B1:
1+1=.....
2+1=.....
3+1=.....
4+1=.....
5+1=.....
B2:
x+1=2
x=
x=
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Bài 1 :
\(\left(3x-2\right)^2+\left(x-1\right)\left(2x+4\right)=9x^2-12x+4+2x^2+4x-2x-4=11x^2-10x=x\left(11x-10\right)\)
Bài 2 :
a) \(\left(2x+5\right)^2-4x\left(x+1\right)=3\)
=> \(4x^2+20x+25-4x^2-4x=3\)
\(\Rightarrow16x=-22\)
\(\Rightarrow x=-\frac{11}{8}\)
b) \(\left(2x-3\right)^2-\left(2x+1\right)\left(x+2\right)=0\)
\(\Rightarrow4x^2-12x+9-2x^2-4x-x-2=0\)
\(\Rightarrow2x^2-17x+7=0\)
..................
\(\frac{x^4+x^3+6x^2+5x+5}{x^2+x+1}=\frac{x^4+x^3+x^2+5x^2+5x+5}{x^2+x+1}=\frac{x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)}{\left(x^2+x+1\right)}=\frac{\left(x^2+x+1\right)\left(x^2+5\right)}{x^2+x+1}=x^2+5\)
\(\frac{x^4+x^3+2x^2+x+1}{x^2+x+1}=\frac{x^4+x^3+x^2+x^2+x+1}{x^2+x+1}=\frac{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x^2+x+1}=\frac{\left(x^2+x+1\right)\left(x^2+1\right)}{x^2+x+1}=x^2+1\)
Bài 1:
a: \(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}=x\)
=>\(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2}\)
=>\(x-\dfrac{1}{2}>=0\)
=>\(x>=\dfrac{1}{2}\)
b: \(\left|1-3x\right|+1=3x\)
=>\(\left|1-3x\right|=3x-1\)
=>\(1-3x< =0\)
=>3x-1>=0
=>3x>=1
=>\(x>=\dfrac{1}{3}\)
Bài 2:
a: \(C=\left|5-x\right|+x=\left|x-5\right|+x\)
TH1: x>=5
\(C=x-5+x=2x-5\)
TH2: x<5
C=5-x+x=5
b: D=|2x-1|-x
TH1: x>=1/2
\(D=2x-1-x=x-1\)
TH2: \(x< \dfrac{1}{2}\)
D=1-2x-x=1-3x
B1: 1+1=2, 2+1=3, 3+1=4, 4+1=5, 5+1=6
B2: x+1=2
X= 2-1
X= 1. HẾT.....
B1 :
1 + 1 = 2
2 + 1 = 3
3 + 1 = 4
4 + 1 = 5
5 + 1 = 6
B2 :
x + 1 = 2
x = 2 - 1
x = 1