Bài 3:
a) \(\dfrac{2x-1}{5}\)-\(\dfrac{x-2}{3}\)
=\(\dfrac{x+7}{15}\)
b) \(\dfrac{x+3}{2}\)-\(\dfrac{x-1}{3}\)
=\(\dfrac{x+5}{6}\)+1
c) \(\dfrac{2\left(x+5\right)}{3}\)+\(\dfrac{x+12}{2}\)
-\(\dfrac{5\left(x-2\right)}{6}\)=\(\dfrac{x}{3}\)+11
d) \(\dfrac{x-4}{5}\)+\(\dfrac{3x-2}{10}\)-x
=\(\dfrac{2x-5}{3}\)-\(\dfrac{7x+2}{6}\)
e) \(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\)
=\(\dfrac{\left(x-4^{ }\right)^2}{6}\)+\(\dfrac{\left(x-2\right)^2}{3}\)
d) \(\dfrac{7x^2-14x-5}{15}\)
=\(\dfrac{\left(2x+1\right)^2}{5}\)-\(\dfrac{\left(x-1\right)^2}{3}\)
e) \(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}\)+\(\dfrac{2}{5}\)
=\(\dfrac{\left(x-2\right)^2}{5}\)+\(\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)
Bài 2:
a) (x+1)(2x-3)-3(x-2)
=2(x-1)\(^2\)
b) (x+1)(x\(^2\)-x+1)-2x
=x(x-1)(x+1)
c) \(\dfrac{x}{3}\)-\(\dfrac{5x}{6}\)-\(\dfrac{15x}{12}\)=\(\dfrac{x}{4}\)-5
d) \(\dfrac{x-1}{2}\)-\(\dfrac{x+1}{15}\)-
\(\dfrac{2x-13}{6}\)=0
e) \(\dfrac{3\left(5x-2\right)}{4}\)-2
=\(\dfrac{7x}{3}\)-5(x-7)
g) \(\dfrac{x-3}{11}\)+\(\dfrac{x+1}{3}\)
=\(\dfrac{x+7}{9}\)-1
h) \(\dfrac{3x-0,4}{2}\)+\(\dfrac{1,5-2x}{3}\)
=\(\dfrac{x+0,5}{5}\)
Bài 1:
a) 5(k+3x)(x+1)-4(1+2x)=80
b) x\(^2\)-4x+6=0
c) (3-x)\(^2\)=x\(^2\)-6x+9
d) x\(^2\)+2+0 và x(x\(^2\)+2)=0
e) x+1=x và x\(^2\)+1=0
f) x+2=0 và \(^{\dfrac{x}{x+2}}\)=0
g) x\(^2\)+\(\dfrac{1}{x}\)=x+\(\dfrac{1}{x}\) và x\(^2\)+x=0
h) x=5=0 và (x+5)(x\(^2\)+1)=0
Chứng minh 2 phương trình của câu d,e tương đương
a) \(\dfrac{x+1}{35}\)+\(\dfrac{x+3}{33}\)=\(\dfrac{x+5}{31}\)+\(\dfrac{x+7}{29}\)Hd: cộng thêm 1 vào các hạng tửb) \(\dfrac{x-10}{1994}\)+\(\dfrac{x-8}{1996}\)+\(\dfrac{x-6}{1998}\)+\(\dfrac{x-4}{2000}\)+\(\dfrac{x-2}{2002}\)=\(\dfrac{x-2002}{2}\)+\(\dfrac{x-2000}{4}\)+\(\dfrac{x-1998}{6}\)+\(\dfrac{x-1996}{8}\)+\(\dfrac{x-1994}{10}\)Hd: trừ đi 1 vào các hạng tử
c) \(\dfrac{x-1991}{9}\)+\(\dfrac{x-1993}{7}\)+\(\dfrac{x-1995}{5}\)+\(\dfrac{x-1997}{3}\)+\(\dfrac{x-1999}{1}\)=\(\dfrac{x-9}{1991}\)+\(\dfrac{x-7}{1993}\)+\(\dfrac{x-5}{1995}\)+\(\dfrac{x-3}{1997}\)+\(\dfrac{x-1}{1999}\)Hd: trừ đi 1 vào các hạng tửd) \(\dfrac{x-8}{15}\)+\(\dfrac{x-74}{13}\)+\(\dfrac{x-67}{11}\)+\(\dfrac{x-64}{9}\)=10Chú ý: 10=1+2+3+4e) \(\dfrac{x-1}{13}\)-\(\dfrac{2x-13}{15}\)=\(\dfrac{3x-15}{27}\)-\(\dfrac{4x-27}{29}\)Hd: thêm hoặc bớt 1 vào các hạng tử