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a)
$2x+6=0$
$2x=-6$
$x=-3$
b) $4x+20=0$
$4x=-20$
$x=-5$
c)
$2(x-1)=5x-7$
$2x-2=5x-7$
$3x=5$
$x=\frac{5}{3}$
d) $2x-3=0$
$2x=3$
$x=\frac{3}{2}$
e)
$3x-1=x+3$
$2x=4$
$x=2$
f)
$15-7x=9-3x$
$6=4x$
$x=\frac{3}{2}$
g) $x-3=18$
$x=18+3=21$
h)
$2x+1=15-5x$
$7x=14$
$x=2$
a) \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow20-4x=0\)
\(\Leftrightarrow4x=20\)
\(\Leftrightarrow x=5\)
Vậy S = {5}
b) ĐKXĐ: \(x\ne0;x\ne-2\)
\(\dfrac{x+2}{x}=\dfrac{\left(x+1\right)\left(x+4\right)}{x^2+2x}+\dfrac{x}{x+2}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{x^2+4x+x+4+x^2}{x\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{x+2}{x}=\dfrac{2x^2+5x+4}{x\left(x+2\right)}\)
\(\Rightarrow x\left(x+2\right)^2=x\left(2x^2+5x+4\right)\)
\(\Leftrightarrow x^3+4x^2+4x=2x^3+5x^2+4x\)
\(\Leftrightarrow x^3+x^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(TM\right)\end{matrix}\right.\)
Vậy S = {-1}
c) Câu này mình không chắc về đề lắm! Bạn dùng ô chữ M bị ngược để viết lại đề nhé!
a) Ta có: \(\left(x-2\right)^2=\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow x^2-4x+4=x^2-16\)
\(\Leftrightarrow x^2-4x+4-x^2+16=0\)
\(\Leftrightarrow-4x+20=0\)
\(\Leftrightarrow-4x=-20\)
hay x=5
Vậy: S={5}
a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
=>-6x-5=0
=>-6x=5
hay x=-5/6
b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)
=>8x+16=0
hay x=-2
c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)
=>9x-10=0
hay x=10/9
d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)
\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)
\(\Leftrightarrow2x^2-6x-4=0\)
\(\Leftrightarrow x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
a. \(\dfrac{-3}{x^2-9}+\dfrac{5}{3-x}=\dfrac{2}{x+3}\)
<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5}{x-3}=\dfrac{2}{x+3}\)
<=> \(\dfrac{-3}{x^2-9}+\dfrac{-5\left(x+3\right)}{x^2-9}=\dfrac{2\left(x-3\right)}{x^2-9}\)
<=> \(-3+\left(-5\right)\left(x+3\right)=2\left(x-3\right)\)
<=> -3 + (-5x) + (-15) = 2x - 6
<=> -5x -2x = 15 - 6 + 3
<=> -7x = 12
<=> x = \(\dfrac{-12}{7}\)
Vậy ........
b. \(\left|x+5\right|=2x-1\)
Nếu x \(\ge\) -5 => \(\left|x+5\right|\) = x + 5
Nếu x < -5 => \(\left|x+5\right|\) = -(x + 5)
TH1: Nếu x \(\ge\) -5
<=> x + 5 = 2x - 1
<=> x - 2x = -1 - 5
<=> -x = -6
<=> x = 6
TH2: Nếu x < -5
<=> -(x + 5) = 2x - 1
<=> -x - 5 = 2x - 1
<=> -5 + 1 = 2x + x
<=> -4 = 3x
<=> x = \(\dfrac{-4}{3}\)
Vậy .........
c. Bạn tự giải câu này nhé (có thể tách các hạng tử rồi tính)
b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1
=>-3/4x+1/6x=2+5-4/5=24/5
=>x=-288/35
c: =>6x^2+3x-30x-15=6x^2+10x-21x-35
=>-27x-15=-11x-35
=>-16x=-20
=>x=5/4
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
\(\dfrac{2x-1}{x+1}=\dfrac{-2x+1}{x-5}\left(x\ne-1;5\right)\)
\(\dfrac{2x-1}{x+1}=\dfrac{2x-1}{5-x}\)
\(x+1=5-x\)
\(2x=4\Rightarrow x=2\)