14 tháng 1 lúc 20:03
Các câu hỏi tương tự
14 tháng 1 lúc 20:22
Tìm \(x\)
\(\dfrac{6}{4}+\dfrac{\dfrac{6}{4}+6}{x}\)
1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
thực hiện phép tính
\(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(\dfrac{6}{x-3}-\dfrac{2x-16}{x^2-9}-\dfrac{4}{x+3}\)
giải các phương trình sau
1, \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
2, \(\dfrac{1}{x-6}-\dfrac{2}{6+x}=\dfrac{3x+6}{x^2-36}\)
a) (X-2)(x+3)-3(4x-2)(x-4)^{^{ }2}b) dfrac{2x^2+1}{8}-dfrac{7x-2}{12}dfrac{x^2-1}{4}-dfrac{x-3}{6}c) x-dfrac{2x-2}{5}+dfrac{x+8}{6}7+dfrac{x-1}{3}d) left(2x+5right)^2left(x+2right)^2e) x^2-5+60g) 2x^3+6x^2x^2+3xh) left(x+dfrac{1}{2}right)^2+2left(x+dfrac{1}{x}right)-80mọi người giúp e với ạ
Đọc tiếp
a) (X-2)(x+3)-3(4x-2)=(x-4)\(^{^{ }2}\)
b) \(\dfrac{2x^2+1}{8}-\dfrac{7x-2}{12}=\dfrac{x^2-1}{4}-\dfrac{x-3}{6}\)
c) \(x-\dfrac{2x-2}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
d) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
e) \(x^2-5+6=0\)
g) \(2x^3+6x^2=x^2+3x\)
h) \(\left(x+\dfrac{1}{2}\right)^2+2\left(x+\dfrac{1}{x}\right)-8=0\)
mọi người giúp e với ạ
thực hiện phép tính
a, \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}\)
b, \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)
a) 6-7x+7=-8x+12
b) \(\dfrac{\text{4x-1}}{8}\)\(\)=\(\dfrac{6-x}{2}\)-\(\dfrac{1}{4}\)
c) 2x(x-3)+7x-21+0
d) \(\dfrac{x}{x-3}\)+\(\dfrac{x-2}{x+3}\)\(\dfrac{2\left(x^2+6\right)}{x^2-9}\)
Giải các phương trình saud) dfrac{1}{x-2}-dfrac{6}{x+3}dfrac{5}{6-x^2-x} e) dfrac{2}{x+2}-dfrac{2x^2+16}{x^3+8}dfrac{5}{x^2-2x+4} f) dfrac{x+1}{x^2+x+1}-dfrac{x-1}{x^2-x+1}dfrac{2left(x+2right)^2}{x^6-1}
Đọc tiếp
Giải các phương trình sau
d) \(\dfrac{1}{x-2}\)-\(\dfrac{6}{x+3}\)=\(\dfrac{5}{6-x^2-x}\)
e) \(\dfrac{2}{x+2}\)-\(\dfrac{2x^2+16}{x^3+8}\)=\(\dfrac{5}{x^2-2x+4}\)
f) \(\dfrac{x+1}{x^2+x+1}\)-\(\dfrac{x-1}{x^2-x+1}\)=\(\dfrac{2\left(x+2\right)^2}{x^6-1}\)