Bài làm
Vì \(\frac{2}{\left(3-x\right)}< 0\) ĐKXĐL x khác 3
=> 2 và ( 3 - x ) trái dấu
Mà 2 là số dương
=> 3 - x < 0
<=> -x < -3
<=> x > 3 ( TM )
Vậy x > 3 là nghiệm bất phương trình
a) \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
b) \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
c) \(x\left(x+3\right)^3-\dfrac{x}{4}\left(x+3\right)=0\)
Giải các phương trình
1,\(3x-1=0\\\) 4, \(\dfrac{x}{3}-\dfrac{x}{5}=4\)
2, \(2-x=3x+1\) 5, \(\dfrac{x-1}{4}+\dfrac{2x+1}{6}=\dfrac{3}{2}\)
3, \(2\left(x-2\right)-1=5x\)
Giải các phương trình
a)3x-2=2x-3 c)11x+42-2x=100-9x-22
b)2x+3=5x+9 d)2x-(3-5x)=4(x+3)
e)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
f)\(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
g)(2x+1)(x-1)=0
h)(x+\(\dfrac{2}{3}\))(x-\(\dfrac{1}{2}\))=0
i)(3x-1)(2x-3)(2x-3)(x+5)=0
k)3x-15=2x(x-5)
m)\(\left|x-2\right|=3\)
n)\(\left|x+1\right|=\left|2x+3\right|\)
j)\(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)
đ)\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
y)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)
a/8(3x-2) - 13x =5(12 - 3x) +7x
b/\(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)
c/\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
GPT: \(\dfrac{x}{2}\)(4x - 3) + 2(3 - x)(x + 4) ≤ 0
giải phương trình\(\dfrac{5}{x-2}\)+\(\dfrac{6}{3-4x}\)=0
Giải các phương trình sau :
a) \(0,25x+1,5=0\)
b) \(6,36-5,3x=0\)
c) \(\dfrac{4}{3}x-\dfrac{5}{6}=\dfrac{1}{2}\)
d) \(-\dfrac{5}{9}x+1=\dfrac{2}{3}x-10\)
a/(x-2)(x2-1)=0
b/\(\dfrac{-1}{1-x}-\dfrac{7}{x+2}=\dfrac{3}{\left(x-1\right)\left(x+2\right)}\)
c/10x-2(2x+4)=2x
a.\(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
b.\(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
c.\(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
d.\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
e.\(\dfrac{x+97}{125}+\dfrac{x-63}{35}=\dfrac{x-7}{21}+\dfrac{x-77}{49}\)
Các cặp phương trình dưới có tương đương ko? Vì sao?
a) x2 - 4 + \(\dfrac{1}{x-2}\) = \(\dfrac{1}{2}\) và x2 - 4 = 0
b) x2 - 5,5x + 3 = 0 và 2x2 - 11x + 6 = 0
c) x2 - 1 = 0 và x2 - 1 + \(\dfrac{1}{x-1}\) = \(\dfrac{1}{x-1}\)
Giúp vs ạ !!!!
