A =5+ 5²+5³+..................................................+5¹⁰⁰

B= 5+5³ +5⁵ +...................................................+5⁹⁹

C= 1+ \(\frac{1}{3}\) + \(\frac{1}{3^2}\) +....................................+ \(\frac{1}{3^{100}}\)

D= 1-\(\frac{1}{3}\)+ \(\frac{1}{3^2}\) - ..........................................- \(\frac{1}{3^{99}}\)+\(\frac{1}{3^{100}}\)

E=\(\frac{1}{2}\)+\(\frac{2}{2^2}\)+\(\frac{3}{2^3}\) +................................... +\(\frac{100}{2^{100}}\)

Thực hiện phép tính:

a)\(\frac{3}{4}\)x\(\frac{16}{9}\)-\(\frac{7}{5}\):\(\frac{-21}{20}\)

b)\(2\frac{1}{3}\)-\(\frac{1}{3}\)x [\(\frac{-3}{2}\)+(\(\frac{2}{3}\)+0,4x5) ]

c) (20+\(9\frac{1}{4}\)):\(2\frac{1}{4}\)

d) (6-\(2\frac{4}{5}\)x\(3\frac{1}{8}\)-\(1\frac{3}{5}\):\(\frac{1}{4}\)

GIÚP MIK VỚI AI NHANH MIK TICK LUÔN CHO.

B= \(\frac{x-4x+1}{x^2}\)= 1-\(\frac{4}{x}\)+\(\frac{1}{x^2}\)

Đặt t = \(\frac{1}{x}\)=> B= 1-4t+t²= t²-4t+1 = t²-2.t.2+4-3= (t-2)²-3 \(\ge\)-3 

Min của A= -3 khi t-2= 0 => t=2 => \(\frac{1}{x}\)= 2 => x = \(\frac{1}{2}\)

Cho a,b,c>0 và a+b+c=3. CMR: \(a^2+b^2+c^2\ge3\)

Ta cần tìm m, n để bđt sau luôn đúng \(a^2\ge ma+n\) (1) 

tương tự: \(b^2\ge mb+n;c^2\ge mc+n\)

cộng 3 bđt lại ta đc: \(a^2+b^2+c^2\ge m\left(a+b+c\right)+3n=3m+3n\)

dự đoán cực trị xảy ra tại a=b=c=1 nên \(3m+3n=\left(a^2+b^2+c^2\right)_{min}=3\)\(\Rightarrow\)\(n=1-m\)

thay n=1-m vào (1) : \(a^2\ge ma-m+1\)(2)\(\Leftrightarrow\)\(\left(a-1\right)\left(a+1\right)\ge m\left(a-1\right)\)

đồng nhất hệ số : \(a+1=m\)\(\Leftrightarrow\)\(m=a+1=1+1=2\) (dấu "=" xảy ra tại a=1)

thay m=2 vào (2) ta có bđt cần CM: \(a^2\ge2a-1\) ( với \(0< a< 3\) ) 

bđt \(\Leftrightarrow\)\(\left(a-1\right)^2\ge0\) luôn đúng 

do đó: \(a^2+b^2+c^2\ge2a-1+2b-1+2c-1=2\left(a+b+c\right)-3=2.3-3=3\)

dấu "=" xảy ra khi a=b=c=1

1.     \(\frac{x^3-10x^2+25x}{x^2-5x}\)\(=0\)  ( đkxđ: \(x\ne0;5\))

<=> \(\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0\)<=>   \(x-5=0\)<=> vô n_o

_{2.       \(A=\)\(\frac{2x^2-2}{x^3-x^2-4x+4}\)\(=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}\)}   ( a, đkxđ: \(x\ne1;\pm2\))

b, \(A=0\)<=> \(\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0\)<=> \(x=-1\)( TM) . Vậy \(A=0\Leftrightarrow x=-1\)

3.    \(B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}\)\(=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}\) ( a,  đkxđ: \(x\ne3,-2\))

\(b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)\). Vậy \(B=0\Leftrightarrow x=2\)

Bai 1: Tinh

a) ( \(\frac{1}{3}\))^-1 - (\(\frac{-6}{7}\))⁰ + ( \(\frac{1}{2}\))^{-2 .} 2

b) \(\frac{2.5^{22}-9.5^{21}}{25^{10}}\): \(\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)

c) ( \(\frac{-4}{9}\))³. (\(\frac{3}{2}\))². (\(\frac{9}{6}\))³

\(ab+bc+ca=1\)\(\Rightarrow\)\(\hept{\begin{cases}a+b+c\ge\sqrt{3}\\a^2+b^2+c^2\ge1\end{cases}}\)

\(\left(a-\frac{1}{\sqrt{3}}\right)^2\ge0\)\(\Leftrightarrow\)\(a\le\frac{\sqrt{3}}{2}a^2+\frac{\sqrt{3}}{6}\)

\(P=\Sigma\frac{a^2\left(1-2b\right)^2}{b\left(1-2b\right)}\ge\frac{\left(a+b+c-2\right)^2}{\left(a+b+c\right)-2\left(a^2+b^2+c^2\right)}\ge\frac{\left(a+b+c-2\right)^2}{\frac{\sqrt{3}-4}{2}\Sigma a^2+\frac{\sqrt{3}}{2}}\ge\sqrt{3}-2\)