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TH
0
16 tháng 6 2015
X x X x X x X x X = X5 = 210.105 = 405
=> X = 40
=> Y = 40 - 30 = 10
=> Z = 20 : 10 = 2
HT
3 tháng 8 2021
4, Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)
xét x \(\ge\) \(-\frac{1}{5}\)
Ta Có Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\) = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\) (1)
xét x \(< -\frac{1}{5}\)
Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x + \(\frac{13}{35}\)
với x \(< -\frac{1}{5}\)
=> -2x \(>\) \(\frac{2}{5}\)
=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)
Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)
5 , D = |x| + |8-x|
D = |x| + |8-x| \(\ge\) |x+8-x| = |8| = 8
Dấu ''='' xảy ra khi x(8-x) \(\ge\) 0 <=> 0\(\le\)x\(\le\) 8
Vậy MinD = 8 khi \(0\le x\le8\)
6,L= |x - 2012| + |2011 - x|
L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x | = |-1| = 1
Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0
HT
3 tháng 8 2021
làm nốt câu 6 nãy ấn nhầm
<=> 2011\(\le\) x \(\le\) 2012
Vậy MinL = 1 khi \(2011\le x\le2012\)
7 , E = | x- \(\frac{2006}{2007}\) | + |x-1|
Ta có :
E = |x-\(\frac{2006}{2007}\) | + |1-x|
E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x | = \(\frac{1}{2007}\)
Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=> \(\frac{2006}{2007}\le x\le1\)
Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\)
8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) |
Ta có :
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) - x |
F = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x | = \(\frac{1}{2}\)
Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0 <=> \(\frac{1}{4}\le x\le\frac{3}{4}\)
Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)
G
5 tháng 9 2016
cậu giải thích giùm mình đoạn này với P(x)=x^7-(x+1)x^6+(x+1)x^5-(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x+15
P(x)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+15
P(x)=x+15=79+15=94
LA
10 tháng 6 2017
hay giai giup mk may phan nay nhe
cmr cac bieu thuc sau ko phu thuoc vao x:
c)C=x(x^3+x^2-3x-2)-(x^2-2)(x^2+x-1)
e)E=(x+1)(x^2-x+1)-(x-1)(x^2+x+1)
tinh gia tri cua da thuc
b)Q(x)=x^14-10x^13=10x^12-10x^11+...+10x^2-10x+10 voi x=9
c)R(x)=x^4-17x^3+17x^2_17x+20 või=16
d)S(x)=x^10-13x^9+13x^8-13X^7+...+13x^2-13x+10 voi 12
29 tháng 3 2019
a. f(x)+g(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)
=2x5-x5-4x4+2x4+3x3-3x3-x2-x2+5x-2x-1+7
=x5-2x4-2x2+3x+6
b. f(x)+h(x)=2x5−4x4+3x3−x2+5x−1+x5−2x4−2x2−x−3
=2x5+x5-4x4-2x4+3x3-x2-2x2+5x-x-1-3
=3x5-6x4+3x3-3x2+6x-4
c. g(x)+h(x)=−x5+2x4−3x3−x2−2x+7+x5−2x4−2x2−x−3
=-x5+x5+2x4-2x4-3x3-x2-2x2-2x-x+7-3
=-3x3-3x2-3x+4
d. f(x)-g(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7
=2x5-x5-4x4-2x4+3x3+3x3-x2+x2+5x+2x-1-7
=x5-6x4+6x3+7x-8
e. f(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1-x5+2x4+2x2+x+3
=2x5-x5-4x4+2x4+3x3-x2+2x2+5x+x-1+3
=x5-2x4+3x3+x2+6x-4
h. g(x)-h(x)=−x5+2x4−3x3−x2−2x+7-(x5−2x4−2x2−x−3)
=−x5+2x4−3x3−x2−2x+7-x5+2x4+2x2+x+3
=-x5-x5+2x4+2x4-3x3-x2+2x2-2x+x+7+3
=-2x5+4x4-3x3+x2-x+10
f. f(x)+g(x)+h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3
=2x5-x5+x5-4x4+2x4-2x4+3x3-3x3-x2-x2-2x2+5x-2x-x-1+7-3
=2x5-4x4-4x2+2x+3
g. f(x)+g(x)-h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-x5+2x4+2x2+x+3
=2x5-x5-x5-4x4+2x4+2x4+3x3-3x3-x2-x2+2x2+5x-2x+x-1+7+3
=4x+9
n. f(x)-g(x)+h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7+x5−2x4−2x2−x−3
=2x5-x5+x5-4x4-2x4-2x4+3x3+3x3-x2+x2-2x2+5x+2x-x-1-7-3
=2x5-8x4+6x3-2x2+6x-11
m. f(x)-g(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)
=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7-x5+2x4+2x2+x+3
=2x5-x5-x5-4x4-2x4+2x4+3x3+3x3-x2+x2+2x2+5x+2x+x-1-7+3
=-4x4+6x3+2x2+8x-5
x=12345678-x
=> x+x=12345678
=> 2x=12345678
=> x=12345678:2
=> x=6172839