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a. (x + 2)(x2 – 3x + 5) = (x + 2)x2
⇔ (x + 2)(x2 – 3x + 5) – (x + 2)x2 = 0
⇔ (x + 2)[(x2 – 3x + 5) – x2] = 0
⇔ (x + 2)(\(x^2\) – 3x + 5 – \(x^2\)) = 0
⇔ (x + 2)(5 – 3x) = 0
⇔ x + 2 = 0 hoặc 5 – 3x = 0
x + 2 = 0 ⇔ x = -2
5 – 3x = 0 ⇔ x = \(\dfrac{5}{3}\)
Vậy phương trình có nghiệm x = -2 hoặc x =\(\dfrac{5}{3}\)
c.\(2x^2\) – x = 3 – 6x
⇔ \(2x^2\) – x + 6x – 3 = 0
⇔ (\(2x^2\) + 6x) – (x + 3) = 0
⇔ 2x(x + 3) – (x + 3) = 0
⇔ (2x – 1)(x + 3) = 0
⇔ 2x – 1 = 0 hoặc x + 3 = 0
2x – 1 = 0 ⇔ x = 1/2
x + 3 = 0 ⇔ x = -3
Vậy phương trình có nghiệm x = \(\dfrac{1}{2}\) hoặc x = -3
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
\(\frac{x-3}{x-2}+\frac{x-2}{x-4}=-1\) (1)
ĐKXĐ \(x\ne2\) và \(x\ne4\)
\(\left(1\right)\Leftrightarrow\frac{x-2-1}{x-2}+\frac{x-4+2}{x-4}=-1\)
\(\Leftrightarrow1-\frac{1}{x-2}+1+\frac{2}{x-4}=-1\)
\(\Leftrightarrow2-\frac{1}{x-2}+\frac{2}{x-4}=-1\)
\(\Leftrightarrow\frac{1}{x-2}-\frac{2}{x-4}=3\)
\(\Leftrightarrow\frac{\left(x-4\right)-2\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{3\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Rightarrow x-4-2x+4=3\left(x^2-6x+8\right)\)
\(\Leftrightarrow-x=3x^2-18x+24\)
\(\Leftrightarrow3x^2-18x+24+x=0\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
Th1 \(x-3=0\Leftrightarrow x=3\) (nhận)
Th2 \(3x-8=0\Leftrightarrow x=\frac{8}{3}\) (nhận)
Vậy Tập nghiệm của phương trình là \(S=\left\{3;\frac{8}{3}\right\}\)
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
\(-537x^2+5054x=-541x^2+5092x\)
\(-537x^2+5054x+541x^2-5092x=0\)
\(4x^2-38x=0\)
\(x\left(2x-19\right)=0\)
\(\orbr{\begin{cases}x=0\\2x=19\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{19}{2}\end{cases}}\)
\(ĐKXĐ:x\ne\pm2\)
\(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-7x-2-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
⇔ (x – 2)(x – 2) – 3(x + 2) = 2x – 22
⇔ x 2 – 2x – 2x + 4 – 3x – 6 = 2x – 22
⇔ x 2 – 2x – 2x – 3x – 2x + 4 – 6 + 22 = 0
⇔ x 2 – 9x + 20 = 0
⇔ x 2 – 5x – 4x + 20 = 0
⇔ x(x – 5) – 4(x – 5) = 0
⇔ (x – 4)(x – 5) = 0
⇔ x – 4 = 0 hoặc x – 5 = 0
x – 4 = 0 ⇔ x = 4
x – 5 = 0 ⇔ x = 5
Vậy phương trình có nghiệm x = 4 hoặc x = 5.