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mk giải từng nha == tại vì mk sợ nhiều qus bị troll
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)
\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)
\(-7+18x^2-6x=x-4\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(18x+9=4x^2-40x+100\)
\(18x+9-4x^2+40x-100=0\)
\(58x-91-4x^2=0\)
\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)
Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath
a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6.2}{2x\left(x+4\right)}+\frac{3x}{2x\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(x+4\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)
c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5.y}{2y\left(y+2\right)}+\frac{2\left(y-2\right)}{2y\left(y+2\right)}=\frac{-5y+2y-4}{2y\left(y+2\right)}=\frac{-3y-4}{2y\left(y+2\right)}\)
d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x\left(x-2y\right)}-\frac{3}{x\left(x-2y\right)}=\frac{x-1-3}{x\left(x-2y\right)}=\frac{x-4}{x\left(x-2y\right)}\)
a)= \(\frac{-1}{xy}\)
b)\(\frac{3}{2x+6}\) - \(\frac{x-6}{2x^2+6x}\)= \(\frac{3x}{2x\left(x+3\right)}\)- \(\frac{x-6}{2x\left(x+3\right)}\)= \(\frac{2x+6}{2x\left(x+3\right)}\)= \(\frac{2\left(x+3\right)}{2x\left(x+3\right)}\)= \(\frac{1}{x}\)
c)\(\frac{1}{xy-x^2}\)- \(\frac{1}{y^2-xy}\)= \(\frac{1}{x\left(x-y\right)}\)- \(\frac{1}{-y\left(x-y\right)}\)= \(\frac{y}{xy\left(x-y\right)}\)- \(\frac{-x}{xy\left(x-y\right)}\)= \(\frac{y+x}{xy\left(x-y\right)}\)
nhớ tick nhé
\(a,35x^2y-14xy+21xy^2=7xy\left(5x+3y-2\right)\)
\(b,x^3-4x^2+4x=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)
\(c,x^2-7x+xy-7y=x\left(x-7\right)+y\left(x-7\right)=\left(x-7\right)\left(x+y\right)\)
\(d,x^2-y^2-10x+25=\left(x-5\right)^2-y^2=\left(x-y-5\right)\left(x+y-5\right)\)
\(e,x^3y+2x^2y^2-xyz^2+xy^3=xy\left(x^2+2xy+y^2-z^2\right)\)
\(=xy\left[\left(x+y\right)^2-z^2\right]=xy\left(x+y-z\right)\left(x+y+z\right)\)
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1.X2-2X-4y2-4y
=x2-2x+1-(4y2+4y+1)
=(x+1)2-(2y+1)2
=>(x+1-2y-1)(x+1+2y+1)
=(x-2y)(x+2y+2)
2.x4+2x3-4x-4
=(x2)2-22+2x3-4x
=(x2-2)(x2+2)+2x(x2-2)
=(x2-2)(x2+2+2x)