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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: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
a) =(x-y)5+(x-y)3=(x-y)3[(x-y)2+1]
b) =33(y-2x)3:-9(y-2x)=-3(y-2x)2
c) =(x-y)2 [3(x-y)3-2(x-y)2+3]:5(x-y)2=[3(x-y)3-2(x-y)2+3]/5
a: \(=2x^2-x+5\)
b: \(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c: \(=-x^3+\dfrac{3}{2}-2x\)
d: \(=-2x^2+4xy-6y^2\)
e: \(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
P/s : Phá ngoặc ra là ok :
a )
\(\left[4x-2\left(x-3\right)\right].\left(-3x\right)\)
\(=\left[4x-2x+6\right]\left(-3x\right)\)
\(=-12x^2+6x^2-18x\)
b )
\(3\left[x-3\left(4-2x\right)+8\right]\)
\(=3\left[x-12+6x+8\right]\)
\(=3\left[7x-4\right]\)
\(=21x-12\)
c )
\(5\left(3x^2-4y^3\right)+9\left(2x^2-y^3\right)\)
\(=15x^2-20y^3+18x^2-9y^3\)
\(=33x^2-29y^3\)
d )
\(3x^2\left(2y-1\right)-2x^2\left(5y-3\right)\)
\(=6x^2y-3x^2-10x^2y+6x^2\)
\(=-4x^2y+3x^2\)
a)\(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=2x^2\left(5x^2-2x+1\right)-3x\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
a. \(\left(2x^2-3x\right)\left(5x^2-2x+1\right)\)
\(=10x^4-4x^3+2x^2-15x^3+6x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
b. \(\left(2x^4-x^3+3x^2\right):\left(\frac{1}{3}x^2\right)\)
\(=\left(2x^4-x^3+3x^2\right).\frac{3}{x^2}\)
\(=0,6x^2-3x+0,9\)
b)\(\frac{9x^4-6x^3+15x^2+2x+1}{3x^2-2x+5}=\frac{3x^2.\left(3x^2-2x+5\right)+2x+1}{3x^2-2x+5}=3x^2+\frac{2x+1}{3x^2-2x+5}\)
=> đa thức dư trong phép chia là 2x+1
\(\frac{x^3+2x^2-3x+9}{x+3}=\frac{x^3+9x^2+27x+27-7x^2-30x-18}{x+3}=\frac{\left(x+3\right)^3-7x^2-30x-18}{x+3}\)
\(\left(x+3\right)^2-\frac{7x^2+21x+9x+18}{x+3}=\left(x+3\right)^2-\frac{7x.\left(x+3\right)+9.\left(x+3\right)-9}{x+3}\)
\(=\left(x+3\right)^2-\frac{\left(7x+9\right).\left(x+3\right)-9}{x+3}=\left(x+3\right)^2-\left(7x+9\right)-\frac{9}{x+3}\)
=> đa thức dư trong phép chia là 9
p/s: t mới lớp 7_sai sót mong bỏ qua :>
Ta có: \(\frac{x^4+2x^3+x-25}{x^2+5}\)
\(=\frac{x^4+5x^2+2x^3+10x-5x^2-25-9x}{x^2+5}\)
\(=\frac{x^2\left(x^2+5\right)+2x\left(x^2+5\right)-5\left(x^2+5\right)-9x}{x^2+5}\)
\(=x^2+2x-5-\frac{9x}{x^2+5}\)
Ta có: x 4 + 2 x 3 + x − 25 x 2 + 5 x 2 +5 x 4 +2x 3 +x−25 = x 4 + 5 x 2 + 2 x 3 + 10 x − 5 x 2 − 25 − 9 x x 2 + 5 = x 2 +5 x 4 +5x 2 +2x 3 +10x−5x 2 −25−9x = x 2 ( x 2 + 5 ) + 2 x ( x 2 + 5 ) − 5 ( x 2 + 5 ) − 9 x x 2 + 5 = x 2 +5 x 2 (x 2 +5)+2x(x 2 +5)−5(x 2 +5)−9x = x 2 + 2 x − 5 − 9 x x 2 + 5 =x 2 +2x−5− x 2 +5 9x