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a) 3x + 4 = 2x + 1 <=> 3x - 2x = 1 - 4 <=> x = -3
b) 5x + 1= 3x - 7 <=> 5x - 3x = -7 -1 <=> 2x = -8 <=> x = -4
c) 1/3x + 3/2 = x + 3 <=> 1/3x - x = 3 - 3/2 <=> -2/3x = 3/2 <=> x= -9/4
d) 2x - 1/2 = 3x - 1/4 <=> 2x -3x = -1/4 +1/2 <=> -x = 1/4 <=> x = -1/4
Vậy phương trình a) có nghiệm là x = -3
\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)
\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)
\(< =>12x-20-14x-21=0\)
\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)
\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)
\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)
\(< =>8x+12+4x-2x+3=0\)
\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)
a) 2x^2 + 3 = 2x(x + 4) - 7
<=> 2x^2 + 3 = 2x^2 + 8x - 7
<=> 2x^2 - 2x^2 - 8x = - 7 - 3
<=> -8x = -10
<=> x = -10/-8 = 5/4
b) 4x^2 - 12x + 5 = 0
<=> 4x^2 - 2x - 10x + 5 = 0
<=> 2x(2x - 1) - 5(2x - 1) = 0
<=> (2x - 5)(2x - 1) = 0
<=> 2x - 5 = 0 hoặc 2x - 1 = 0
<=> x = 5/2 hoặc x = 1/2
c) |5 - 2x| = 1 - x
<=> \(\hept{\begin{cases}5-2x\text{ nếu }5-2x\ge0\Leftrightarrow x\ge\frac{5}{2}\\-\left(5-2x\right)\text{ nếu }5-2x< 0\Leftrightarrow x< \frac{5}{2}\end{cases}}\)
+) nếu x >= 5/2, ta có:
5 - 2x = 1 - x
<=> -2x + 1 = 1 - 5
<=> -x = -4
<=> x = 4 (tm)
+) nếu x < 5/2, ta có:
-(5 - 2x) = 1 - x
<=> -5 + 2x = 1 - x
<=> 2x + 1 = 1 + 5
<=> 3x = 6
<=> x = 2 (ktm)
d) \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}-\frac{2x+3}{x^2+x+1}\) ; ĐKXĐ: x # 1
<=> \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x+3}{x^2+x+1}\)
<=> \(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
<=> 2(x^2 + x + 1) = (2x - 1)(2x + 1) - (2x + 3)(x - 1)
<=> 2x^2 + 2x + 2 = 2x^2 - x + 2
<=> 2x^2 - 2x^2 + 2x - x = 2 - 2
<=> x = 0
mạn phép vô đây để kiếm câu trả lời
\(2x^2+3=2x\left(x+4\right)-7\)
\(< =>2x^2+3=2x.x+4.2x-7\)
\(< =>2x^2+3=2x^2+8x-7\)
\(< =>2x^2+3-2x^2=8x-7\)
\(< =>\left(2x^2-2x^2\right)-8x=-7-3\)
\(< =>-8x=-10< =>8x=10\)
\(< =>x=10:8=\frac{10}{8}=\frac{5}{4}\)
Ta có:
Chọn C