tinh gia tri cua bieu thuc a+b*c,voi a = 213, b= 205 ,c = 152
cho bieu thuc A=[x+2/x^2-x+x-2/x^2+x].x^2-1/x^2+2
a) tim dieu kien cua x de gia tri cua bieu thuc A duoc xac dinh
b) tinh gia tri cua bieu thuc A voi x = -200
a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)
\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)
b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
cho bieu thuc ;A =500 +X va B = X - 500 , voi X = 8075
A) tinh gia tri cua A va B
B) tinh gia tri cua A+ B
a,500+8075=8575
b,8075-500=7575
**** minh nhe !
bai 1 tinh gia tri cua bieu thuc 78 x m + 22 x m voi m =135
bai 2 tinh gia tri cua bieu thuc 78 x m + 42 x m -20 x m voi m =1035
bai 3 cho bieu thuc B = 119 x n - n x 9 bieu thuc B co gia tri bang 8470 khi n =............
CHO bieu thuc
B=17,58*43+57*17,58/293*A
a)tim gia tri cua bieu thuc Bkhi a=2
b)tinh gia tri a khi B=2
c)tim gia tri so tu nhien ad biet bieu thuc B co gia tri lon nhat ,gia tri lon do la bao nhieu
cho a R b =a*b/ a+b . Tinh gia tri bieu thuc cua : 13R7 - 12r4 . TINH gia tri cua bieu thuc tren ( 30 giay )
\(\frac{31}{20}\)đáp số đúng 1000000000000000000000000000000000000000000000000000% đó nha
h and kb nhé
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
Tinh gia. Tri cua bieu thuc sau
A,a+(+62) voi a=123
B,(-75) +b voi b =-325
C, |-326|+c voi c=-56
D,-135+d voi |d|=35
So nguyen. Nha
Tinh gia tri cua cac bieu thuc sau voi /a/=1,5;b=-0,75
M= a+2ab-b
\(\left|a\right|=1,5\Rightarrow\orbr{\begin{cases}a=1,5\\a=-1,5\end{cases}}\)
\(Th1:a=1,5;b=-0,75\)hay \(a=\frac{3}{2};b=\frac{-3}{4}\)
\(\Rightarrow M=\frac{3}{2}+2.\frac{3}{2}.\frac{-3}{4}-\frac{-3}{4}\)
\(=\frac{6}{4}+\frac{-9}{4}+\frac{3}{4}=0\)
\(Th2:a=-1,5;b=0,75\)hay \(a=\frac{-3}{2};b=\frac{-3}{4}\)
\(\Rightarrow M=\frac{-3}{2}+2.\frac{-3}{2}.\frac{-3}{4}-\frac{-3}{4}\)
\(=\frac{-6}{4}+\frac{9}{4}+\frac{3}{4}=\frac{3}{2}\)
Vì |a| = 1,5
\(\Rightarrow\orbr{\begin{cases}a=1,5\\a=-1,5\end{cases}}\)
Thay a;b vào M ta có :
\(\orbr{\begin{cases}M=1,5+2.\left(1,5\right).\left(-0,75\right)+0,75\\M=-1,5+2.\left(-1,5\right).\left(-0,75\right)+0,75\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=1,5+3.\frac{-3}{4}+0,75\\M=-1,5+2.\left(1,5\right).\left(0,75\right)+0,75\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=1,5-2,25+0,75\\M=-1,5+3.\frac{3}{4}+0,75\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=1,5-\left(2,25+0,75\right)\\M=-1,5+2,25+0,75\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=1,5-3\\M=-1,5+\left(2,25+0,75\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=-1,5\\M=-1,5+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}M=-1,5\\M=1,5\end{cases}}\)
Vậy M = -1,5 hoặc M = 1,5
tinh gia tri cua bieu thuc sau voi A=1,5 ;B=-0,75
N=A:2-2:B
N=1,5 : 2 - 2 : -0,75
= 3/2 X 1/2 - 2 x -4/3
= 3/4 x -8/3
=-2
Thay A= 1,5 ; B=-0,75 vào biểu thức N ta được
N=1,5/2-2/-0,75=0,75-2,75= (-2)