Tìm x:a) |x+3|=x+3
b) |x-2|=2-x
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a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)
\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)
\(\Leftrightarrow x=15\)
b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)
\(\Leftrightarrow x^2-36=x^2\)(vô lý)
a. (x - 5)(x + 3) = x(x - 3)
<=> x2 + 3x - 5x - 15 = x2 - 3x
<=> x2 - x2 + 3x - 5x + 3x - 15 = 0
<=> x = 15
b. (x + 2)2 = (x - 1)(x + 2)
<=> x2 + 4x + 4 = x2 + 2x - x - 2
<=> x2 - x2 + 4x - 2x + x = -2 - 4
<=> 3x = -5
<=> \(x=\dfrac{-5}{3}\)
c. (x - 6)(x + 6) = x2
<=> x2 - 36 - x2 = 0
<=> x2 - x2 = 36
<=> 0 = 36 (vô lí)
Vậy nghiệm của PT là \(S=\varnothing\)
d. (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 - 4x2 + 8 = 0
<=> 4x2 - 4x2 - 12x = -8 - 9
<=> -12x = -17
<=> \(x=\dfrac{17}{12}\)
\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)
\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)
a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)
\(\Leftrightarrow2\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
a) \(\dfrac{x.2}{-15}=\dfrac{-5}{3}\)
\(\dfrac{x.2}{-15}=\dfrac{25}{-15}\)
x.2=25
x=12,5
b) \(\dfrac{x-1}{-12}=\dfrac{-3}{x-1}\)
(x-1)2=-3.(-12)
(x-1)2=36
⇒(x-1)2\(\Rightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
a) \(\dfrac{3,5}{15}=\dfrac{-2}{x}\)
\(\Rightarrow x=\dfrac{15.-2}{3,5}\)
\(\Rightarrow x=-8,57\)
b) \(2\left(3x-2\right)-3\left(x-2\right)-=-1\)
\(\Rightarrow6x-4-3x+6=-1\)
\(\Rightarrow6x-3x=-1+4-6\)
\(\Rightarrow3x=-3\)
\(\Rightarrow x=-\dfrac{3}{3}=-1\)
\(a,x-\dfrac{5}{7}=\dfrac{3}{5}\times\dfrac{1}{2}\)
\(x-\dfrac{5}{7}=\dfrac{3}{10}\)
\(x=\dfrac{3}{10}+\dfrac{5}{7}\)
\(x=\dfrac{21}{70}+\dfrac{50}{70}\)
\(x=\dfrac{71}{70}\)
\(Vayx=\dfrac{71}{70}\)
\(b,x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{3}{4}\)
\(x\times\dfrac{2}{3}=\dfrac{5}{8}+\dfrac{6}{8}\)
\(x\times\dfrac{2}{3}=\dfrac{11}{8}\)
\(x=\dfrac{11}{8}:\dfrac{2}{3}\)
\(x=\dfrac{11}{8}\times\dfrac{3}{2}\)
\(x=\dfrac{33}{16}\)
\(Vayx=\dfrac{33}{16}\)
a: \(\Leftrightarrow8x\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{0;3;-3\right\}\)
b: \(\Leftrightarrow x^2-4x+4-x^2-2x+3=12\)
=>-6x=5
hay x=-5/6
a. 25 + 3(x-8) = 106
3(x-8) = 106 - 25 = 81
x -8 = 81 : 3 = 27
x = 27 +8 = 35
b. 32(x +4) - 52 = 5.52
9(x+4) - 25 = 5.25 = 125
9(x+4) = 125 + 25 = 150(Đề sai)
\(a,25+3\left(x-8\right)=106\\ 3\left(x-8\right)=81\\ x-8=27\\ x=35\)
a: x=2/3-4/5=10/15-12/15=-2/15
b: 1/2-x=7/12
=>x=1/2-7/12=-1/12
c: =>7/2:x=-7/2
=>x=-1
d: =>1/6x=3/8-5/2=3/8-20/8=-17/8
=>x=-17/8*6=-102/8=-51/4
e: =>1,5x=-1,5
=>x=-1