giá trị biểu thức
C = 1 / 1 x 3 + 1/ 3 x 5 + 1/ 5 x 7 + ... + 1/ 99 x 101
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A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
Sửa đề:
A=/x+5/+10
Ta có: /x+5/>= 0 với mọi x>=0
=> A=/x+5/+10 >= 10
=> Amin=10. Dấu "=" xảy ra <=> x+5=0<=> x=-5
Vậy...
\(\text{a) }A=\left|x+5\right|+10\)
\(\text{Vì }\left|x+5\right|\ge0\forall x\)
\(\Rightarrow A=\left|x+5\right|+10\ge10\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left|x+5\right|=0\)
\(\Rightarrow x=-5\)
\(\text{Vậy Min}_A=10\Leftrightarrow x=-5\)
\(\text{b) }\left|3-x\right|+5\)
\(\text{Vì }\left|3-x\right|\ge0\forall x\)
\(\Rightarrow\left|3-x\right|+5\ge5\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left|3-x\right|=0\)
\(\Rightarrow x=3\)
\(\text{Vậy Min}_B=5\Leftrightarrow x=3\)
\(\text{d) }D=\left(x+2\right)^2+15\)
\(\text{Vì ( x + 2 )}^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+15\ge15\)
\(\text{Dấu ''='' xảy ra khi :}\)
\(\left(x+2\right)^2=0\)
\(\Rightarrow x+2=0\)
\(\Rightarrow x=-2\)
1/2+1/6+1/12+1/20=4/5
Tính nhanh
18,5 : x + 14,8 : x + 12 ,7 : x = 10
x:(18,5+14,8+12,7)=10
x:46=10
=>x=460
Ta cố bdt \(|a|+|b|\ge|a+b|\), dễ dàng chứng mình bằng bình phương 2 vế. Dấu = sảy ra <=>IaI.IbI=a.b <=> a.b>=0
áp dụng vào từng câu
a)A=Ix+1I+Ix+2I+Ix+3I+I-x-4I+I-x-5I ( vì Ix+4I=I-x=4I, Ix+5I=I-x-5I
A>=I(x+1)+(-x-5)I+I(x+2)+(-x-4)I +Ix+3I=4+2+Ix+3I=6+Ix+3I>=6
Dấu bằng khi (x+1)(-x-5)>=0;(x+2)(-x-4)>=0;Ix+3I=0 =>x=-3
b) LÀm tương tự MinB=18
Dấu = khi (2x+1)(-2x-11)>=0;(2x+3)(-2x-9)>=0;(2x+5)(-2x-7)>=0 <=>-7/2<=x<=-5/2
1) 201,5 - 36,4 : 2,5 x 0,9
= 201,5 - 14,56 x 0,9
= 201,5 - 13,104
= 188,396
2)
a) X - 31,27 = 101
X = 101 + 31,27
X = 132,27
b) X + 1031 = 14,5
X = 14,5 - 1031
X = -1016,5
c) 87,5 x X = 4725
X = 4725 : 87,5
X = 54
d) 186 : X = 3720
X = 186 : 3720
X = 1/20
3)
a) 1,47 x 3,6 + 1,47 x 6,4
= 1,47 x (3,6 + 6,4)
= 1,47 x 10
= 14,7
b) 25,8 x 1,02 - 25,8 x 1,01
= 25,8 x ( 1,02 - 1,01)
= 25,8 x 0,01
=0,258
1) 188,396
2) Tìm X
a) X = 132,27
b) X = - 1016,5
c) X = 54
d) X = 20
3. Tính thuận tiện nhất
a) 1,47 x 3,6 + 1,47 x 6,4 = 1,47 x (3,6 + 6,4)
= 1,47 x 10
= 14,7
b) 25,8 x 1,02 - 25,8 x 1,01 = 25,8 x (1,02 - 1,01)
= 25,8 x 0,01
= 0,258
Nha bạn
Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
Vì \(\left|2x+1\right|\ge0;\left|x+y-\frac{1}{2}\right|\ge0\)
Mà \(\left|2x+1\right|+\left|x+y-\frac{1}{2}\right|\le0\Rightarrow\orbr{\begin{cases}2x+1=0\\x+y-\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{4}\end{cases}}\)(1)
Thế (1) vào A
\(\Rightarrow A=4.\left(-\frac{1}{2}\right)^3.\left(\frac{1}{4}\right)^2-\frac{1}{4}.\left(-\frac{1}{2}\right)+2.\frac{1}{4}-5\)
\(\Rightarrow A=-\frac{1}{2}+\frac{1}{8}+\frac{1}{2}-5\)
\(\Leftrightarrow A=\frac{1}{8}-5=\frac{1}{8}-\frac{40}{8}=-\frac{39}{8}\)
a: =11/2*4*5/3
=22*5/3
=110/3
b: =30/12-3/12+20/12
=47/12
c: =28/15+5
=28/15+75/15
=103/15
\(C=\dfrac{1}{1 \times 3}+\dfrac{1}{3 \times 5}+\dfrac{1}{5 \times 7}+...+\dfrac{1}{99 times 101}\)
\(C=\dfrac{1}{2} \times (\dfrac{2}{1 \times 3}+\dfrac{2}{3 \times 5}+\dfrac{2}{5 \times 7}+...+\dfrac{2}{99 times 101})\)
\(C=\dfrac{1}{2} \times (1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101})\)
\(C=\dfrac{1}{2} \times (1-\dfrac{1}{101})\)
\(C=\dfrac{1}{2} \times \dfrac{100}{101}\)
\(C=\dfrac{50}{101}\)
\(C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+.....+\dfrac{1}{99.101}\)
`=>` \(2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+......+\dfrac{2}{99.101}\)
\(2C=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.......+\dfrac{1}{99}-\dfrac{1}{101}\)
`2C = 1-1/101`
`2C=101/101 - 1/101`
`2C = 100/101`
`2C=100/101 +1/101`
`C=101/101 xx 2`
`C=(101xx2)/101`
`C=201101`