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e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
Câu a đơn giản
b)
\(A=\frac{x^4-x^3-x+1}{x^4+x^3+3x^2+2x+2}=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+\frac{x^2}{4}\right)+\left(\frac{11}{4}x^2+2x+\frac{4}{11}\right)+1-\frac{4}{11}}\)
\(=\frac{\left(x-1\right)\left(x^3-1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left(x^2+x+1\right)}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x^2+x+0,25\right)+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
\(=\frac{\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]}{\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}}\)
Vì \(\left(x-1\right)^2\left[\left(x+0,5\right)^2+0,75\right]>0\)và \(\left(x^2+\frac{x}{2}\right)^2+\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{11}}\right)^2+\frac{7}{11}>0\)
nên \(A>0\)hay A ko âm
Nhớ k nha !
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
\(=\dfrac{2x+y}{2\left(x+y\right)}-\dfrac{x+2y}{x-y}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-2xy+xy-y^2}{2\left(x+y\right)\left(x-y\right)}-\dfrac{2\left(x+2y\right)\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-xy-y^2-2\left(x^2+xy-2y^2\right)}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{2x^2-xy-y^2-2x^2-2xy+4y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-3xy+3y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9xy+9y^2-8x}{6\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9x^2y+9xy^2-8x^2+30\left(x^2-y^2\right)}{6x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-9x^2y+9xy^2+22x^2-30y^2}{6x\cdot\left(x-y\right)\left(x+y\right)}\)
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
\(A=\frac{3x^2+5xy-2y^2}{3x^2-7xy+2y^2}=\frac{6xy-2y^2+3x^2-xy}{2y^2-6xy-xy+3x^2}\)
\(=\frac{2y\left(3x-y\right)+x\left(3x-y\right)}{2y\left(y-3x\right)-x\left(y-3x\right)}\)
\(=\frac{\left(3x-y\right)\left(2y+x\right)}{\left(y-3x\right)\left(2y-x\right)}=\frac{-1\left(3x-y\right)\left(2y+x\right)}{\left(y-3x\right)\left(-1\right)\left(2y-x\right)}\)
\(=\frac{\left(-3x+y\right)\left(2y+x\right)}{\left(y-3x\right)\left(-2y+x\right)}=\frac{\left(y-3x\right)\left(2y+x\right)}{\left(y-3x\right)\left(x-2y\right)}=\frac{2y+x}{x-2y}\)
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)