g \(2\left(x+2\right)^2\ge2x+\frac{7}{2}\)
h \(\frac{x^2+x-1}{1-x}>-x\)
\(|x^2+x+12|>x^2+x+12\)
Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
tìm m để bpt vô nghiệm
\(\left|\frac{2\left(x+\frac{1}{x}\right)^2-\left(x-\frac{1}{x}\right)-7}{3\left(x+\frac{1}{x}\right)^2+\left(x-\frac{1}{x}\right)+m-12}\right|>2\)
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
Giari các phương trình sau.
a. \(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
b. \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)
c. \(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
d. \(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
e. \(\frac{x}{2x+6}-\frac{x}{2x+2}=\frac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
f. \(\frac{x}{x+1}-\frac{2x-3}{1-x}=\frac{3x^2+5}{x^2-1}\)
g. \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\)
h. \(\frac{x-1}{x}-\frac{1}{x+1}=\frac{2x-1}{x^2+x}\)
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
e)
ĐKXĐ: \(x\neq -1; x\neq -3\)
PT \(\Leftrightarrow \frac{x(2x+2)-x(2x+6)}{(2x+6)(2x+2)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Leftrightarrow \frac{-4x}{2(x+3).2(x+1)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Leftrightarrow \frac{-x}{(x+3)(x+1)}=\frac{3x+2}{(x+1)(x+3)}\)
\(\Rightarrow -x=3x+2\Rightarrow x=-\frac{1}{2}\) (thỏa mãn)
Vậy..............
f)
ĐKXĐ: \(x\neq \pm 1\)
PT \(\Leftrightarrow \frac{x}{x+1}+\frac{2x-3}{x-1}=\frac{3x^2+5}{x^2-1}\)
\(\Leftrightarrow \frac{x(x-1)+(2x-3)(x+1)}{(x-1)(x+1)}=\frac{3x^2+5}{x^2-1}\)
\(\Leftrightarrow \frac{3x^2-2x-3}{x^2-1}=\frac{3x^2+5}{x^2-1}\)
\(\Rightarrow 3x^2-2x-3=3x^2+5\)
\(\Leftrightarrow x=-4\) (thỏa mãn)
Vậy.........
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
a, (x-1)3 - x(x-1)2 = 5(2-x) - 11(x+2)
b, (x-2)3 + (3x-1)(3x+1) = (x+1)3
c, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
d, \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
e, \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)
⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)
⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)
⇔\(9x-10=0\)
hay 9x=10
⇔\(x=\frac{10}{9}\)
Vậy: \(x=\frac{10}{9}\)
c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)
⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)
⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)
⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)
⇔\(6x-3-5x+10-3x-21=0\)
⇔\(-2x-14=0\)
⇔\(-2x=14\)
hay x=-7
Vậy: x=-7
d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)
⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
⇔\(6x-18+7x-35-13x-4=0\)
⇔\(-21\ne0\)
Vậy: x∈∅
e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)
⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)
⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)
⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)
⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)
⇔\(-12x+96=0\)
⇔\(-12x=-96\)
hay x=8
Vậy: x=8
Giải phương trình:
1. \(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
2. \(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
3. \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
4. \(\frac{x+3}{2}-\frac{x-1}{3}=\frac{x+5}{6}+1\)
5. \(\frac{x-4}{5}-\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
6. \(\frac{\left(x+2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)
7. \(\frac{\left(x+2\right)^2}{8}-2\left(2x-1\right)=25+\frac{\left(x-2\right)^2}{8}\)
8.\(\frac{7x^2-14x-5}{5}=\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}\)
9. \(\frac{\left(2x-3\right)\left(2x+3\right)}{8}=\frac{\left(x-4\right)^2}{6}+\frac{\left(x-2\right)^2}{3}\)
10. \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
1. Xét dấu các biểu thức sau :
a, f(x) = \(\frac{\left(7-4x\right)\left(x^2+x-2\right)}{2x^2-3x+2}\)
b, g(x) = \(\frac{\left(25-x^2\right)\left(x^2+6x+9\right)}{-x^2-2x+8}\)
c, h(x) = \(\frac{x\left(x^2-4x-12\right)}{\sqrt{6}x^2-3x+\sqrt{2}}\)
d, k(x) = \(\frac{-x^3-5x^2+4}{x^4+4x^3-8x-5}\)
Tìm x, biết:
a) \(\left(x-\frac{-7}{8}\right)+2,3=3,24\) ; b) \(x.\left(4,6+\frac{3}{5}\right)=7,2-8,15\) ; c) |2x-3|+3=16 ; d) \(42+\frac{3}{7}.\left|3x-1\right|=12\)
e) \(\left(x-\frac{1}{2}\right).\left(1+5x\right)=0\) ; f) |2x-1|=|x+2| ; g) \(2.3^x.3^2=18\left(x\in N\right)\) ; h) \(\frac{x+2}{5}=\frac{2-3x}{3}\)