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\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)
\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
\(\dfrac{5x-1+x+1}{3x^2y}=\dfrac{6x}{3x^2y}=\dfrac{2}{xy}\)
\(\dfrac{21x^2+22y}{36x^3y^2}\)
\(\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x-8}{4x-7}=1-\dfrac{1}{4x-7}\)
\(C=\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+7}{\left(x+2\right)\left(\left(4x+7\right)\right)}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4x+8}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}\\ C=\dfrac{4}{4x+7}\)
\(D=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\\ D=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\\ D=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{\left(3x-2\right)2x}{\left(2x-1\right)2x}-\dfrac{3x-2}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(3x-2\right)2x-\left(3x-2\right)}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left[\left(3x-2\right)2x-\left(3x-2\right)\right]}{2x\left(2x-1\right)}\\ C=\dfrac{\left(1-3x\right)\left(2x-1\right)+\left(3x-2\right)\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=\dfrac{\left[\left(1-3x\right)+\left(3x-2\right)\right]\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\\ C=-\dfrac{1}{2x}\)
a: \(=6x^4-9x^3+3x^2-4x^3+6x^2-2x+10x^2-15x+5\)
\(=6x^4-13x^3+19x^2-17x+5\)
b: \(=6x^4-\dfrac{9}{4}x^3-\dfrac{9}{2}x^2-\dfrac{8}{3}x^3+x^2+2x-\dfrac{20}{3}x^2+\dfrac{5}{2}x+5\)
\(=6x^4-\dfrac{59}{12}x^3-\dfrac{67}{6}x^2+\dfrac{9}{2}x+5\)
c: \(=3x^4-\dfrac{9}{8}x^3-\dfrac{3}{4}x^2+8x^3-3x^2-6x-\dfrac{4}{3}x^2+\dfrac{1}{2}x+1\)
\(=3x^4-\dfrac{55}{8}x^3-\dfrac{25}{12}x^2-\dfrac{11}{2}x+1\)
\(=\dfrac{4x+8}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\)
\(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ =\dfrac{4x+7}{\left(x+2\right)\left(4x+7\right)}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\\ =\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}\\ =\dfrac{4x+8}{\left(x+2\right)\left(4x+7\right)}\\ =\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}\\ =\dfrac{4}{4x+7}\)
Sửađề: \(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\)
\(=\dfrac{4x+7+1}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\)