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a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)
b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)
c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)
a) \(\dfrac{1}{x^3-8}=\dfrac{1}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{2}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\dfrac{3}{4-2x}=\dfrac{-3}{2\left(x-2\right)}=\dfrac{-3\left(x^2+2x+4\right)}{2\left(x-2\right)\left(x^2+2x+4\right)}\)
b) \(\dfrac{x}{x^2-1}=\dfrac{x}{\left(x+1\right)\left(x-1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2\left(x-1\right)}\)
\(\dfrac{1}{x^2+2x+1}=\dfrac{1}{\left(x+1\right)^2}=\dfrac{x-1}{\left(x+1\right)^2\left(x-1\right)}\)
c) \(\dfrac{1}{x+2}=\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{1}{x^2-4x+4}=\dfrac{1}{\left(x-2\right)^2}=\dfrac{x+2}{\left(x+2\right)\left(x-2\right)^2}\)
\(\dfrac{5}{2-x}=\dfrac{-5}{x-2}=\dfrac{-5\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)^2}\)
d) \(\dfrac{1}{3x+3y}=\dfrac{1}{3\left(x+y\right)}=\dfrac{\left(x-y\right)^2}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{2x}{x^2-y^2}=\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=\dfrac{6x\left(x-y\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
\(\dfrac{x^2-xy+y^2}{x^2-2xy+y^2}=\dfrac{x^2-xy+y^2}{\left(x-y\right)^2}=\dfrac{3\left(x^2-xy+y^2\right)\left(x+y\right)}{3\left(x+y\right)\left(x-y\right)^2}=\dfrac{3\left(x^3+y^3\right)}{3\left(x+y\right)\left(x-y\right)^2}\)
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
a/ pt đãcho tương đương với
6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16
<=>18x=18
=> x=1
b/ pt đã cho tương đương với
10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8
<=> -4x=5
<=.> x=-\(\frac{5}{4}\)
c/ pt đã cho tương đương với
21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0
<=>42x=41
<=> x= \(\frac{41}{42}\)
d/ pt đã cho tương đương với
( x\(^2\)+x )(x+6)-x\(^3\)=5x
<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x
<=> 8x\(^2\)+6x-5x=0
<=>8x\(^2\)+16x-10x-5x=0
<=> (x+2)2x-5(x+2)=0
<=> (x+2)(2x-5)=0
<=>x+2=0 hoặc 2x+5=0
=> x=-2 hoặc x= -\(\frac{5}{2}\)
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
\(a,=x+x^2-x^3+x^4-x^5+1+x-x^2+x^3-x^4-x-x^2+x^3-x^4+x^5+1+x-x^2+x^3-x^4\\ =2x-2x^2+2x^3-2x^4\)
a: |3x-1|>5
=>\(\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x>6\\3x< -4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
b: \(\left|x^3+1\right|>=x+1\)
=>\(\left|x^3+1\right|-x-1>=0\)(1)
TH1: x>=-1
BPT (1) sẽ tương đương với: \(x^3+1-x-1>=0\)
=>\(x^3-x>=0\)
=>\(x\left(x^2-1\right)>=0\)
=>\(x\left(x-1\right)\left(x+1\right)>=0\)
=>x(x-1)>=0
=>\(\left[{}\begin{matrix}x>=1\\x< =0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x>=1\\-1< =x< =0\end{matrix}\right.\)
TH2: x<-1
BPT (1) sẽ tương đương với:
\(-x^3-1-x-1>=0\)
=>\(x^3+x+2< =0\)
=>\(x^3+x^2-x^2-x+2x+2< =0\)
=>\(\left(x+1\right)\left(x^2-x+2\right)< =0\)
=>x+1<=0
=>x<=-1
c: \(\left|x+1\right|>\left|x-2\right|\)
=>|x+1|-|x-2|>0(2)
TH1: x<-1
BPT (2) sẽ tương đương:
-x-1-(2-x)>0
=>-x-1-2+x>0
=>-3>0(vô lý)
=>\(x\in\varnothing\)
TH2: -1<=x<2
BPT (2) sẽ tương đương:
\(x+1-\left(2-x\right)>0\)
=>x+1-2+x>0
=>2x-1>0
=>2x>1
=>\(x>\dfrac{1}{2}\)
=>\(\dfrac{1}{2}< x< 2\)
TH3: x>=2
BPT (2) sẽ tương đương:
\(x+1-\left(x-2\right)>0\)
=>x+1-x+2>0
=>3>0(luôn đúng)
=>x>=2
d: |x-1|>|x+2|-3
=>|x-1|-|x+2|+3>0(3)
TH1: x<-2
BPT (3) sẽ tương đương:
1-x-(-x-2)+3>0
=>4-x+x+2>0
=>6>0(luôn đúng)
=>x<-2
TH2: -2<=x<1
BPT (3) sẽ tương đương:
\(1-x-\left(x+2\right)+3>0\)
=>\(4-x-x-2>0\)
=>-2x+2>0
=>-2x>-2
=>x<1
=>-2<=x<1
TH3: x>=1
BPT (3) sẽ tương đương:
\(x-1-x-2+3>0\)
=>0>0(sai)
=>\(x\in\varnothing\)
e: |x-1|+|x-5|>8(4)
TH1: x<1
BPT (4) sẽ tương đương:
1-x+5-x>8
=>6-2x>8
=>-2x>-2
=>x<1
TH2: 1<=x<5
BPT (4) sẽ tương đương:
x-1+5-x>8
=>4>8(vô lý)
=>\(x\in\varnothing\)
TH3: x>=5
BPT (4) sẽ tương đương:
\(x-1+x-5>8\)
=>2x>8+6=14
=>x>7
f: |x-3|+|x+1|<8(5)
TH1: x<-1
BPT (5) sẽ tương đương:
3-x-x-1<8
=>-2x<6
=>x>-3
=>-3<x<-1
TH2: -1<=x<3
BPT (5) sẽ tương đương:
x+1+3-x<8
=>4<8(luôn đúng)
=>-1<=x<3
TH3: x>=3
BPT (5) sẽ tương đương:
x-3+x+1<8
=>2x-2<8
=>2x<10
=>x<5
=>3<=x<5