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A, \(4^{2x-6}=1\)
\(\Rightarrow4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
B, \(5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}=5^2.3+2.5^2\)
\(\Rightarrow5^{2x-3}=5^2.\left(3+2\right)\)
\(\Rightarrow5^{2x-3}=5^2.5\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
a) \(5^4:\left(2x-1\right)^2=25\\ \Leftrightarrow5^4:\left(2x-1\right)^2=5^2\\ \Leftrightarrow5^4:5^2=\left(2x-1\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^2=5^2\\\left(2x-1\right)^2=\left(-5\right)^2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b) \(17^3.x=17^4\\ \Rightarrow x=17^4:17^3=17\)
c) \(\left(2x+1\right)^3:49=7\\ \Leftrightarrow\left(2x-1\right)^3:7^2=7\\ \Leftrightarrow\left(2x-1\right)^3=7.7^2=7^3\\ \Leftrightarrow2x-1=7\\ \Leftrightarrow x=4\)
a)54 : (2x - 1)2 =25
54 : (2x - 1)2 =52
(2x - 1)2 =54 : 52
(2x - 1)2 = 52
2x - 1 = 5
2x = 5+1
2x = 6
x = 6: 2
x = 3
b)173.x = 174
x = 174 : 173
x = 17
c)(2x + 1)3 : 49 =7
(2x + 1)3 = 7 . 49
(2x + 1)3 = 343
(2x + 1)3 = 73
2x +1 = 7
2x = 7- 1
2x = 6
x = 6:2
x = 3
a: =2/5-3/5+3/7=3/7-1/5
=15/35-7/35
=8/35
b: =>5/7:x=4/3
=>x=5/7:4/3=5/7*3/4=15/28
c: =>x-1/3=15/8:4/5=15/8*5/4=75/32
=>x=75/32+1/3=257/96
d: =>2x+1/8=2/7
=>2x=9/56
=>x=9/112
e: =>2x=10/3-5/4-3/4=10/3-2=4/3
=>x=2/3
\(a,\dfrac{2}{5}+\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{3}{5}\\=\left(\dfrac{2}{5}-\dfrac{3}{5}\right)+\dfrac{3}{7}\\ =-\dfrac{1}{5}+\dfrac{3}{7}\\ =-\dfrac{7}{35}+\dfrac{15}{35}\\ =\dfrac{8}{35}\\ b,1-\dfrac{5}{7}:x=-\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=1-\left(-\dfrac{1}{3}\right)\\ =>\dfrac{5}{7}:x=1+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{3}{3}+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{4}{3}\\ =>x=\dfrac{5}{7}:\dfrac{4}{3}\\ =>x=\dfrac{5}{7}.\dfrac{3}{4}\\ =>x=\dfrac{15}{28}\\ c,\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)=\dfrac{15}{8}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}:\dfrac{4}{5}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}.\dfrac{5}{4}\\ =>x-\dfrac{1}{3}=\dfrac{75}{32}\\ =>x=\dfrac{75}{32}+\dfrac{1}{3}\\ =>x=\dfrac{257}{96}\)
\(d,\dfrac{2}{3}:\left(2x+\dfrac{1}{8}\right)=\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}:\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}.\dfrac{3}{7}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{7}\\ =>2x=\dfrac{2}{7}-\dfrac{1}{8}\\ =>2x=\dfrac{16}{56}-\dfrac{7}{56}\\ =>2x=\dfrac{9}{56}\\ =>x=\dfrac{9}{56}:2\\ =>x=\dfrac{9}{112}\\ e,2x+\dfrac{3}{4}=\dfrac{10}{3}-\dfrac{5}{4}\\ =>e,2x+\dfrac{3}{4}=\dfrac{40}{12}-\dfrac{15}{12}\\ =>2x+\dfrac{3}{4}=\dfrac{25}{12}\\ =>2x=\dfrac{25}{12}-\dfrac{3}{4}\\ =>2x=\dfrac{25}{12}-\dfrac{9}{12}\\ =>2x=\dfrac{16}{12}\\ =>2x=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}:2\\ =>x=\dfrac{4}{6}\\ =>x=\dfrac{2}{3}\)
a: (x-2)(y-3)=5
=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)
=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)
b: (2x-1)*(y-4)=-11
=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)
=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)
c: xy-2x+y=3
=>\(x\left(y-2\right)+y-2=1\)
=>\(\left(x+1\right)\left(y-2\right)=1\)
=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)
=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)
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 !
a) \(\left(2x+3\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(2x+3\right)^2=\left(\frac{3}{5}\right)^2\) hoặc \(\left(2x+3\right)^2=\left(-\frac{3}{5}\right)^2\)
\(\Rightarrow2x+3=\frac{3}{5}\) hoặc \(2x+3=-\frac{3}{5}\)
\(\Rightarrow2x=-\frac{12}{5}\) hoặc \(2x=-\frac{18}{5}\)
\(\Rightarrow x=-\frac{6}{5}\) hoặc \(x=-\frac{9}{5}\)
Vậy.......
b) \(\left(3x-1\right)^3=\frac{1}{27}\)
\(\left(3x-1\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow3x-1=\frac{1}{3}\)
\(\Rightarrow3x=\frac{4}{3}\)
\(\Rightarrow x=\frac{4}{9}\)
Vậy.......
a) (2x+3)2=\(\frac{9}{25}\)
2x+3=\(\sqrt{\frac{9}{25}}\)=\(\frac{3}{5}\)
2x=\(\frac{-12}{5}\)
x=\(\frac{-6}{5}\)
b) (3x-1)3=\(\frac{-1}{27}\)
3x-1=\(\frac{-1}{3}\)
3x=\(\frac{2}{3}\)
x=\(\frac{2}{9}\)
g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)
\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)
\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)
\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
a) 2/3 + 1/3(2x-1) =-5
1/3(2x-1) = -5 - 2/3
1/3(2x-1) = -17/3
2x-1 = -17/3 : 1/3
2x-1 = -17
2x = -17 + 1
2x = -16
x = -16 : 2
x = -8
b) 25% .x + 2x = 4
x .( 25%+2 ) = 4
x . 9/4 = 4
x = 4 : 9/4
x = 16/9