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a: TH1: x<1
A=1-x+2-x=3-2x
TH2; 1<=x<2
A=x-1+2-x=1
TH3: x>=2
A=x-1+x-2=2x-3
b: TH1: x<5/2
B=5-2x+3-x+x-2=-2x+6
TH2: 5/2<=x<3
B=2x-5+3-x+x-2=2x-4
TH3: x>=3
B=x-3+2x-5+x-2=4x-10
c: TH1: x<-3/2
C=-2x-3-(5-x)+2x
=-2x-3-5+x+2x
=x-8
TH2: -3/2<=x<5
C=2x+3-(5-x)+2x=4x+3-5+x=5x-2
TH3: x>=5
C=2x+3-(x-5)+2x=4x+3-x+5=3x+8
a) Ta có:
\(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k\left(90-10^2+10\right)\)
\(=10^k.0=0\)
b) Ta có:
\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=2,5.10.5^{n-3}+5^n-6.5^{n-1}\)
\(=5.5.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^2.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^{n-3+2}+5^n-6.5^{n-1}\)
\(=5^{n-1}\left(1+5-6\right)\)
\(=5^{n-1}.0=0\)
a) Rút gọn biểu thức:
\(90\times10^k-10^{k+2}+10^{k+1}=90\times10^k-10^k\times10^2+10^k\times10\) \(=10^k\times\left(90-10^2+10\right)\) \(=10^k\times\left(90-100+10\right)\) \(=10^k\times0=0\)
b) Rút gọn biểu thức:
\(2,5\times5^{n-3}\times10+5^n-6\times5^{n-1}=2,5\times\dfrac{5^n}{5^3}\times10+5^n-6\times\dfrac{5^n}{5}\) \(=2,5\times\dfrac{5^n}{125}\times10+5^n-\dfrac{6}{5}\times5^n\) \(=0,2\times5^n+5^n-1,2\times5^n\) \(=5^n\times\left(0,2+1-1,2\right)=5^n\times0=0\)
Mk nhầm nha câu đầu chỉ có 1 cái x-1 + x -2 thôi ko có cái đằng sau nhé ! giá trị tuyệt đối thì vẫn giữ nguyên !
a) \(10^{n+1}-6.10^n\)
\(=10^n.10-6.19^n\)
\(=10^n.\left(10-6\right)\)
\(=10^n.4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n.2^3+2^n.2^2-2^n.2+2^n.1\)
\(=2^n.\left(2^3+2^2-2+1\right)\)
\(=2^n.11\)
c) \(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k.\left(90-10^2+10\right)\)
\(=0\)
d) \(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=\dfrac{2,5.5^n.10}{5^3}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n+5^{n+1}-6.5^n}{5}=\dfrac{5^n+5^n.5-6.5^n}{5}=\dfrac{5^n\left(1+5-6\right)}{5}=\dfrac{0}{5}=0\)
a) 5A = 5 + 5^2 + 5^3 + 5^4 +...+ 5^51
=> 5A - A = 4A = 5^51 - 1
=> A = \(\frac{5^{51}-1}{4}\)
b) 3B = 3^100 - 3^99 -...- 3
=> 3B - B = 2B = 3^100 - 2.3^99 + 1
=> B = \(\frac{3^{100}-2\times3^{99}+1}{2}\)
a, 1+5+52+.....+550
=> 5(1+5+52+.....+550)=5+52+53.....+551
=>4(1+5+52+.....+550)=551-1
=>1+5+52+.....+550=(551-1):4
b,399-398-...-3-1
=399-(398+...+3+1)
=399-(399-1):2
1.
\(A=\sqrt{1+2+...+\left(n-1\right)+n+\left(n-1\right)+...+2+1}\)
\(=\sqrt{\frac{\left(n-1\right).n}{2}\cdot2+n}=\sqrt{n^2-n+n}=\sqrt{n^2}=n\)
2.
\(A=\frac{x^2-3}{x+2}=\frac{x\left(x+2\right)-3}{x+2}=\frac{x^2+2x-2x-3}{x+2}=x+\frac{-2x-3}{x+2}\)
\(=x+\left(-1\right)+\frac{1}{x+2}\le\frac{1}{1}\)
Vậy GTLN của A = 1 tại x = -1
\(90\cdot10-10+10=900-10+10=900\)