a) \(\dfrac{2x+3}{-4}\)≥\(\dfrac{4-x}{-3}\)
=> 3(2x+3)≥4(4-x)
<=> 6x+9≥16-4x
<=> 10x ≥ 7
<=> x ≥\(\dfrac{7}{10}\)
b) \(\dfrac{7x-1}{6}+2x\) ≤ \(\dfrac{16-x}{5}\)=> 5(7x-1) + 30*2x ≤ 6(16-x)<=> 35x - 5 +60x ≤ 96 -6x<=> 101x ≤ 101<=> x ≤ 1
Giai các bpt sau
a,\(\dfrac{x-1}{2}-\dfrac{7x+3}{15}\le\dfrac{2x+1}{3}+\dfrac{3-2x}{5}\)
b,\(\dfrac{2x+1}{-3}-\dfrac{2x^2+3}{-4}>\dfrac{x\left(5-3x\right)}{-6}-\dfrac{4x+1}{-5}\)
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/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
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
BT10: Thực hiện phép tính
\(a,\dfrac{4}{5}y^2x^5-x^3.x^2y^2\)
\(b,-xy^3-\dfrac{2}{7}y^2.xy\)
\(c,\dfrac{5}{6}xy^2z-\dfrac{1}{4}xyz.y\)
\(d,15x^4+7x^4-20x^2.x^2\)
\(e,\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+xy.x^4\)
\(f,13x^2y^5-2x^2y^5+x^6\)
Giari bptr sau
\(\dfrac{7x-1}{6}+2x\ge\dfrac{16-x}{5}\)
Giải bất phương trình sau
a)\(\dfrac{2-x}{3}\)\(-x-2\le\dfrac{x-17}{2}\)
b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)
\(\dfrac{x-1}{2}\)-\(\dfrac{7x+3}{15}\)\(\le\)\(\dfrac{2x+1}{3}\)+\(\dfrac{3-2x}{5}\)
Giải phương trình sau :
a,\(\dfrac{7-3x}{12}+\dfrac{5x+2}{7}=x+13\)
b,\(\dfrac{3\left(x+3\right)}{4}-\dfrac{1}{2}=\dfrac{5x+9}{7}-\dfrac{7x-9}{4}\)
c,\(\dfrac{2x+1}{3}-\dfrac{5x+2}{7}=x+3\)
d,\(\dfrac{2x-3}{3}-\dfrac{2x+3}{7}=\dfrac{4x+3}{5}-17\)
Giải các phương trình sau:
\(e.\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
\(f.\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
\(g.\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
\(h.\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)
