1. Tìm x, biết :
a, | x | = 1/5
b, | x | = 0,37
c, | x | = 0
d, | x | = 1 2/3
Câu d là câu hỗn số ạ
Mấy ac giúp e zới
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)
a, \(\left|2x+1\right|=3\)
=> 2x + 1 = 3 hoặc 2x + 1 = -3
=> 2x = 3 - 1 hoặc 2x = -3 - 1
=> 2x = 2 hoặc 2x = -4
=> x = 1 hoặc x = -2
b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)
=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)
=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)
=> \(\frac{5}{2}x=\frac{16-9}{6}\)
=> \(\frac{5}{2}x=\frac{7}{6}\)
=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)
c, \(3\cdot5^{x-3}+1=16\)
=> 3 . 5x-3 = 16 - 1
=> 3 . 5x-3 = 15
=> 5x-3 = 15 : 3
=> 5x-3 = 5
=> x - 3 = 5 : 5
=> x - 3 = 1
=> x = 1 + 3 = 4
d, \((x-1)^2=25\)
=> \((x-1)^2=5^2\)
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)
=> 4 + |3x + 1| = -196
=> |3x + 1| = -196 - 4 = -200
=> |3x + 1| = -200
Không thỏa mãn điều kiện
Bài 1 :
Khẳng định b) sai vì |-2,5| > 0
Các khẳng định a) và c) đúng
Bài 2 :
Nếu |x| = y thì x = y hoặc x = -y. theo công thức trên thì :
a) x = \(\frac{1}{5}\) hoặc x = \(-\frac{1}{5}\)
b) x = 0,37 hoặc x = -0,37
c) x = 0
d) x = \(1\frac{2}{3}\) hoặc x = \(-1\frac{2}{3}\)
a)\(|a|=\frac{1}{5}\)
\(\Rightarrow\hept{\begin{cases}a=\frac{1}{5}\\a=-\frac{1}{5}\end{cases}}\)
Vậy \(a\in\left\{\frac{1}{5};-\frac{1}{5}\right\}\)
b) \(|x|=0,37\)
\(\Rightarrow\hept{\begin{cases}x=0,37\\x=-0,37\end{cases}}\)
Vậy \(x\in\left\{0,37;-0,37\right\}\)