\(\frac{-3x^2-x+4}{x^2+3x+5}>0\)
Giải các phương trình.
a) \(\frac{2.\left(1-3x\right)}{5}-\frac{2+3x}{10}=7-\frac{3.\left(2x+1\right)}{4}\)b) \(\frac{3x+2}{2}-\frac{3x+1}{6}=2x+\frac{5}{3}\)
c) 3x-5=7
d) \(\frac{5}{x+3}=\frac{3}{x-1}\)
e) -2x+14=0
f) 2x.(x-3)+5.(x-3)=0
g) (x2-4)-(x-2).(3-2x)=0
h) 2x3+6x2=x2+3x
Ai giúp vs !!!
\(a.\frac{3x-7}{5}=\frac{2x-1}{3}\\ b.\frac{4x-7}{12}-x=\frac{3x}{8}\\ c.\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\\ d.\frac{5x-8}{3}=\frac{1-3x}{2}\\ e.\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\\ f.\frac{x-1}{\frac{2}{5}}-3-\frac{3x-2}{\frac{5}{4}}-2=1\)
\(\frac{3x-7}{5}=\frac{2x-1}{3}\)
\(\Leftrightarrow9x-21=10x-5\)
\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)
\(\frac{4x-7}{12}-x=\frac{3x}{8}\)
\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)
\(\Leftrightarrow-56-64x=36x\)
\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)
\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)
Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0
Vậy x = 2019
\(\frac{5x-8}{3}=\frac{1-3x}{2}\)
\(\Leftrightarrow10x-16=3-9x\)
\(\Leftrightarrow19x=19\Leftrightarrow x=1\)
\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)
\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)
\(\Rightarrow-16x-272=120x+312\)
\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)
bài 1: giải các bất phương trình sau:
1) (x-3)(4-x)≥0
2) \(\frac{1+2x}{3x-4}< 0\)
3) (x+1)(x-1)(3x-6)>0
4) 3x(2x+7)(9-3x)≥0
5) \(\frac{\left(2x-5\right)\left(x+2\right)}{-4x+3}>0\)
6) \(\frac{2}{x-1}\le\frac{5}{2x-1}\)
7) \(\frac{x-3}{x+1}>\frac{x+5}{x-2}\)
8) \(\frac{2x^2+x}{1-2x}\ge1-x\)
Giải phương trình:
2x-2=8-3x
x2-3x+1=x+x2
(x2+1)(2x+4)=0
(4x+1)(x2+2)=0
\(\frac{x}{2}=3-\frac{x+4}{3}\)
\(\frac{3-x}{4}=1-\frac{3x-5}{6}\)
\(\frac{2x+5}{9}=2+\frac{x-3}{6}\)
\(\frac{x+5}{3}=1+\frac{x-3}{9}\)
\(\frac{2x-5}{x+5}=3\)
\(\frac{x^2-6}{x}=x+\frac{3}{2}\)
\(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\frac{5}{3x+2}=2x-1\)
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy...
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)
\(\Leftrightarrow\)\(-4x=-1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy...
mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))
\(2x-2=8-3x\)
\(\Leftrightarrow\)\(2x+3x=8+2\)
\(\Leftrightarrow\)\(5x=10\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(x^2-3x+1=x+x^2\)
\(\Leftrightarrow\)\(4x-1=\left(x^2+x\right)-\left(x^2+x\right)\)
\(\Leftrightarrow\)\(4x-1=0\)
\(\Leftrightarrow\)\(4x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{4}\)
Vậy \(x=\frac{1}{4}\)
\(\left(x^2+1\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+1=0\\2x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\2x=-4\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{-1}\\x=\frac{-4}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x\in\left\{\varnothing\right\}\\x=-2\end{cases}}}\)
Vậy \(x=-2\)
Chúc bạn học tốt ~
giải phương trình:\(\frac{x^2+x}{x^2+3}+\frac{3x^2-x+15}{x^2+4}+\frac{x^2+x+2}{x^2+5}+x^3-3x^2+1=0\)
Bài 4 Giải các bất phương trình sau :
31 , \(\frac{-3x^2-x+4}{x^2+3x+5}>0\)
32 , \(\frac{4x^2+3x-1}{x^2+5x+7}>0\)
33 , \(\frac{5x^2+3x-8}{x^2-7x+6}< 0\)
34 , \(\frac{2x-5}{x^2-6x-7}< \frac{1}{x-3}\)
35 , \(\frac{x^2-5x+6}{x^2+5x+6}\ge\frac{x+1}{x}\)
Rút gọn thừa số chung :
\(.\frac{x}{3x-1}+\frac{1}{3x-1}=\frac{x}{3x+2}+\frac{3}{3x+2}\)
Đơn giản biểu thức :
\(-\frac{x}{3x+2}-\frac{3}{3x+2}+\frac{x}{3x-1}+\frac{1}{3x-1}=0\)
Giải phương trình
\(-\frac{3x-5}{\left(3x-1\right)\left(3x+2\right)}=0\)
Giải phương trình :
3x=5
\(\frac{1}{3x-1}=0\)
\(\frac{1}{3x+2}=0\)
giải các pt sau
a)10x2-x-11=0 b)2x2-3x-2=0 c)2x2-8=0 d)3x2-5x=0 e)x2-2x+1=0 f)3x4-12x2+9=0 g)x4-4x2-5=0
n)x3-3x2-x+3=0 m)x4-3x2-4=0 h)\(\frac{12}{x-1}-\frac{8}{x+1}=1\)
i)x3+6x2+5x=0 k)3x2-x-6=0
bài 1 giải phương trình chứa ẩn ở mẫu
a\(\frac{3x^2+7x-10}{x}=0\)
b \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)
c \(\frac{x-3}{x-2}+\frac{3x-5}{x-1}=0\)
d \(\frac{x+1}{x-2}-\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}\)
e \(\frac{x-6}{x-4}=\frac{x}{x-2}\)
f\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)