\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{19.21}\)

\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right)\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=2\left(1-\dfrac{1}{21}\right)=2.\dfrac{20}{21}=\dfrac{40}{21}\)

Giải:

\(a^2=bc\Rightarrow\dfrac{a}{c}=\dfrac{b}{a}\) hay \(\dfrac{a}{b}=\dfrac{c}{a}\)

a, Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{a}=\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrowđpcm\)

b, \(\dfrac{a}{b}=\dfrac{c}{a}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{a^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a^2}{b^2}=\dfrac{c^2}{a^2}=\dfrac{a^2+c^2}{b^2+a^2}\)

\(\Rightarrow\dfrac{a^2+c^2}{b^2+a^2}=\dfrac{a^2}{b^2}=\dfrac{a}{b}.\dfrac{c}{a}=\dfrac{c}{b}\)

\(\Rightarrowđpcm\)

1,

\(A=\left|x+5\right|+\left|x-5\right|=\left|x+5\right|+\left|5-x\right|\)

Áp dụng tính chất dãy tỉ số bằng nhau có:
\(A=\left|x+5\right|+\left|5-x\right|\ge\left|x+5+5-x\right|=\left|10\right|=10\)

Dấu " = " khi \(\left\{{}\begin{matrix}x+5\ge0\\5-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-5\\x\le5\end{matrix}\right.\)

Vậy \(MIN_A=10\) khi \(-5\le x\le5\)

2,

a, \(M=1,5-\left|x+1,1\right|\le1,5\)

Dấu " = " khi \(\left|x+1,1\right|=0\Rightarrow x=-1,1\)

Vậy \(MAX_M=1,5\) khi x = -1,1

b, \(N=-3,7-\left|1,7-x\right|\le-3,7\)

Dấu " = " khi \(\left|1,7-x\right|=0\Rightarrow x=1,7\)

Vậy \(MAX_N=-3,7\) khi x = 1,7

Ta có: \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|\ge0\\\left|y-\dfrac{1}{4}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\Rightarrow\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|\ge0\)

Mà \(\left|x-\dfrac{1}{4}\right|+\left|y-\dfrac{1}{4}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{4}\right|=0\\\left|y-\dfrac{1}{4}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{1}{4}\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{-1}{2}\right)\)

Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{-2,5}{4,5}=\dfrac{-5}{9}\Rightarrow\dfrac{a}{-5}=\dfrac{b}{9}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{-5}=\dfrac{b}{9}=\dfrac{a+b}{-5+9}=\dfrac{1,44}{4}=0,36\)

\(\Rightarrow\left\{{}\begin{matrix}a=-1,8\\b=3,24\end{matrix}\right.\)

Vậy...

a, Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(A=\left|x-1\right|+\left|5-x\right|+3\ge\left|x-1+5-x\right|+3\)

\(=\left|4\right|+3=7\)

Dấu " = " khi \(\left\{{}\begin{matrix}x-1\ge0\\5-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le5\end{matrix}\right.\)

Vậy \(MIN_A=7\) khi \(1\le x\le5\)

b, \(B=4x^2-5\ge-5\)

Dấu " = " khi \(4x^2=0\Leftrightarrow x=0\)

Vậy \(MIN_B=-5\) khi x = 0

c, \(C=x^2+2x+5=x^2+2x+1+4\)

\(=\left(x+1\right)^2+4\ge4\)

Dấu " = " khi \(\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy \(MIN_C=4\) khi x = -1

\(\overline{ba}+\overline{ab}=10b+a+10a+b=11a+11b\)

\(=11\left(a+b\right)⋮11\)

\(\Rightarrow\overline{ba}+\overline{ab}⋮11\)

\(\Rightarrowđpcm\)

Ta có: \(8x^2-4x+1⋮2x+1\)

\(\Rightarrow\left(8x^2+4x\right)-\left(8x-1\right)⋮2x+1\)

\(\Rightarrow4x\left(2x+1\right)-\left(8x+4\right)+5⋮2x+1\)

\(\Rightarrow4x\left(2x+1\right)-4\left(2x+1\right)+5⋮2x+1\)

\(\Rightarrow\left(4x-4\right)\left(2x+1\right)+5⋮2x+1\)

\(\Rightarrow5⋮2x+1\)

\(\Rightarrow2x+1\in\left\{1;-1;5;-5\right\}\)

\(\Rightarrow x\in\left\{0;-1;2;-3\right\}\)

Vậy...

Gọi số đó là a

a chia 40; 140;28 đều dư 5 

=> a - 5 chia hết cho 40; 140; 28 

=> a - 5 \(\in\) BC (40;140; 28). Vì a nhỏ nhất nên a - 5 nhỏ nhất 

=> a - 5 \(\in\) BCNN (40;140;28) 

40 = 2³.5; 140 = 2².5.7; 28 = 2².7 => BCNN (40;140;28) = 2³.5.7 = 280 => a - 5 = 280 => a = 285

Vậy.....