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a) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2+2\left(x-1\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
b) \(\dfrac{9}{x^3-9x}-\dfrac{-1}{x+3}\)
\(=\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\)
\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\)
c) \(\dfrac{x^3-8}{5x+10}.\dfrac{x^2+4x}{x^2+2x+4}\)
\(=\dfrac{x\left(x-2\right)\left(x^2+2x+4\right)\left(x+4\right)}{5\left(x+2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{x\left(x-2\right)\left(x+4\right)}{5\left(x+2\right)}\)
d) \(\dfrac{5x+10}{4x-8}.\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}.\dfrac{2\left(2-x\right)}{x+2}\)
\(=-\dfrac{10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}\)
\(=-\dfrac{5}{2}\)
e) \(\dfrac{\left(x-13\right)^2}{2x^5}.\dfrac{-3x^2}{x-13}\)
\(=\dfrac{x-13}{2x^3}.\dfrac{-3}{1}\)
\(=\dfrac{-3\left(x-13\right)}{2x^3}\)
g) \(\dfrac{x^2+6x+9}{1-x}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2}{x-1}.\dfrac{\left(x-1\right)^2}{2\left(x+3\right)^2}\)
\(=-\dfrac{\left(x+3\right)^2\left(x-1\right)^2}{2\left(x-1\right)\left(x+3\right)^2}\)
\(=-\dfrac{x-1}{2}\).
\(A=-x^2+6x-10=-\left(x^2-6x+9\right)-1=-\left(x-3\right)^2-1\le-1\)
Vậy GTLN của A là -1 khi x = 3
\(B=-2x^2-4x-10=-2\left(x^2+2x+1\right)-8=-2\left(x+1\right)^2-8\le-8\)
Vậy GTLN của B là -8 khi x = -1
\(C=-2x^2+3x-10=-2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-\frac{71}{8}=-2\left(x-\frac{3}{4}\right)^2-\frac{71}{8}\le-\frac{71}{8}\)
Vậy GTLN của C là \(-\frac{71}{8}\)khi x = \(\frac{3}{4}\)
\(D=-x^2-y^2+2x-4y-10\)
\(D=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)-5\)
\(D=-\left(x-1\right)^2-\left(y+2\right)^2-5\le-5\)
Vậy GTLN của D là -5 khi x = 1; y = -2
\(A=-3x^2+6x-4\)
\(A=-\left(3x^2-6x+4\right)\)
\(A=-3\left(x^2-2x+4\right)\)
\(A=-3\left(x^2-2x+1+3\right)\)
\(A=-3\left(x-1\right)^2-9\)
Vì \(-3\left(x-1\right)^2\le0\) với mọi x
\(\Rightarrow-3\left(x-1\right)^2-9\le-9\)
\(\Rightarrow Amin=-9\Leftrightarrow x=1\)
\(B=-x^2-4x-y^2+2y\)
\(B=-x^2-4x-2-y^2+2y-1+3\)
\(B=-\left(x^2+4x+2\right)-\left(y^2-2y+1\right)+3\)
\(B=-\left(x+2\right)^2-\left(y-1\right)^2+3\)
Vì \(-\left(x+2\right)^2\le0\) với mọi x
\(-\left(y-1\right)^2\le0\) với mọi y
\(\Rightarrow-\left(x+2\right)^2-\left(y-1\right)^2+3\le3\) với mọi x,y
\(\Rightarrow Bmin=3\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)
Sửa đề \(C=-x^2-8x-y^2+2y\)
\(C=-x^2-8x-16-y^2+2y-1+17\)
\(C=-\left(x^2+8x+16\right)-\left(y^2-2y+1\right)+17\)
\(C=-\left(x+4\right)^2-\left(y-1\right)^2+17\)
Vì \(-\left(x+4\right)^2-\left(y-1\right)^2\le0\) với mọi x,y
\(\Rightarrow-\left(x+4\right)^2-\left(y-1\right)^2+17\le17\)
\(\Rightarrow Cmin=17\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)
\(D=\left(x^2+2\right)^2-2\left(x^2-2\right)\left(x^2+2\right)-10\)
\(D=\left(x^2+2\right)^2-2\left(x^2-2\right)\left(x^2+2\right)+\left(x^2-2\right)^2-\left(x^2-2\right)^2-10\)
\(D=\left(x^2+2-x^2-2\right)^2-\left(x^2-2\right)^2-10\)
\(D=-\left(x^2-2\right)^2-10\)
Vì \(-\left(x^2-2\right)^2\le0\)
\(\Rightarrow-\left(x^2-2\right)^2-10\le-10\)
\(\Rightarrow Dmin=-10\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
Bài 1
a) (x5 + 4x3 - 6x2) : 4x2
= 4x2(\(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)) : 4x2
= \(\dfrac{1}{4}\)x3 + x - \(\dfrac{3}{2}\)
b) (x3 - 8) : (x2 + 2x + 4)
= (x - 2)(x2 + 2x + 4) : (x2 + 2x + 4)
= x - 2
c) (3x2 - 6x) : (2 - x)
= -(6x - 3x2) : (2 - x)
= -3x(2 - x) : (2 - x)
= -3x
d) (x3 + 2x2 - 2x - 1) : (x2 + 3x + 1)
= [(x3 - 1) + (2x2 - 2x)] : (x2 + 3x + 1)
= [(x - 1)(x2 + x + 1) + 2x(x - 1)] : (x2 + 3x + 1)
= (x - 1)(x2 + x + 1 + 2x) : (x2 + 3x + 1)
= (x - 1)(x2 + 3x + 1) : (x2 + 3x + 1)
= x - 1
Bài 2
a) (x - 4)2 - (x - 2)(x + 2) = 6
x2 - 8x + 16 - (x2 - 4) = 6
x2 - 8x + 16 - x2 + 4 = 6
-8x + 20 = 6
\(\Rightarrow\) -8x = - 14
\(\Rightarrow\) x = \(\dfrac{7}{4}\)
b) 9(x + 1)2 - (3x - 2)(3x + 2) = 10
9(x2 + 2x + 1) - (9x2 - 4) = 10
9x2 + 18x + 9 - 9x2 + 4 = 10
18x + 13 = 10
\(\Rightarrow\) 18x = -3
\(\Rightarrow\) x = \(\dfrac{-1}{6}\)
Nhớ tik mik nha 
không lần sau mik ko giúp đâu 
AK... có j ko hiểu thì bn cứ bình luận bên dưới 
a)<=>(x^2+x-3)(x^2+x-2)-12=(x-2)(x+3)(x^2+x+1)
TH1:=>x-2=0
=>x=2
TH2:x+3=0
=>x=-3
dựa vô bệt thức ta thấy
D<0=> phương trình ko có nghiệm thực
=>x=-3 hoặc 2
nhớ tick nhé
b)Ta có:\(B=\left(0,5x^2+x\right)^2-3\left|0,5x^2+x\right|\)
\(B=\left|0,5x^2+x\right|^2-3\left|0,5x^2+x\right|+\dfrac{9}{4}-\dfrac{9}{4}\)
\(B=\left(\left|0,5x^2+x\right|-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
"="<=>\(\left|0,5x^2+x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
g)Ta có:\(G=\left(x^2+x-6\right)\left(x^2+x+2\right)\)
Đặt \(x^2+x-2=t\)
\(\Rightarrow G=\left(t-4\right)\left(t+4\right)\)
\(G=t^2-16\ge-16\)
"="<=>\(x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
E=\(x^4-6x^3+9x^2+x^2-6x+9\)
\(=x^2\left(x^2-6x+9\right)+x^2-6x+9\\ =x^2\left(x-3\right)^2+\left(x-3\right)^2\ge0\forall x\\ E_{min}=0\Leftrightarrow x=3\)
\(a,A=-x^2+6x-10\)
\(=-x^2+6x-9-1\)
\(=-\left(x^2-6x+9\right)-1\)
\(=-\left(x-3\right)^2-1\)
Ta có: \(-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2-1\le-1\forall x\)
=> Max A =-1 tại \(-\left(x-3\right)^2=0\Rightarrow x=3\)
cn lại lm tg tự
=.= hok tốt!!
chắc bn nảy hỏi lun cả bài tâp về nhà quá, làm km 1 câu
a) = a+a+a + a +a +1 -a -a -a = a(a+a+1) +(a+a+1) - a(a+a+1)= (a+a+1)(a-a+1)
tự bn thêm mũ 4;3;2 vào được là bn làm dc cac câu sau
1/ \(A=3\left(x+1\right)^2-\left(x+3\right)^2\)
\(=3\left(x^2+2x+1\right)-\left(x^2+6x+9\right)\)
\(=3x^2+6x+3-x^2-6x-9\)
\(=2x^2-6\)
Vậy biểu thức A vẫn phụ thuộc vào biến -_-
2/ \(B=\left(x-2\right)^2-\left(x-4\right)x\)
\(=x^2-4x+4-x^2-4x\)
\(=4\)
Vậy biểu thức B không phụ thuộc vào biến (đpcm)
3/ \(C=3\left(x+2\right)^2-3\left(x^2-4x\right)\)
\(=3\left(x^2+4x+4\right)-3x^2+12x\)
\(=3x^2+12x+12-3x^2+12x\)
\(=24x+12\)
Vậy biểu thức C vẫn phụ thuộc vào biến -_-
4/ \(D=3x\left(x-2\right)\left(x+2\right)-x\left(3x+3\right)\)
\(=3x\left(x^2-4\right)-3x^2-3x\)
\(=3x^3-12x-3x^2-3x\)
\(=3x^3-3x^2-15x\)
Vậy biểu thức D vẫn phụ thuộc vào biến -_-
5/ \(E=x^2-\left(x+1\right)\left(x-1\right)+5\)
\(=x^2-\left(x^2-1\right)+5\)
\(=x^2-x^2+1+5\)
\(=6\)
Vậy biểu thức E không phụ thuộc vào biến.
\(A=19-3x^2-24x\)
\(\Leftrightarrow A=-3\left(x^2+8x-\dfrac{19}{3}\right)\)
\(\Leftrightarrow A=-3\left(x^2+2x.4+16-\dfrac{67}{3}\right)\)
\(\Leftrightarrow A=-3\left[\left(x+4\right)^2-\dfrac{67}{3}\right]\)
\(\Leftrightarrow A=-3\left(x+4\right)^2+67\le67\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-3\left(x+4\right)^2=0\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
Vậy Max A là : 67 \(\Leftrightarrow x=-4\)
\(B=-x^2+6x-23\)
\(\Leftrightarrow B=-\left(x^2-6x+9\right)-14\)
\(\Leftrightarrow B=-\left(x-3\right)^2-14\le-14\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy Max B là : \(-14\Leftrightarrow x=3\)
\(C=4\left(x-1\right)^2-9\left(x+2\right)^2\)
\(\Leftrightarrow C=4x^2-8x+4-9x^2-36x-36\)
\(\Leftrightarrow C=-5x^2-44x-32\)
\(\Leftrightarrow C=-5\left(x^2+\dfrac{44}{5}x+\dfrac{32}{5}\right)\)
\(\Leftrightarrow C=-5\left(x^2+2x.\dfrac{22}{5}+\dfrac{484}{25}\right)+64,8\)
\(\Leftrightarrow C=-5\left(x+\dfrac{22}{5}\right)^2+64,8\le64,8\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-5\left(x+\dfrac{22}{5}\right)^2=0\Leftrightarrow\left(x+\dfrac{22}{5}\right)^2=0\Leftrightarrow x+\dfrac{22}{5}=0\)
\(\Leftrightarrow x=-\dfrac{22}{5}\)
Vậy Max C là : 64 , 8 \(\Leftrightarrow x=-\dfrac{22}{5}\)
\(E=\left(x+2\right)^2-2x^2+8\)
\(\Leftrightarrow E=x^2+4x+4-2x^2+8\)
\(\Leftrightarrow E=-x^2+4x+12\)
\(\Leftrightarrow E=-\left(x^2-4x+4\right)+16\)
\(\Leftrightarrow E=-\left(x-2\right)^2+16\le16\forall x\)
Dấu " = " xảy ra
\(\Leftrightarrow-\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy Max E là : \(16\Leftrightarrow x=2\)