37.
\(a.\left(x+2\right)^2-2\left(x+4\right)=\left(x-3\right)\left(x+1\right)\\ \Leftrightarrow x^2+4x+4-2x-8=x^2-3x+x-3\\ \Leftrightarrow x^2+2x-4-x^2+2x+3=0\\ \Leftrightarrow4x-1=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\\ b.\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\\ \Leftrightarrow\left(2x^2-3x+2x-3\right)-3x+6=2\left(x^2-2x+1\right)\\ \Leftrightarrow2x^2-x-3-3x+6=2x^2-4x+2\\ \Leftrightarrow2x^2-4x+3=2x^2-4x+2\\ \Leftrightarrow-4x+3=-4x+2\\ \Leftrightarrow3=2\)
=> Pt vô nghiệm
\(c.2x\left(x+2\right)=\left(x+1\right)\left(2x-5\right)\\ \Leftrightarrow2x^2+4x=2x^2-5x+2x-5\\ \Leftrightarrow4x=-3x-5\\ \Leftrightarrow4x+4x=-5\\ \Leftrightarrow7x=-5\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d.3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\ \Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-12x+12+9x=3x^2+3x\\ \Leftrightarrow-12x+12+9x=3x\\ \Leftrightarrow-3x+12=3x\\ \Leftrightarrow3x+3x=12\\ \Leftrightarrow6x=12\\ \Leftrightarrow x=2\)
Bài 38:
a: \(\dfrac{x}{2}-\dfrac{3x}{4}=x-\dfrac{7x}{8}\)
=>\(\dfrac{4x}{8}-\dfrac{6x}{8}=\dfrac{8x}{8}-\dfrac{7x}{8}\)
=>-2x=x
=>-3x=0
=>x=0
b: \(\dfrac{8x-3}{4}-\dfrac{3x-2}{2}=\dfrac{2x-1}{2}+\dfrac{x+3}{4}\)
=>\(\dfrac{8x-3-2\left(3x-2\right)}{4}=\dfrac{2\left(2x-1\right)+x+3}{4}\)
=>8x-3-6x+4=4x-2+x+3
=>5x+1=2x+1
=>3x=0
=>x=0
c: \(\dfrac{x-1}{2}-\dfrac{x+1}{15}-\dfrac{2x-13}{6}=0\)
=>\(\dfrac{15\left(x-1\right)}{30}-\dfrac{2\left(x+1\right)}{30}-\dfrac{5\left(2x-3\right)}{30}=0\)
=>15(x-1)-2(x+1)-5(2x-3)=0
=>15x-15-2x-2-10x+15=0
=>3x-2=0
=>3x=2
=>\(x=\dfrac{2}{3}\)
d: \(\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{x+7}{6}\)
=>\(\dfrac{6\left(x+5\right)+3\left(3-2x\right)}{12}=\dfrac{12x}{12}-\dfrac{2\left(x+7\right)}{12}\)
=>6x+30+9-6x=12x-2x-14
=>10x-14=39
=>10x=14+39=55
=>\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Mn giúp có thể giúp mình câu C bài 4 và bài 5 được ko ạ, giải chi tiết 1 chút với ạ. Mình cảm ơn
Giúp mình bài 2b với bài 3 với ạ
