Lời giải:

$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$

$=1$

`# \text {DNamNgV}`

\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}\)

\(\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)=\dfrac{1}{99}\)

\(\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}\)

\(\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}\)

\(\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}\)

\(\Rightarrow x=\dfrac{99}{99}\)

\(\Rightarrow x=1\)

Vậy, `x = 1.`

Lời giải:

$\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$

$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12})=(x+1)(\frac{1}{13}+\frac{1}{14})$

$\Rightarrow (x+1)(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14})=0$

Hiển nhiên $\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0$ 

$\Rightarrow x+1=0$

$\Rightarrow x=-1$

Lời giải:

Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$

Đề bài 

$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow 12=x$ (thỏa mãn)

Trước tiên ta cần chứng minh : Với a<b thì : \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\) với c là số nguyên dương.

\(\Leftrightarrow a.\left(b+c\right)< b.\left(a+c\right)\)

\(\Leftrightarrow ab+ac< ab+bc\)

\(\Leftrightarrow ac< bc\)

\(\Leftrightarrow a< b\left(LĐ\right)\)

Áp dụng bổ đề đó ta có : \(\dfrac{a}{b+c}< \dfrac{a+a}{a+b+c}=\dfrac{2a}{a+b+c}\)

\(CMTT:\dfrac{b}{c+a}< \dfrac{2b}{a+b+c}\)

\(\dfrac{c}{a+b}< \dfrac{2c}{a+b+c}\)

Do đó : \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< \dfrac{2a+2b+2c}{a+b+c}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\left(đpcm\right)\)

Lời giải:

Vì $b+c> a\Rightarrow 2(b+c)> a+b+c$

$\Rightarrow b+c> \frac{a+b+c}{2}$

$\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}(1)$

Hoàn toàn tương tự ta có:
$\frac{b}{c+a}< \frac{2b}{a+b+c}(2)$

$\frac{c}{a+b}< \frac{2c}{a+b+c}(3)$

Từ $(1); (2); (3)\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{2(a+b+c)}{a+b+c}=2$

Ta có đpcm. 

Tổng A có 1000 số hạng.

$A>\genfrac{}{}{0ex}{}{1001}{1000^2+1000}.1000=\genfrac{}{}{0ex}{}{1001.1000}{1000\left(1000+1\right)}=1$

$A<\genfrac{}{}{0ex}{}{1001}{1000^2}.1000=\genfrac{}{}{0ex}{}{1001}{1000}=1+\genfrac{}{}{0ex}{}{1}{1000}<2$

Vậy $1<A<2\Rightarrow 1^2<A^2<2^2\Rightarrow 1<A^2<4$

Chúc bạn học tốt.

Tổng A có 1000 số hạng

A>(1001/1000^2+1000)*1000=1001*1000/1000*(1000+1)=1

A<(1001/1000^2)*1000=1001/1000=1+1/1000<1

Vậy 1<A<2 nên 1<A^2<4

1 dm³ = 1l nước
Thể tích nước cho phép: 
     1 . 0,5 . 0,4= 0,2 (m³)
Đổi: 0,2 m³ = 200 dm³
       200 dm³ = 200 l nước
Số can nước cần dùng:
      200 : 10 = 20 (can)
Vậy cần 20 can nước như vậy để đổ vào bể mực nước cho phép.

Thể tích bể cá là: 1 . 0,5 . 0,4 = 0,2 (m³) = 200l

Cần số can nước là: 200 : 10 = 20 (can)

Vậy cần 20 can nước như vậy đổ vào bể để được mực nước cho phép.

g, 2|\(x+1\)| +(y + \(x\))² = 0

       |\(x\) + 1| ≥ 0 ∀ \(x\)

       (y + \(x\))² ≥ 0 ∀ \(x\); y 

⇒2.|\(x\) +1| + (y + \(x\))² = 0 ⇔  \(\left\{{}\begin{matrix}x+1=0\\y+x=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

h, -2\(x^3\) + 4\(x\) = 0

    2\(x\).( -\(x^2\) + 2) =0

     \(\left[{}\begin{matrix}2x=0\\-x^2+2=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)

Vậy \(x\in\) {-\(\sqrt{2}\); 0; \(\sqrt{2}\)}

\(a,2\left|3x-1\right|+1=5\\ \Rightarrow2\left|3x-1\right|=5-1\\ \Rightarrow2\left|3x-1\right|=4\\ \Rightarrow\left|3x-1\right|=4:2\\ \Rightarrow\left|3x-1\right|=2\\ \Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(b,\left|\dfrac{x}{2}-1\right|=3\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-1=3\\\dfrac{x}{2}-1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{x}{2}=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

\(c,\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}\\ \Rightarrow\left|-x+\dfrac{2}{5}\right|=3\\ \Rightarrow\left[{}\begin{matrix}-x+\dfrac{2}{5}=3\\-x+\dfrac{2}{5}=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-x=\dfrac{13}{5}\\-x=-\dfrac{17}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{13}{5}\\x=\dfrac{17}{5}\end{matrix}\right.\)

\(d,\left|x-\dfrac{1}{3}\right|=2\dfrac{3}{5}\\ \Rightarrow\left|x-\dfrac{1}{3}\right|=\dfrac{13}{5}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{13}{5}\\x-\dfrac{1}{3}=-\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}+\dfrac{1}{3}\\x=-\dfrac{13}{5}+\dfrac{1}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{44}{15}\\x=-\dfrac{34}{15}\end{matrix}\right.\)