tìm giá trị của các biểu thức sau:
1) A= 1+3+3^3+3^3+...+3^2020
2) B=6^2+6^4+6^6+...+6^1010
3) C=1-2+2^2-2^3+...-2^2011+2^2012
4) D=-2+2^2-2^3+2^4-...-2^2019+2^2020
giúp mk vs nha
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Bài 3 :
Vì \(\left(x-2\right)^2\ge0\forall x\)
Nên : \(A=\left(x-2\right)^2-4\ge-4\forall x\)
Vậy \(A_{min}=-4\) khi x = 2
B1: lấy máy tính mà tính thôi bạn (nhớ lm theo từng bước)
B2:
a, \(\left|x-\frac{2}{3}\right|-\frac{1}{2}=\frac{5}{6}\)
\(\left|x-\frac{2}{3}\right|=\frac{4}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{4}{3}\\x-\frac{2}{3}=\frac{-4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)
b, \(\frac{\left(-2\right)^x}{512}=-32\Rightarrow\left(-2\right)^x=-16384\Rightarrow x\in\varnothing\)
B3:
Vì \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-4\ge-4\)
Dấu "=" xảy ra khi x = 2
Vậy GTNN của A = -4 khi x = 2
a/\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
= \(\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{1-3}{1+5}=\frac{-2}{6}=-3\)
\(C=2^2+4^2+...+2010^2\)
\(=2\left(1+1\right)+4\left(3+1\right)+...+2010\left(2009+1\right)\)
\(=2+1.2+4+3.4+...+2010+2009.2010\)\
\(=\left(2+4+...+2010\right)+\left(1.2+3.4+...+2009.2010\right)\)
Đặt A = 2+4+...+2010 = \(\frac{\left(2010+2\right).1005}{2}=1011030\)
Đặt B=1.2+3.4+...+2009.2010
3B=1.2.3+3.4.3+...+2009.2010.3
3B=1.2(3-0)+3.4(5-2)+...+2009.2010(2011-2008)
3B=1.2.3-0.1.2+3.4.5-2.3.4+...+2009.2010.2011-2008.2009.2010
3B=2009.2010.2011
B=\(\frac{2009.2010.2011}{3}=2706866330\)
Thay A và B vào C ta có:
\(C=1011030+2706866330=2707877360\)
B=1.2+2.3+...+2010.2011
3B=1.2.3+2.3.3+...+2010.2011.3
3B=1.2.(3-0)+2.3.(4-1)+...+2010.2011.(2012-2009)
3B=1.2.3-0.1.2+2.3.4-1.2.3+...+2010.2011.2012-2009.2010.2011
3B=(1.2.3+2.3.4+...+2010.2011.2012)-(0.1.2+1.2.3+...+2009.2010.2011)
3B=2010.2011.2012-0.1.2
3B=2010.2011.2012
B=\(\frac{2010.2011.2012}{3}=2710908440\)
nhìu dữ
a)3/2
b)-1/3
c)-5/6
d)0
e)-1/2
Bài 2
a=3
b=1/2
c=-1/3
d=0
e=9
f=-2/3
a: \(A=\dfrac{2}{3}-4\cdot\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b: \(B=\dfrac{-2+5}{6}\cdot11-7=\dfrac{11}{2}-7=-\dfrac{3}{2}\)
c: \(=1+\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{27}{4}+5-1\)
\(=\dfrac{1}{3}+\dfrac{27}{4}+5=\dfrac{145}{12}\)
d: \(D=\dfrac{2}{3}\cdot\dfrac{2}{5}-\dfrac{1}{5}\cdot\dfrac{1}{6}=\dfrac{4}{15}-\dfrac{1}{30}=\dfrac{7}{30}\)
Answer:
a) \(\frac{5x}{2x+2}+1=\frac{6}{x+1}\)
\(\Rightarrow\frac{5x}{2\left(x+1\right)}+\frac{2\left(x+1\right)}{2\left(x+1\right)}=\frac{12}{2\left(x+1\right)}\)
\(\Rightarrow5x+2x+2-12=0\)
\(\Rightarrow7x-10=0\)
\(\Rightarrow x=\frac{10}{7}\)
b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\left(ĐK:x\ne0\right)\)
\(\Rightarrow x^2-6=x^2+\frac{3}{2}x\)
\(\Rightarrow\frac{3}{2}x=-6\)
\(\Rightarrow x=-4\)
c) \(\frac{3x-2}{4}\ge\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\ge0\)
\(\Rightarrow9x-6-6x-6\ge0\)
\(\Rightarrow3x-12\ge0\)
\(\Rightarrow x\ge4\)
d) \(\left(x+1\right)^2< \left(x-1\right)^2\)
\(\Rightarrow x^2+2x+1< x^2-2x+1\)
\(\Rightarrow4x< 0\)
\(\Rightarrow x< 0\)
e) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Rightarrow\frac{2x-3+5\left(x^2-2x\right)}{35}\le\frac{5x^2-7\left(2x-3\right)}{35}\)
\(\Rightarrow2x-3+5x^2-10x\le5x^2-14x+21\)
\(\Rightarrow6x\le24\)
\(\Rightarrow x\le4\)
f) \(\frac{3x-2}{4}\le\frac{3x+3}{6}\)
\(\Rightarrow\frac{3\left(3x-2\right)-2\left(3x+3\right)}{12}\le0\)
\(\Rightarrow9x-6-6x-6\le0\)
\(\Rightarrow3x\le12\)
\(\Rightarrow x\le4\)