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Ta có :
\(VT=\left(\dfrac{1}{2}xy-\dfrac{1}{3}y\right)\left(\dfrac{1}{4}x^2y^2+\dfrac{1}{6}xy^2+\dfrac{1}{9}y^2\right)\)
\(=\dfrac{1}{8}x^3y^3+\dfrac{1}{12}x^2y^3+\dfrac{1}{18}xy^3-\dfrac{1}{12}x^2y^3-\dfrac{1}{18}xy^3-\dfrac{1}{27}y^3\)
\(=\dfrac{1}{8}x^3y^3-\dfrac{1}{27}y^3=VT\)
\(\Rightarrow dpcm\)
Vậy : ..............
a)Ta có : \(\dfrac{x+1}{1-x}\)( giữ nguyên )
\(\dfrac{x^2-2}{1-x}\)( giữ nguyên )
\(\dfrac{2x^2-x}{x-1}=\dfrac{x-2x^2}{1-x}\)
b)Ta có : \(\dfrac{1}{x-1}=\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x+1}{x^3-1}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2-2x}{x^3-1}\)
\(\dfrac{2x-3x^2}{x^3-1}\)(giữ nguyên )
c) MTC = ( x+ 2)2(x - 2)2
Do đó , ta có : \(\dfrac{1}{x^2+4x+4}=\dfrac{1}{\left(x+2\right)^2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)^2\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}\)
\(\dfrac{x}{x^2-4}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x^2-2^2\right)}{\left(x+2\right)^2\left(x-2\right)^2}=\dfrac{x^3-4x}{\left(x+2\right)^2\left(x-2\right)^2}\)
d) MTC = xyz( x - y)( y - z)( x - z)
Do đó , ta có : \(\dfrac{1}{x\left(x-y\right)\left(x-z\right)}=\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{y\left(y-x\right)\left(y-z\right)}=\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(\dfrac{1}{z\left(z-x\right)\left(z-y\right)}=\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
Cộng các phân thức lại ta có :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
a )
Để A \(⋮\) B thì \(x^n\ge x^3\) \(\Rightarrow n\ge3\)
Để M \(⋮\) N thì \(y^n\ge y^2\Rightarrow n\ge2\)
a, A= 5\(x^ny^3\)
B= 4\(x^3y\)
=> A\(⋮\)B -> n \(\ge\)3
b, làm tương tự như trên
c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
a, \(A=-x^2+2x+2\)
\(=-\left(x^2-2x-2\right)=-\left(x^2-2x+1-3\right)\)
\(=-\left(x-1\right)^2+3\le3\)
Dấu " = " khi \(-\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_A=3\) khi x = 1
b, \(B=-x^2-8x+17\)
\(=-\left(x^2+8x-17\right)\)
\(=-\left(x^2+8x+16-33\right)\)
\(=-\left(x+4\right)^2+33\le33\)
Dấu " = " khi \(-\left(x+4\right)^4=0\Leftrightarrow x=-4\)
Vậy \(MAX_B=33\) khi x = -4
c, \(C=-x^2+7x+15\)
\(=-\left(x^2-\dfrac{7}{2}x.2+\dfrac{49}{4}-\dfrac{109}{4}\right)\)
\(=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{109}{4}\le\dfrac{109}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{7}{2}\right)^2=0\Leftrightarrow x=\dfrac{7}{2}\)
Vậy \(MAX_C=\dfrac{109}{4}\) khi \(x=\dfrac{7}{2}\)
d, \(D=-x^2-5x+11\)
\(=-\left(x^2+\dfrac{5}{2}.x.2+\dfrac{25}{4}-\dfrac{69}{4}\right)\)
\(=-\left(x+\dfrac{5}{2}\right)^2+\dfrac{69}{4}\le\dfrac{69}{4}\)
Dấu " = " khi \(-\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)
Vậy \(MAX_D=\dfrac{69}{4}\) khi \(x=\dfrac{-5}{2}\)
f, sai đề à?
g, \(G=-x^2-x-y^2-3y+13\)
\(=-\left(x^2+x+y^2+3y-13\right)\)
\(=-\left(x^2+\dfrac{1}{2}x.2.+\dfrac{1}{4}+y^2+\dfrac{3}{2}.x.2+\dfrac{9}{4}-15,5\right)\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\left(y+\dfrac{3}{2}\right)^2+15,5\le15,5\)
Dấu " = " khi \(\left\{{}\begin{matrix}-\left(x+\dfrac{1}{2}\right)^2=0\\-\left(y+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(MAX_G=15,5\) khi \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
a) x = 1
b) x = 6; x = -3
c) x = 5,5; x = 1,5
d) x = 1; x = -1
e) x = -2; x = -1,000000371....
a, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
\(D=5x^2-10x-2\)
\(=5\left(x^2-2x+1\right)-7\)
\(=5\left(x-1\right)^2-7\ge-7\)
Vậy \(min_D=-7\)
Để D = -7 thì \(x-1=0\Rightarrow x=1\)
\(E=x^2-2xy+2y^2+y-3\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)-\dfrac{13}{4}\)
\(=\left(x-y\right)^2+\left(y+\dfrac{1}{2}\right)^2-\dfrac{13}{4}\ge\dfrac{13}{4}\)
Vậy \(min_E=\dfrac{-13}{4}\)
Để \(E=-\dfrac{13}{4}\) thì \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y=-\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
a)(x-1)(x+1)(x+2)
=(x2-1)(x+2)
=x3-x+2x2-2
b)\(\dfrac{1}{2}\)x2y(2x+y)(2x-y)
=\(\dfrac{1}{2}\)x2y(4x2-y2)
=2x4y-\(\dfrac{1}{2}\)x2y3
c)(x-\(\dfrac{1}{2}\))(x+\(\dfrac{1}{2}\))(4x-1)
=(x2-\(\dfrac{1}{4}\))(4x-1)
=4x3-x2-x+\(\dfrac{1}{4}\)
a: \(=-x^2y\cdot x+x^2y\cdot y=x^2y\left(-x+y\right)\)
b: \(=-xy^2\cdot x^2-xy^2\cdot z=-xy^2\left(x^2+z\right)\)
c: x^2y^3-xy^2
=xy^2*xy-xy^2
=xy^2(xy-1)
d: -x^3z-z
=z(-x^3-1)
=-z(x+1)(x^2-x+1)
e: =x(x-y)+(x-y)
=(x-y)(x+1)
n: =x^2(x-1)-(x-1)
=(x-1)(x^2-1)
=(x-1)^2(x+1)