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a) 13 - 2x = x- 2 b)2x-15+8x=7-2x+14 c)12-4y+3y=4y-10-8y
<=>13 + 2 = x+2x <=>2x +2x+8x =7+14+15 <=>12+10 =4y-8y+4y-3y
<=> 15 =3x <=>12x =36 <=> 22 =-3y
<=> x=5 <=>x=3 <=>y=-22/3
vậy S=[5] vậy S=[3] vậy S=[-22/3]
a) Ta có: -7x+13>-7y+13
\(\Leftrightarrow-7x>-7y\)
hay x<y
b) Ta có: 11x-1>11y+1
mà 11x+1>11x-1
nên 11x+1>11y+1
\(\Leftrightarrow11x>11y\)
hay x>y
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)\(=\left(x-y\right)x^2-2x+y-x^3+x^2y\)
\(=x^3-x^2y-2x+y-x^3+x^2y\)
\(=-2x+y\)
x = 9 => 10 = x + 1 thay vào F ta có
F = \(x^{14}-\left(x+1\right)x^{13}+...+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
F = \(x^{14}-x^{14}+x^{13}+...+x^3+x^2-x^2-x+x+1\)
=>F = 1
Ta có: x = 9 => x - 9 = 0
\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-9x^{13}-x^{13}+9x^{12}+x^{12}-9x^{11}+...-x^3+9x^2+x^2-9x-x+9+1\)
\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+...-x^2\left(x-9\right)+x\left(x-9\right)-\left(x-9\right)+1\)
\(=0+1=1\)
\(A=6xy\left(xy-y^2\right)-8x^2.\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(A=6x^2y^2-6xy^3-8x^3+8x^2y^2+5y^2x^2-5xy^3\)
\(A=19x^2y^2-11xy^3-8x^3\)
Tại x=1/2, y=2
\(A=19.\frac{1}{4}.2^2-11.\frac{1}{2}.2^3-8\left(\frac{1}{2}\right)^3=19-44-1=-26\)
Bài 1:
a) Ta có: 22x-13=x-6
\(\Leftrightarrow22x-13-x+6=0\)
\(\Leftrightarrow21x-7=0\)
\(\Leftrightarrow21x=7\)
hay \(x=\frac{1}{3}\)
Vậy: \(x=\frac{1}{3}\)
b) Ta có: (x-7)(2x+10)=0
\(\Leftrightarrow\left(x-7\right)\cdot2\cdot\left(x+5\right)=0\)
mà \(2\ne0\)
nên \(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;7\right\}\)
c) ĐKXĐ: \(x\ne14\)
Ta có: \(\frac{12x+9}{x-14}=7\)
\(\Leftrightarrow12x+9=7\left(x-14\right)\)
\(\Leftrightarrow12x+9=7x-98\)
\(\Leftrightarrow12x+9-7x+98=0\)
\(\Leftrightarrow5x+107=0\)
\(\Leftrightarrow5x=-107\)
hay \(x=\frac{-107}{5}\)(tm)
Vậy: \(x=\frac{-107}{5}\)
d) Ta có: \(\frac{x+2}{4}+\frac{3x-4}{6}=\frac{x-14}{24}\)
\(\Leftrightarrow\frac{6\left(x+2\right)}{24}+\frac{4\left(3x-4\right)}{24}=\frac{x-14}{24}\)
Suy ra: \(6\left(x+2\right)+4\left(3x-4\right)=x-14\)
\(\Leftrightarrow6x+12+12x-16-x+14=0\)
\(\Leftrightarrow17x+10=0\)
\(\Leftrightarrow17x=-10\)
hay \(x=\frac{-10}{17}\)
Vậy: \(x=\frac{-10}{17}\)