=5/18

CHO MÌNH CÁCH GIẢI VỚI Ạ  BẠN MẶC THIÊN PHONG ƠI. MÌNH CẢM ƠN

K={10;a}

K={10;b}

K={12;a}

K={12;b}

đúng rồi !!!

\(A=x^3+3x^2+3x\)

\(A=x^3+3x^2+3x+1-1\)

\(A=\left(x^3+3x^2+3x+1\right)-1\)

\(A=\left(x+1\right)^3-1\)

Thay x = 29

\(\Rightarrow A=\left(29+1\right)^3-1\)

\(\Rightarrow A=30^3-1=27000-1=26999\)

Thay x= 29

=> A= 29^3+ 3.29^2 + 3.29=  24389+ 3. 841+ 3. 29= 24389+ 2523+ 87= 26999

-(x - 2) = 1 + 2x - (-4)

-x + 2 = 1 + 2x + 4

2 - x = 2x + 5

-x = 2x + 5 - 2

-x = 2x + 3

-x - 2x = 3

-3x = 3

x = -1

\(\frac{x-y}{x+y}=\frac{z-x}{z+x}\Rightarrow\left(x-y\right)\left(z+x\right)=\left(z-x\right)\left(x+y\right).\)

\(\Rightarrow xz+x^2-yz-xy=zx+zy-x^2-xy\)

\(\Rightarrow x^2-yz=zy-x^2\)

\(\Rightarrow2x^2=2yz\Rightarrow x^2=yz\)

Ta có : \(\frac{x-y}{x+y}=\frac{z-x}{z+x}=k\)

=> \(\hept{\begin{cases}x-y=k(x+y)\\z-x=k(z+x)\end{cases}}\)

=> \(\hept{\begin{cases}x(1-k)=-y(1+k)\\z(1-k)=-x(1+k)\end{cases}}\)

=> \(\frac{x(1-k)}{z(1-k)}=\frac{-y(1+k)}{-x(1+k)}\)

=> \(\frac{x}{z}=\frac{y}{x}\Rightarrow x^2=yz\)

Giúp mình với ạ !!! Mình cần gấp :)

\(DK:x\in R\)

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+...+\frac{1}{50.56}\)

\(A=\frac{1}{6}.\left(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+...+\frac{6}{50.56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+...+\frac{1}{50}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\left(\frac{1}{8}-\frac{1}{56}\right)\)

\(A=\frac{1}{6}.\frac{3}{28}\)

\(A=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(B=5.\left(\frac{9}{12.21}+\frac{9}{21.30}\right)-4.\left(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64}\right)\)

\(B=5.\left(\frac{1}{12}-\frac{1}{30}\right)-4.\left(\frac{1}{24}-\frac{1}{64}\right)\)

\(B=5.\frac{1}{20}-4.\frac{5}{192}\)

\(B=\frac{1}{4}-\frac{5}{48}\)

\(B=\frac{7}{48}\)

Ta có \(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}=\frac{1}{56}\times\frac{48}{7}=\frac{6}{49}\)

Lấy \(\frac{6}{49}-\frac{1}{8}=-\frac{1}{392}< 0\)

\(\Rightarrow\frac{6}{49}< \frac{1}{8}\) hay \(\frac{A}{B}< \frac{1}{8}\)

\(A=\frac{1}{8.14}+\frac{1}{14.20}+\frac{1}{20.26}+....+\frac{1}{50.56}\)

\(=\frac{1}{6}.(\frac{6}{8.14}+\frac{6}{14.20}+\frac{6}{20.26}+....+\frac{6}{50.56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{14}+\frac{1}{14}-\frac{1}{20}+\frac{1}{20}-\frac{1}{26}+....+\frac{1}{50}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{1}{8}-\frac{1}{56})\)

\(=\frac{1}{6}.(\frac{7}{56}-\frac{1}{56})\)

\(=\frac{1}{6}.\frac{6}{56}\)

\(=\frac{1}{56}\)

\(B=\frac{45}{12.21}+\frac{45}{21.30}-\frac{40}{24.34}-\frac{40}{34.44}-\frac{40}{44.54}-\frac{40}{54.64}\)

\(=5(\frac{9}{12.21}+\frac{9}{21.30})-4(\frac{10}{24.34}+\frac{10}{34.44}+\frac{10}{44.54}+\frac{10}{54.64})\)

\(=5(\frac{1}{12}-\frac{1}{21}+\frac{1}{21}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{34}+\frac{1}{34}-\frac{1}{44}+\frac{1}{44}-\frac{1}{54}+\frac{1}{54}-\frac{1}{64})\)

\(=5(\frac{1}{12}-\frac{1}{30})-4(\frac{1}{24}-\frac{1}{64})\)

\(=5(\frac{5}{60}-\frac{2}{60})-(\frac{4}{24}-\frac{4}{64})\)

\(=5.\frac{1}{20}-(\frac{1}{6}-\frac{1}{16})\)

\(=\frac{1}{4}-(\frac{8}{48}-\frac{3}{48})\)

\(=\frac{1}{4}-\frac{5}{48}\)

\(=\frac{12}{48}-\frac{5}{48}=\frac{7}{48}\)

\(\frac{A}{B}=\frac{1}{56}\div\frac{7}{48}\)

\(=\frac{1}{56}.\frac{48}{7}\)

\(=\frac{6}{49}=\frac{48}{392}\)bé hơn \(\frac{49}{392}=\frac{1}{8}\)

Vậy \(\frac{A}{B}\)bé hơn \(\frac{1}{8}\)

Chúc bạn học tốt

a, \(M=1+6+6^2+6^3+...+6^{99}\)

\(M=6\cdot(1+6)+6^2(1+6)+6^3(1+6)+...+6^{99}(1+6)\)

\(M=6\cdot7+6^2\cdot7+6^3\cdot7+...+6^{99}\cdot7\)

\(M=7\cdot\left[6+6^2+6^3+...+6^{99}\right]⋮7(đpcm)\)

b, \(M=1+6+6^2+6^3+...+6^{99}\)

\(M=6\cdot\left[1+6+6^2+6^3\right]+...+6^{96}\left[1+6+6^2+6^3\right]\)

\(M=6\cdot\left[7+36+216\right]+...+6^{96}\left[7+36+216\right]\)

\(M=6\cdot259+...+6^{96}\cdot259\)

\(M=259\cdot\left[6+...+6^{96}\right]⋮259\)

Vậy \(M⋮259(đpcm)\)

bai nao vay