3 x 8 = 21

5 x 2 = 10

3.8=24

5.2=10

1 + 1 + 1 + 1 = 4

1 + 1 + 1 + 1 = 4

a) \(A=2x+\sqrt{x^2-2x+1}=2x+\sqrt{\left(x-1\right)^2}=2x+\left|x-1\right|\)

với x \(\ge\)1 thì A = 2x + x - 1 = 3x - 1

với x < 1 thì A = 2x + 1 - x = x + 1

b) A = \(2x+\left|x-1\right|=1\)

TH1 : x \(\ge\)1 thì A = 3x - 1 = 1 \(\Rightarrow\)x = \(\frac{2}{3}\)( ko t/m )

TH2 : x < 1 thì A = x + 1 = 1 \(\Rightarrow\)x = 0 ( t/m )

vậy x = 0

\(A=2x+\sqrt{x^2-2x+1}=2x+\sqrt{\left(x-1\right)^2}=2x+|x-1|\)

Để A=1 thì  \(2x+|x-1|=1\)\(\left(1\right)\)

Với  \(x\ge1\)thì  (1)   trở thành   \(2x+x-1=1\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)(loại)

Với  \(x< 1\)thì  (1)  trở thành   \(2x-x+1=1\Leftrightarrow x=0\)(chọn)

Vậy   \(S=0\)

P=1/(x+y)(x^2-xy+y^2)+1/xy

P=1/(x^2-xy+y^2)+1/xy ( vĩ+y=1)

P=1/(x^2-xy+y^2)+3/xy

Đến đây áp dụng bất đẳng thức Svac có

P>=(√3+1)^2/(x+y)^2

P>=(√3+1)^2 (vì x+y=1)

hay P>=4+2√3(đpcm)

Ta có : \(8^x+8^x+8^2\ge3\sqrt[3]{8^x.8^x.8^2}=12.4^x\)

\(8^y+8^y+8^2\ge3\sqrt[3]{8^y.8^y.8^2}=12.4^y\)

\(8^z+8^z+8^2\ge3\sqrt[3]{8^z.8^z.8^2}=12.4^z\)

\(8^x+8^y+8^z\ge3\sqrt[3]{8^x.8^y.8^z}=3\sqrt[3]{8^6}=192\)

Cộng các vế , ta được :

\(3\left(8^x+8^y+8^z+64\right)\ge3\left(4^{x+1}+4^{y+1}+4^{z+1}+64\right)\)

hay \(8^x+8^y+8^z\ge4^{x+1}+4^{y+1}+4^{z+1}\)

Ta có : \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\ge xy\left(x+y\right)\)

\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)

\(\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

Tương tự : \(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3yz}}{yz}=\sqrt{\frac{3}{yz}}\);  \(\frac{\sqrt{1+x^3+z^3}}{xz}\ge\frac{\sqrt{3xz}}{xz}=\sqrt{\frac{3}{xz}}\)

\(\Rightarrow A\ge\sqrt{3}\left(\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{xz}}\right)\ge3\sqrt{3}\sqrt{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

50000+505-12:24= 50504,5

~ Hok tốt ~
#JH

50000+505-12:24=50505-0,5=50504,5

Chúc bạn học tốt :) :)

10000 : 2000 = 5

~ Hok tốt ~
#JH

10000 : 2000 = 5

~ Kb rồi tk mik nha ~
​#JH

Đặt : \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

Ta thấy :

\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}\)

\(.......................\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

Vì \(\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)nên \(\frac{1}{6}< A< \frac{1}{4}\)hay \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)

~ Hok tốt ~

Bài 1:

Đặt  \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

Ta có: 

\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Ta có:

\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đ}pcm\right)\)

Bài 2:

\(a)\)Tách tổng A thành ba nhóm:

\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)

\(A>\frac{1}{30}\cdot20+\frac{1}{50}\cdot20+\frac{1}{70}\cdot20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}\)

\(A>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\left(\text{đ}pcm\right)\)

\(b)\)Tách tổng A thành sáu nhóm:

\(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)

\(A< \frac{1}{11}\cdot10+\frac{1}{21}\cdot10+\frac{1}{31}\cdot10+\frac{1}{41}\cdot10+\frac{1}{51}\cdot10+\frac{1}{61}\cdot10\)

\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\left(\text{đ}pcm\right)\)

#Sakura

= 1

2 : 2 + 2 - 2 x 2 = 1 + 2 - 4

                        = 3 - 4 = -1