tim gia tri :(1-1/1+2).(1-1/1+2+3).(1-1/1/2/3/4)....(1-1/1+2+3+...+2006)
tinh gia tri bieu thuc: A=(1-1/(1+2))(1-1/(1+2+3))...(1-1/(1+2+3+...+2006))
Ta có
(1-1/(1+2))=(0/(1+2))=0
(1-1/(1+2+3))=(0/(1+2+3))=0
..........................................
.(1-1/(1+2+3+...+2006))=.(0/(1+2+3+...+2006))=0
=>0.0.0.0.....0=0
tim x de :
1) -2 / 3x - 1 co gia tri duong
2) 3 - 1/2x / 2x^2 + 11 co gia tri am
3) 2/3 - 2x / -x^2 - 1 co gia tri khong am
C=(1+1/1*3)*(1+1/2*4)*(1+1/3*5)...(1+1/2014*2016).tim gia tri cua c
Tinh gia tri bieu thuc:
A,3/1.6+3/6.11+...+3/96.101. B,1+1/2+1/2^2+1/2^3+....+1/2^2006. C,-1/3+1/3^2+1/3^3+...+1/3^100-1/3^101. D,12-12/7-12/289-12/85 phần 4-4/7-4/289-4/85:3+3/13+3/169+3/91 phần 7+7/13+7/169+7/91
a) tim gia tri nho nhat
A= (2x+1/3)^4 -1
b) tim gia tri lon nhat
B= - (4/9x-2/15)^6 +3
\(A=\left(2x+\frac{1}{3}\right)^4-1\) . Có: \(\left(2x+\frac{1}{3}\right)\ge0\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\)
Dấu = xảy ra khi: \(2x+\frac{1}{3}=0\)
\(\Rightarrow2x=-\frac{1}{3}\)
\(\Rightarrow x=-\frac{1}{3}:2=-\frac{1}{6}\)
Vậy: \(Min_A=-1\) tại \(x=-\frac{1}{6}\)
a) Tinh gia tri bieu thuc:
3/5 : 4/5 + 1/2 x 2/3
b) Tim x:
5/4 nhan x = 3/8 + 1/4
A
\(\frac{3}{5}\div\frac{4}{5}+\frac{1}{2}\times\frac{2}{3}\)
\(=\frac{3}{4}+\frac{1}{3}\)
\(=\frac{13}{12}\)
B
\(\frac{5}{4}\times x=\frac{3}{8}+\frac{1}{4}\)
\(\frac{5}{4}\times x =\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{5}{4}\)
\(x=\frac{4}{8}=\frac{1}{2}\)
1.tim gia tri cac bieu thuc sau:
a) 4 va 3/5 - 1 va 2/7 - 1 va 1/7
b) 27/12 - 5/4 - 1/2
\(4\frac{3}{5}-1\frac{2}{7}-1\frac{1}{7}=\frac{23}{5}-\frac{1}{7}=\frac{156}{35}\)
\(\frac{27}{12}-\frac{5}{4}-\frac{1}{2}=\frac{27}{12}-\frac{15}{12}-\frac{6}{12}=\frac{6}{12}=\frac{1}{2}\)
1) Tim gia tri nho nhat cua cac bieu thuc sau
a) (2x+1)^4 -1
b) (x^2-16)^2 +/y-3/ -2
NT:(2x+1)^4>=0.Dấu ''='' xảy ra khi x=-1/2
=>(2x+1)^4-1>=-1.Dấu"=" xẩy ra khi x=-1/2
Vậy Min của biểu thức trên là -1
1) tim gia tri nho nhat cua cac bieu thuc sau
a) (2x+1)^4-1
b) (x^2-16)^2+/y-3/-2
a: \(\left(2x+1\right)^4-1\ge-1\)
Dấu '=' xảy ra khi x=-1/2
b: \(\left(x^2-16\right)^2+\left|y-3\right|-2\ge-2\)
Dấu '=' xảy ra khi \(\left(x,y\right)\in\left\{\left(4;3\right);\left(-4;3\right)\right\}\)