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17 tháng 7 2019
a) \(\frac{4}{x+5}=\frac{3}{2x-1}\)
=> 4(2x - 1) = 3(x + 5)
=> 8x - 4 = 3x + 15
=> 8x - 3x = 15 + 4
=> 5x = 19
=> x = 19/5
b) \(\frac{x+11}{19}+\frac{x+12}{20}+\frac{x+13}{21}=3\)
=> \(\left(\frac{x+11}{19}-1\right)+\left(\frac{x+12}{20}-1\right)+\left(\frac{x+13}{21}-1\right)=0\)
=> \(\frac{x-8}{19}+\frac{x-8}{20}+\frac{x-8}{21}=0\)
=> \(\left(x-8\right)\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}\right)=0\)
=> x - 8 = 0
=> x = 8
c) \(\left(2x-1\right)^2=\left(2x-1\right)^3\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^3=0\)
=> \(\left(2x-1\right)^2.\left[1-\left(2x-1\right)\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\1-2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2-2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
17 tháng 7 2019
a) 4/x + 3 = 3/2x - 1
<=> 4.(2x - 1) = (x + 3).3
<=> 8x - 4 = 3x + 9
<=> 8x = 3x + 9 + 4
<=> 8x = 3x + 13
<=> 8x - 3x = 13
<=> 5x = 13
<=> x = 13/5
=> x = 13/5
c) (2x - 1)2 = (2x - 1)3
<=> 4x2 - 4x + 1 = 8x3 - 12x2 + 6x - 1
<=> 8x3 - 12x2 + 6x - 1 = 4x2 - 4x + 1
<=> 8x3 - 12x2 + 6x - 1 - 1 = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x = 4x2 - 4x
<=> 8x3 - 12x2 + 6x - 2x - 4x = 4x2
<=> 8x3 - 12x2 + 10x - 2 = 4x2
<=> 8x3 - 12x2 + 10x - 2 - 4x2 = 0
<=> 8x2 - 16x2 + 10x - 2 = 0
<=> 2(x - 1)(2x - 1)2 = 0
<=> x - 1 = 0 hoặc 2x - 1 = 0
x = 0 + 1 2x = 0 + 1
x = 1 2x = 1
x = 1/2
=> x = 1 hoặc x = 1/2
9 tháng 3 2016
\(\frac{\left(x+1\right)^2-\frac{x}{2}}{4}=\frac{\left(2x-3\right)^2}{3}-\frac{\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}}{4}\)
\(\Rightarrow3\left[\left(x+1\right)^2-\frac{x}{2}\right]=4\left(2x-3\right)^2-3\left[\frac{x+1}{4}-\frac{x\left(3-2x\right)}{3}\right]\)
\(\Rightarrow3\left(x+1\right)^2-\frac{3x}{2}=4\left(2x-3\right)^2-\frac{3\left(x+1\right)}{4}+\frac{3x\left(3-2x\right)}{3}\)
\(\Rightarrow36\left(x+1\right)^2-18x=48\left(2x-3\right)^2-9\left(x+1\right)+12x\left(3-2x\right)\)
=> 36.(x2 + 2x + 1) - 18x = 48.(4x2 - 12x + 9) - 9(x + 1) + 12x(3 - 2x)
=> 36x2 + 72x + 36 - 18x - 192x2 + 576x - 432 + 9x + 9 - 36x + 24x2 = 0
=> -132x2 + 603x - 387 = 0
Có: \(\Delta=603^2-4.\left(-387\right)\left(-132\right)=159273\Rightarrow\sqrt{\Delta}=\sqrt{159273}\)
\(\Rightarrow x=\frac{-603+\sqrt{159273}}{-264}\) hoặc \(x=\frac{-603-\sqrt{159273}}{-264}\)
Vậy phương trình có 2 nghiệm : x = \(\left\{\frac{-603+\sqrt{159273}}{-264};\frac{-603-\sqrt{159273}}{-264}\right\}\)
Câu này không có nghiệm nguyên nha bạn.
DH
8 tháng 10 2019
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
DH
8 tháng 10 2019
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
VM
8 tháng 10 2019
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
P1
6 tháng 8 2017
Câu b thôi các bạn nhé, câu a mình ko cần nx với cả mình ghi sai dữ liệu câu a r
DP
6 tháng 8 2017
a, \(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot...\cdot\frac{30}{62}\cdot\frac{31}{64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=2x\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{2\cdot2\cdot3\cdot2\cdot4\cdot2\cdot5\cdot2\cdot....\cdot31\cdot2\cdot32\cdot2}=2x\)
\(\Leftrightarrow\frac{1}{2\cdot2\cdot2\cdot2\cdot....\cdot2\cdot2\cdot32}=2x\)
Có : (31 - 1) : 1 + 1 = 31 (thừa số 2)
\(\Rightarrow\frac{1}{2^{31}.32}=2x\)
\(\Rightarrow x=\frac{1}{2^{31}.32}\div2\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+1=x+4\)
\(\Leftrightarrow0=3\text{ (vô lý) }\)
Ta có:
\(\frac{4^{x+2}+4^{x+1}+4^x}{21}=\frac{4^x\cdot\left(4^2+4+1\right)}{21}=\frac{4^x\cdot21}{21}=4^x\)
\(\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{31}=\frac{9^x\cdot\left(1+3+3^2\right)}{31}=\frac{9^x\cdot13}{31}\)
Xét \(4^x=\frac{9^x\cdot13}{31}\)
=> \(\frac{4^x}{9^x}=\frac{13}{31}\)
Vì \(\hept{\begin{cases}\left(4;9\right)=1\\13\notin B\left(4\right)\\31\notin B\left(9\right)\end{cases}\Rightarrow x\in\varnothing}\)
Vậy x không tồn tại