rút gọn :
a) a.(b+1)-b-1/ b.(a-1)+a-1
b) 2a+2ab-b-1 / 3b . ( 2a-1) + 6a - 3
Rút gọn biểu thức: M = 2a+2ab−b−1/3b(2a−1)+6a−3 (a,b∈Q;a≠12;b≠−1)
A. M=2a/3b
B. M=a/b
C. M=−1
D. M=1/3
`M=(2a+2ab-b-1)/(3b(2a-1)+6a-3)`
`=(2a-1+b(2a-1))/(3(2a-1)(b+1))`
`=((2a-1)(b+1))/(3(2a-1)(b+1))`
`=1/3`
`=>` CHọn D
rút gọn :
a) \(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}\)
b) \(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\)
a, \(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}=\frac{a\left(b+1\right)-\left(b+1\right)}{b\left(a-1\right)+\left(a-1\right)}=\frac{\left(b+1\right)\left(a-1\right)}{\left(b+1\right)\left(a-1\right)}=1\)
b, \(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}=\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}=\frac{\left(b+1\right)\left(2a-1\right)}{\left(2a-1\right)\left(b+1\right)3}=\frac{1}{3}\)
Rút gọn: \(S=\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\) \(\left(a,b\in Q;a\ne\frac{1}{2};b\ne1\right)\)
\(S=\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\\ =\frac{2a\left(b+1\right)-\left(b+1\right)}{3b\left(2a-1\right)+3\left(2a-1\right)}\\ =\frac{\left(2a-1\right)\left(b+1\right)}{3\left(b+1\right)\left(2a-1\right)}\\=\frac{1}{3}\)
rút gọn
a)\(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}\left(a,b\in Q;a\ne1;b\ne-1\right)\)
b)\(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\left(a,b\in Q,a\ne\frac{1}{2};b\ne-1\right)\)
các bạn giúp mình nha. Mình cảm ơn nhiều
1.Tìm GTNN của \(B=\frac{|x|+2020}{2019}\)
2.Rút gọn
a,\(\frac{a\left(b+1\right)-b-1}{b\left(a-1\right)+a-1}\)(a,b\(\in Q;a\ne1;b\ne-1)\)
b,\(\frac{2a+2ab-b-1}{3b\left(2a-1\right)+6a-3}\)\(\left(a,b\in Q;a\ne\frac{1}{2};b\ne-1\right)\)
bài 3: rút gọn
A= x.(y+1)-y-1/ y.(x-1)+x-1
B= 2a + 2ab - b-1/ 3b.(2a-1)+6a-3
bài 4: tìm GTLN của
A=2018/ |x|+2006
B=|x| + 2018/ -2005
Làm nhanh giúp tớ nhé!!! Tớ sắp phải đi hok
Câu 3:
\(A=\dfrac{\left(y+1\right)\left(x-1\right)}{\left(x-1\right)\left(y+1\right)}=1\)
\(B=\dfrac{2a\left(1+b\right)-\left(b+1\right)}{3a\left(2a-1\right)+3\left(2a-1\right)}=\dfrac{\left(b+1\right)\left(2a-1\right)}{3\left(a+1\right)\left(2a-1\right)}=\dfrac{b+1}{3a+3}\)
Câu 4:
\(\left|x\right|+2006>=2006\)
=>A<=1009/1003
Dấu '=' xảy ra khi x=0
\(\left|x\right|+2018>=2018\)
=>B>=-2018/2005
Dấu '=' xảy ra khi x=0
Cho biểu thức : A= ( -2a + 3b - 4c ) - ( -2a - 3b - 4c )
a) Rút gọn A b) Tính giá trị của A khi a=2012 ; b=-1 ; c=-2013
\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(=-2a+3b-4c+2a+3b+4c\)
\(=6b\)
b) Khi \(a=2012,b=-1,c=-2013\) ta có :
\(A=6b=6\cdot\left(-1\right)=-6\)
Vậy \(A=-6\) khi \(a=2012,b=-1,c=-2013\)
Giải:
a) \(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(A=-2a+3b-4c+2a+3b+4c\)
\(A=\left(-2a+2a\right)+\left(3b+3b\right)+\left(-4c+4c\right)\)
\(A=0+2.3b+0\)
\(A=6b\)
b) Ta thay: \(a=2012;b=-1;c=-2013\)
Ta có:
\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)
\(A=\left(-2.2012+-3.1--4.2013\right)-\left(-2.2012--3.1--4.2013\right)\)
\(A=\left(-2.2012-3.1+4.2013\right)-\left(-2.2012+3.1+4.2013\right)\)
\(A=-2.2012-3.1+4.2013+2.2012-3.1-4.2013\)
\(A=\left(-2.2012+2.2012\right)+\left(-3.1-3.1\right)+\left(4.2013-4.2013\right)\)
\(A=0+2.-3.1+0\)
\(A=-6\)
1,Cho biểu thức:A=(-2a+3b-4c)-(-2a-3b-4c)
a,Rút gọn A
b,Tính giá trị của A khi a=2012;b=-1;c=-2013
\(A=\left(\dfrac{1}{2a-b}-\dfrac{a^2-1}{2a^3-b+2a-a^2b}\right)\div\left(\dfrac{4a+2b}{a^3b+ab}-\dfrac{2}{a}\right)\)
a) rút gọn biểu thức A
b)tính giá trị biểu thức A biết 4a^2+b^2=5ab a>b>0