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a) \(\left(-12x^{13}y^{15}+6x^{10}y^{14}\right):\left(-3x^{10}y^{14}\right)\)
\(=-12x^{13}y^{15}:-3x^{10}y^{14}+6x^{10}y^{14}:-3x^{10}y^{14}\)
\(=4x^3y-2\)
b) \(\left(x-y\right)\left(x^2-2x+y\right)-x^3+x^2y\)
\(=x^3-2x^2+xy-x^2y+2xy-y^2-x^3+x^2y\)
\(=-2x^2+3xy-y^2\)
a) \(-12x^{13}\)\(y^{15}\)+\(6x^{10}\)\(y^{14}\):\(-3x^{10}\)\(y^{14}\)
=\(-12x\)\(^{13}\)\(y^{15}\)\(:\)\(-3x^{10}y^{14}\)\(+6x^{10}y^{14}:-3x^{10}y^{14}\)
\(=4x^3y-2\)
b)\(=\left(x-y\right)x^2-2x+y-x^3+x^2y\)
\(=x^3-x^2y-2x+y-x^3+x^2y\)
\(=-2x+y\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
a) Chỗ sai trong phương trình là: \(5 - x + 8 = 3x + 3x - 27\) (dòng thứ 2) vì khi phá ngoặc đã không đổi dấu của số 8.
Sửa lại:
\(\begin{array}{l}5 - \left( {x + 8} \right) = 3x + 3\left( {x - 9} \right)\\\,\,\,\,5 - x - 8 = 3x + 3x - 27\\\,\,\,\,\,\,\, - 3 - x = 6x - 27\\\,\,\,\, - x - 6x = - 27 + 3\\\,\,\,\,\,\,\,\,\,\,\,\, - 7x = - 24\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \left( { - 24} \right):\left( { - 7} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{24}}{7}\end{array}\)
Vậy phương trình có nghiệm \(x = \frac{{24}}{7}.\)
b) Chỗ sai trong phương trình là: \(4x + 5x = 9 - 18\) (dòng thứ 3) vì khi chuyển \( - 18\) từ vế trái sang vế phải đã không đổi dấu thành \( + 18\).
Sửa lại:
\(\begin{array}{l}3x - 18 + x = 12 - \left( {5x + 3} \right)\\\,\,\,\,\,\,\,4x - 18 = 12 - 5x - 3\\\,\,\,\,\,\,\,4x + 5x = 9 + 18\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x = 27\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 27:9\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3.\end{array}\)
Vậy phương trình có nghiệm \(x = 3.\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
Đặt \(A=x^{20}+x^{10}+1\)
\(x^{50}+x^{10}+1\)
\(=x^{50}-x^{20}+A\)
\(=x^{20}\left(x^{30}-1\right)+A\)
\(=x^{20}\left(x^{10}-1\right)A+A\)
\(=\left(x^{30}-x^{20}+1\right)A\)
mà \(\left(x^{30}-x^{20}+1\right)A⋮A\)
\(\Rightarrow\left(x^{50}+x^{10}+1\right)⋮\left(x^{20}+x^{10}+1\right)\)