300000000:2=150000000

tk mk nha bn

chúc bạn học giỏi

=150000000

300000000:2 = 1500000000

mik nghĩ vậy

Giả sử trong 100 số đó không có 2 số nào bằng nhau.

\(\Rightarrow\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+...+\dfrac{1}{\sqrt{x_{100}}}\le\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}\)

\(< 1+\dfrac{2}{\sqrt{2}+\sqrt{1}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}+...+\dfrac{2}{\sqrt{100}+\sqrt{99}}\)

\(=1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(=1+2\left(\sqrt{100}-\sqrt{1}\right)=19< 20\)

Vậy trong 100 số đã cho có ít nhất 2 số bằng nhau

Giả sử 100 số nguyên dương đã cho ko tồn tại \(x_i=x_k\)

Ko mất tính tổng quát giả sử \(x_1< x_2< x_3< ...< x_{100}\)

Vì \(x_1;x_2;x_3;...;x_{100}\) đều là các số nguyên dương suy ra \(x_1\ge1;x_2\ge2;....;x_{100}\ge100\)

Tức là có: \(VT< \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{100}}< 10< VP\)

Mâu thuẫn với giả thiết suy ra điều giả sử sai

Tức tồn tại \(x_i=x_k\) với \(i\ne k\) và \(i,k\in\left\{1;2;...;100\right\}\)

3333333333

                                       Bài giải

    33333333333 . Học tốt nha !

\(33333333333\)

\(\left|2\dfrac{1}{5}-x\right|\)\(+\left|x-\dfrac{1}{5}\right|\)\(+8\dfrac{1}{5}\)\(=1,2\)

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{6}{5}-\dfrac{41}{5}\)

\(\Rightarrow\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|=\dfrac{-36}{5}\) (vô lý vì \(\left|2\dfrac{1}{5}-x\right|+\left|x-\dfrac{1}{5}\right|\ge0\))

Vậy: Không tìm được giá trị x thoả mãn.

a:=>3x=15

=>x=5

b: =>x+3=0,96

=>x=-2,04

c: =>x^2=36

=>x=6 hoặc x=-6

`a, 3/4=(3x)/20`

`3x*4=3*20`

`3x*4=60`

`3x=60 \div 4`

`3x=15`

`x=15 \div 3`

`x=5`

`b, (1,2)/(x+3)=5/4`

`1,2*4=(x+3)*5`

`4,8=(x+3)*5`

`x+3= 4,8 \div 5`

`x+3=0,96`

`x=0,96-3`

`x=-2,04`

`c, (x^2)/32=9/8`

`x^2*8=32*9`

`x^2*8=288`

`x^2=288 \div 8`

`x^2=36`

`x^2=(+-6)^2`

`-> \text {x= 6 hoặc -6}`

tích trung tỉ bằng tihs ngoại tỉ là ra ý mà

\(\dfrac{x+1}{x+2}=\dfrac{0,8}{1,2}\) \(\Leftrightarrow0,8\left(x+2\right)=1,2\left(x+1\right)\)

\(\Leftrightarrow0,8x+1,6=1,2x+1,2\) \(\Leftrightarrow0,4x=0,4\Leftrightarrow x=1\)

vậy \(x=1\)

\(\dfrac{x+1}{x+2}=\dfrac{0,8}{1,2}\)

\(\Rightarrow\left(x+1\right)1,2=\left(x+2\right)0,8\)

\(1,2x+1,2=0,8x+1,6\)

\(1,2x-0,8x=1,6-1,2\)

\(0,4x=0,4\)

\(x=1\)

\(\dfrac{x-2}{-1,2}=\dfrac{-5}{2}\Rightarrow x=\dfrac{-5.\left(-1,2\right)}{2}+2=\dfrac{6}{2}+2=3+2=5\\ \dfrac{-6}{x+1}=\dfrac{1,8}{9}\Rightarrow x=\dfrac{-6.9}{1,8}-1=\dfrac{-54}{1,8}-1=-30-1=-31\\ \dfrac{-3}{x}=\dfrac{x}{-12}\Rightarrow x=\sqrt{\left(-12\right).\left(-3\right)}=\sqrt{36}=\sqrt{\left(\pm6\right)^2}=\pm6\)

\(\dfrac{x-4}{x-1}=\dfrac{3}{5}\\ \Rightarrow5\left(x-4\right)=3\left(x-1\right)\\ \Leftrightarrow5x-20=3x-3\\ \Leftrightarrow5x-3x=-3+20\\ \Leftrightarrow2x=17\\ \Leftrightarrow x=\dfrac{17}{2}\\ ---\\ \dfrac{1,12}{-10}=\dfrac{11,2}{x}\Rightarrow x=\dfrac{11,2.\left(-10\right)}{1,12}=\dfrac{10.1,12.\left(-10\right)}{1,12}=-100\)

Bài 1: Tìm x

a) Ta có: \(\dfrac{4}{3}:0.8=\dfrac{2}{3}:\left(0.1\cdot x\right)\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}:\dfrac{4}{5}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(\dfrac{1}{10}\cdot x\right)=\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow x\cdot\dfrac{1}{10}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\cdot\dfrac{3}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}:\dfrac{1}{10}=\dfrac{2}{5}\cdot10=\dfrac{20}{5}=4\)

Vậy: x=4

b) Ta có: \(\left|x\right|=-1.2\)

mà \(\left|x\right|\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

Bài 2: Tính

a) Ta có: \(\left(-2.5\right)\cdot\left(-4\right)\cdot\left(-7.9\right)\)

\(=\left(2.5\cdot4\right)\cdot\left(-7.9\right)\)

\(=-7.9\cdot10=-79\)

b) Ta có: \(\left(-0.375\right)\cdot\dfrac{13}{3}\cdot\left(-2\right)^3\)

\(=\dfrac{3}{8}\cdot8\cdot\dfrac{13}{3}\)

\(=3\cdot\dfrac{13}{3}=13\)

b. \(\dfrac{13+x}{20}\)= \(\dfrac{15}{20}\)

=>  13+x=15

            x=15-13

            x=2

$#Shả$

`x xx1,2+x xx1,8=45`

`<=>x xx(1,2+1,8)=45`

`<=> x xx 3 =45`

`<=>x=45:3=15`

`(13+x)/20=3/4`

`<=>4xx(13+x)=20xx3`

`<=>4xx(13+x)=60`

`<=>13+x=60:4=15`

`<=>x=15-13=2`

a. x x(1,2+1,8)=45

    x x 3= 45

    x= 45:3

    x=15

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)