{-2}{3}\)
\(\Leftrightarrow x=\frac{3}{20}\)
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
a)\(\frac{x+1}{3}=\frac{2}{6}\)
=> \(\frac{x+1}{3}=\frac{1}{3}\)
=> x + 1 = 1 => x = 0
b) \(\frac{4-x}{-5}=\frac{-5}{4-x}\)
=> (4 - x)2 = 25
=> (4 - x)2 = 52
=> 4 - x = 5 hoặc 4 - x = -5
=> x = -1 hoặc x = 9
c) \(\frac{3}{x+2}=\frac{5}{2x+1}\)
=> 3(2x + 1) = 5(x + 2)
=> 6x + 3 = 5x + 10
=> 6x + 3 - 5x - 10 = 0
=> 6x - 5x + 3 - 10 = 0
=> 6x - 5x = 7
=> x = 7
d) \(\frac{1}{2}=\frac{x+1}{3x}\)
=> 3x = 2(x + 1)
=> 3x = 2x + 2
=> 3x - 2x - 2 = 0
=> x - 2 = 0
=> x = 2
e) \(\frac{-3}{x+1}=\frac{4}{\left(2-2x\right)}\)
=> -3(2 - 2x) = 4(x + 1)
=> -6 + 6x = 4x + 4
=> -6 + 6x - 4x - 4 = 0
=> -10 + 2x = 0
=> 2x = 10
=> x = 5
a) \(2^{2x}.2^4=1024\)
\(2^{2x}=1024:2^4\)
\(2^{2x}=1024:16\)
\(2^{2x}=64\)
\(2^{2x}=2^6\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
vay \(x=3\)
b) \(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=2.5.3^{12}+2^3.\left(3^3\right)^4\)
\(2.3^x=2.5.3^{12}+2^3.3^{12}\)
\(2.3^x=2.3^{12}.\left(5+2^2\right)\)
\(2.3^x=2.3^{12}.9\)
\(2.3^x=2.3^{12}.3^2\)
\(2.3^x=2.3^{14}\)
\(\Rightarrow x=14\)
vay \(x=14\)
c) \(5^8.25^x+1=5^{17}\)
\(5^8.\left(5^2\right)^x+1=5^{17}\)
\(5^8.5^{2x}+1=5^{17}\)
\(5^{8+2x}=5^{17}-1\)
e) \(\left(2x-4\right)^5=\left(2x-4\right)^3\)
\(\left(2x-4\right)^5-\left(2x-4\right)^3=0\)
\(\left(2x-4\right)\left[\left(2x-4\right)^2-1\right]=0\)
\(\left(2x-4\right)\left(2x-4-1\right)\left(2x-4+1\right)=0\)
\(\left(2x-4\right)\left(2x-5\right)\left(2x-3\right)=0\)
\(\Rightarrow2x-4=0\)hoac \(\orbr{\begin{cases}2x-5=0\\2x-3=0\end{cases}}\)
\(\Rightarrow2x=4\)hoac \(\orbr{\begin{cases}2x=5\\2x=3\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
vay \(x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)