\(\Leftrightarrow2x-\dfrac{1}{2}x-\dfrac{1}{4}x-\dfrac{3}{4}=6-\left(\dfrac{3-6+x}{3}\right)\cdot\dfrac{1}{2}\)
\(\Leftrightarrow x\cdot\dfrac{5}{4}-\dfrac{3}{4}=6-\dfrac{x-3}{3}\cdot\dfrac{1}{2}\)
\(\Leftrightarrow x\cdot\dfrac{5}{4}-\dfrac{3}{4}=6-\dfrac{1}{6}x+\dfrac{1}{2}\)
=>17/12x=29/4
=>x=29/4:17/12=87/17
Giải các phương trình sau:
1. \(a,\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{2x-6}\)
\(b,\dfrac{1}{x-2}+\dfrac{5}{x+1}=\dfrac{3}{2-x}\)
\(c,\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
2. \(a,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(b,2x^2-6x+1\)
1) Cho P = \(\left(\dfrac{4x-x^3}{1-4x^2}-x\right):\left(\dfrac{4x^2-x^4}{1-x^2}+1\right)\)
a) rút gọn b) tìm x để P > 0
2) Cho Q = \(\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{x^3+27}+\dfrac{1}{x+3}\right):\dfrac{x^2-1}{x+3}\)
a) rút gọn b) tìm GTLN
3) Cho A = \(\dfrac{1}{\left(x-y\right)^3}\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)+\dfrac{3}{\left(x-y\right)^4}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{6}{\left(x-y\right)^5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
chứng minh A là lập phương một số hữu tỉ
Giải phương trình:
b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)
c) (x+2) . (x-5). (x-6) (x+3) = 180
d) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{2}-x-1\)
e) \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+........+\dfrac{1}{10.110}\right).\left(x-3\right)=\dfrac{1}{1.11}+\dfrac{1}{2.12}+.......+\dfrac{1}{100.110}\)
4.Giải phương trình
a) \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
b)\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)
c)\(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
d)\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\)
e)\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
f)\(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
g)\(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\)
h)\(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
i)\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
k)\(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\)
l)\(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
m)\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
( \(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\) ). \(\dfrac{4x^2-4}{5}\)
\(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}.\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\)
\(\left(\dfrac{x+1}{x}\right)^2:\left(\dfrac{x^2+3}{x^2}+\dfrac{2}{x+1}.\left(\dfrac{1}{x}+1\right)\right)\)
Thực hiện phép tính:
\(a,\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(b,\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
Chứng minh rằng với x > 0 thì: \(\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+\dfrac{1}{x^3}}\ge6\)
tìm x,y viết dưới dạng phân số
a. \(5+\dfrac{x}{5+\dfrac{2}{5+\dfrac{3}{5+\dfrac{4}{5}}}}=\dfrac{x}{1+\dfrac{5}{2+\dfrac{4}{3+\dfrac{3}{5+\dfrac{1}{6}}}}}\)
b.
\(\dfrac{y}{3+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{3}}}}}}+\dfrac{y}{7+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{4}}}}\)
= 2
c,
\(x.\left(\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+1}}}}}}}}\right)=\)\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}}}}\)+\(x.\left(3+\dfrac{1}{3-\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3-\dfrac{1}{3}}}}}\right)\)
Giair bằng máy tính casio
Bài 1:cho phương trình
a,\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
b,\(\dfrac{\left(x+10\right)\left(x+4\right)}{12}-\dfrac{\left(x+4\right)\left(2-x\right)}{4}=\dfrac{\left(x+10\right)\left(x-2\right)}{3}\)
c,\(\dfrac{2\left(x-3\right)}{7}+\dfrac{x-5}{3}=\dfrac{13x+4}{21}\)
d,\(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{5}\)
e,\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
