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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, Vì x2 ≥ 0 , 2y2 ≥ 0 với mọi x,y
=>x2+2y2+ 1 ≥ 1
=>Phân thức trên luôn có nghĩa
a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)
b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)
c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)
d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)
\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)
e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)
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 :
\(\dfrac{yz\left(y-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{-xz\left(x-z\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)+\(\dfrac{xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
= \(\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
\(D=5x^2-10x-2\)
\(=5\left(x^2-2x+1\right)-7\)
\(=5\left(x-1\right)^2-7\ge-7\)
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+\(\dfrac{1}{4}\)
x11+x4+1
= x11+x10+x9-x10-x9-x8+x8+x7+x6-x7-x6-x5+x5+x4+x3-x3-x2-x+x2+x+1
= x9(x2+x+1)-x8(x2+x+1)+x6(x2+x+1)-x5(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
= (x2+x+1)(x9-x8+x6-x5+x3-x+1)
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 \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
hepl me Toshiro KiyoshiTrần Đăng NhấtHồng Phúc NguyễnT.Thùy Ninh
Nguyễn Huy TúAkai HarumaXuân Tuấn Trịnh