a, x( x - 3) - ( x + 2)( x - 1) = 3
<=> x2 - 3x - x2 + x - 2x + 2 = 3
<=> x2 - 3x - x2 + x - 2x = 3 - 2
<=> -4x = 1
<=> x =\(-\frac{1}{4}\)
Vậy_
b, \(x-\frac{x-1}{5}+\frac{x+2}{6}=4+\frac{x+1}{3}\)
\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)
=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10
<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10
<=> 19x = 114
<=> x = 6
Vậy _
$\frac{4x+3}{5}$ -$\frac{6x-2}{7}$ =$\frac{5x+4}{3}$ +3
b.
$\frac{x+4}{5}$ -x+4=$\frac{x}{3}$ -$\frac{x-2}{2}$
c.$\frac{5x+2}{6}$ -$\frac{8x-1}{3}$ =$\frac{4x+2}{5}$ -5
d.$\frac{2x+3}{3}$ =$\frac{5-4}{2}$
e. $\frac{5x+3}{12}$ =$\frac{1+2x}{9}$
f.$\frac{7x-1}{6}$ =$\frac{16-x}{5}$
g. $\frac{x-3}{5}$ =6-$\frac{1-2x}{3}$
h. $\frac{3x-2}{6}$ -5=$\frac{3-2(x+7)}{4}$
giúp vs ạ, cần gấp
tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
giải phương trình sau:
a. \(\frac{6x+1}{x^2-7x+10} +\frac{5}{x-2}=\frac{3}{x-5}\)
b.\(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\)
c. \(\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)
d.\(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
e.\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x-3}\)
f. \(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{2\left(x+2\right)^2}{x^6-1}\)
giải phương trình chứa ẩn
a) \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
b)\(\frac{3}{x+1}-\frac{2}{x+2}=\frac{4x+5}{x^2+3x+2}\)
c)\(\frac{2\cdot\left(x^2+x+6\right)}{x^3-8}+\frac{2}{2-x}=\frac{3}{x^2+2x+4}\)
d)\(\frac{6}{x^3+1}-\frac{1-x}{^{x^2}-x+1}=\frac{5}{x+1}\)
e)\(\frac{1}{x^23x+2}-\frac{3}{x^2-x-2}=\frac{-1}{x^2-4}\)
1) \(\frac{X+2}{X+3}+\frac{X-1}{X+1}=\frac{2}{X^2+4X+3}+1\)
2)\(\frac{X+1}{X-2}+\frac{2X-1}{X-1}=\frac{2}{X^2-3X+2}+\frac{11}{2}\)
3) Tìm GTLN CỦA -2X2+4X+3
4)\(\frac{X+1}{X-2}+\frac{X}{X+1}-\frac{2X+5}{X^2-X-2}=2\)
5)\(\frac{2X-1}{X+2}+\frac{X}{X+3}-\frac{2X^2+X+1}{X^2+5X+6}=\frac{-9}{2}\)
Tính:
1. \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2\text{x}}{x^2-1}\)
2. \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2\text{x}}{x^2+x+1}-\frac{6}{x-1}\)
3. \(\frac{5}{2\text{x}^2+6\text{x}}-\frac{4-3\text{x}^2}{x^2-9}-3\)
4. \(\frac{7}{8x^2-18}+\frac{1}{2\text{x}^2+3\text{x}}-\frac{1}{4\text{x}-6}\)
5. \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
1 . \(\frac{x-1}{12}-\frac{2x-12}{14}=\frac{3x-14}{25}-\frac{4x-25}{27}\)
2 . \(\frac{3}{x-1}+\frac{4}{x-2}=\frac{5}{x-3}+\frac{6}{x-4}\)
3 . \(\frac{2x-1}{2}+\frac{2x+1}{3}=\frac{2x+7}{6}+\frac{2x+9}{7}\)
4 . \(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)
bài tập. Giải các pt
1, \(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
2,\(\frac{x+1}{x-2}=\frac{1}{x^2-4}\)
3,\(\frac{x+2}{x}-\frac{x^2+5x+4}{x\left(x+2\right)}=\frac{x}{x+2}\)
4,\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
5,\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
6,\(\frac{x+1}{x}+\frac{1}{x+1}=\frac{2x-1}{2x^2+2}\)
7,\(\frac{2}{x+1}-\frac{3x+1}{\left(x+1\right)}=\frac{1}{\left(x+1\right)\left(x-2\right)}\)
8,\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)
9,\(\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\)
\(\frac{x^2-x}{x^2-x+1}-\frac{x^2-x+2}{x^2-x-2}=1.\)
\(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
\(\frac{1}{x^2-2x+2}+\frac{1}{x^2-2x+3}=\frac{9}{2\left(x^2-2x+4\right)}\)
\(\frac{1}{x^2-2x+3}+\frac{1}{x^2-2x+2}=\frac{6}{x^2-2x+4}\)
