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\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
a) = \(12a^2b\left(a^2-b^2\right)\)
= \(12a^4b-12a^2b^3\)
b)nhân ra :
= \(2x^4-16x^3+4x^2-3x^3+24x^2-6x+5x^2-40x+10\)
= \(2x^4-19x^3+33x^2-46x+10\)
Tìm x:
a) \(\frac{1}{4}x^2-\left(\frac{1}{4}x^2-2x\right)=-14\)
= \(\frac{1}{4}x^2-\frac{1}{4}x^2+2x=-14\)
=\(2x=-14=>x=-7\)
b) \(x^3+27-x\left(x^2-1\right)=27\)
= \(x^3+27-x^3+x=27\)
= \(27+x=27=>x=0\)
a, \(\left(x+2\right)^9:\left(x+2\right)^6=\left(x+2\right)^3\)
b, mình xin lỗi nhưng đề có nhầm chỗ nào không thế :(
c, \(\left(x^2+2x+4\right)^5:\left(x^2+2x+4\right)=\left(x^2+2x+4\right)^4\)
d, \(2\left(x^2+1\right)^3:\dfrac{1}{3}\left(x^2+1\right)=6\left(x^2+1\right)^2\)
a) (-x2 +6x3 - 26x + 21) : (3-2x)
= -3x2 + 5x + 11/2 ( dư 37/1/2)
b) (2x4 - 13x3 - 15 + 5x + 21x2) : (4x-x2 -3)
= -2x2 + 5x + 5
\(\frac{3\left(x+1\right)}{x+2}-\frac{3x-6}{x^2-4}\)
\(=\frac{3\left(x+1\right)}{x+2}-\left(\frac{3x-6}{x^2-4}\right)\)
\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)
\(=\frac{3\left(x+2\right)\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x+2\right)\left(x-2\right)}\)
\(=\frac{3x-6}{x+2}\)
\(\frac{x^2+4x+4}{1-x}.\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\)
\(=\frac{x^2+4x+4}{1-x}.\left[\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\right]\)
\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)
\(=\frac{\left(x-1\right)\left(x-1\right)\left(x+2\right)\left(x+2\right)}{3\left(-x+1\right)\left(x+2\right)\left(x+2\right)\left(x+2\right)}\)
\(=\frac{-x+1}{3x+6}\)
Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
a) \((x-1).(x^5+x^4+x^3+x^2+x+1)\)
\(=x.x^5+x.x^4+x.x^3+x.x^2+x.x+x.1+\left(-1\right).x^5+\left(-1\right).x^4+\left(-1\right).x^3+\left(-1\right).x^2+\left(-1\right).x+\left(-1\right).1\)
\(=x^6+x^5+x^4+x^3+x^2+x-x^5-x^4-x^3-x^2-x\)
\(=x^6\)
b) \(\left(x+1\right).\left(x^6-x^5+x^4-x^3+x^2-x+x\right)\)
\(=x.x^6+x.\left(-x^5\right)+x.x^4+x.\left(-x^3\right)+x.x^2+x.\left(-x\right)+x.x+1.x^6+1.\left(-x^5\right)+1.x^4+1.\left(-x^3\right)+1.x^3+1.\left(-x\right)+1.x\)
\(=x^7-x^6+x^5-x^4+x^3-x^2+x^2+x^6-x^5+x^4-x^3+x^2-x+x\)
\(=x^7\)