Bài 2: Cộng, trừ số hữu tỉ

Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow5x=2\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy ...

Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x=2\left(1-x\right)\)

\(\Leftrightarrow3x=2-2x\)

\(\Leftrightarrow3x+2x=2\)

\(\Leftrightarrow5x=2\)

hay \(x=\dfrac{2}{5}\)

Vậy: \(x=\dfrac{2}{5}\)

x:2=y:3->x/2=y/3

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

x/2=y/3=x+y/2+3=10/5=2

từ: x/2=2->x=2.2=4

y/3=2->y=2.3=6

vậy...

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\) (theo tính chất củadãy tỉ số bằng nhau)

\(\Rightarrow x=4,y=6\)

Ta có: \(x:2=y:3\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=10

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=2\\\dfrac{y}{3}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

Vậy: x=4; y=6

Ta có  \(\dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\), \(\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\),.., \(\dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Xét hiệu \(A-\dfrac{7}{12}=\dfrac{99}{100}-\dfrac{7}{12}=\dfrac{99\cdot3-7\cdot25}{300}=\dfrac{122}{300}>0\) \(\Rightarrow A>\dfrac{7}{12}\)(1)

\(A-\dfrac{5}{6}=\dfrac{99}{100}-\dfrac{5}{6}=\dfrac{99\cdot3-125\cdot5}{300}=\dfrac{-328}{300}< 0\) \(\Rightarrow A< \dfrac{5}{6}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)

\(=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}\right)+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)\)

\(=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)\)

\(=\dfrac{7}{12}+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)>\dfrac{7}{12}\)

\(\Leftrightarrow\dfrac{7}{12}< A\)(1)

Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)

\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{70}\right)+...+\left(\dfrac{1}{91}+\dfrac{1}{92}+\dfrac{1}{93}+...+\dfrac{1}{100}\right)\)

\(\Leftrightarrow A< \dfrac{1}{50}\cdot10+\dfrac{1}{60}\cdot10+\dfrac{1}{70}\cdot10+\dfrac{1}{80}\cdot10+\dfrac{1}{90}\cdot10\)

\(\Leftrightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)

\(\Leftrightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\cdot3\)

\(\Leftrightarrow A< \dfrac{167}{210}< \dfrac{175}{210}=\dfrac{5}{6}\)

hay \(A< \dfrac{5}{6}\)(2)

Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)(đpcm)

Ta có: \(A=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+\dfrac{5}{14\cdot19}+...+\dfrac{5}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\left(\dfrac{49-4}{4\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(\Leftrightarrow A=\dfrac{9}{196}\cdot\dfrac{-623}{89}=-\dfrac{9}{28}\)

ĐK: \(x \ne -2\)

\(\dfrac{x+2}{2}=\dfrac{8}{x+2} \\ \Leftrightarrow (x+2)^2=8.2 \\ \Leftrightarrow (x+2)^2=4^2=(-4)^2 \\ \Leftrightarrow\left[ \begin{array}{l}x+2=4\\x+2=-4\end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l}x=2\\x=-6\end{array} \right. \)

Vậy \(x=2;x=-6\).

Giúp mik vs mik đang cần gấp

cái....đánh máy xong công thức xong rồi đến lúc chèn vào nó lỗi, cáu:vv

\(\dfrac{x+2}{2}=\dfrac{8}{x+2}\Rightarrow\left(x+2\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)^2=4^2\\\left(x+2\right)^2=\left(-4\right)^2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy...

tʊS⚽☛Gåmʌʌ❕ŋG

=>3^x(1+3^2)=3^34+3^36

=>3^x.10=3^34.10

=>3^x=3^34

=>x=34

4 mũ7 nhân 8 mũ 3 phần 2 mũ 20

(x-2020)^2021=1

(x-2020)^2021=1^2021

=>x-2020=1

x=1+2020

x=2021

Vậy x=2021

Ta có: \(\left(x-2020\right)^{2021}=1\)

\(\Leftrightarrow\left(x-2020\right)^{2021}=1^{2021}\)

\(\Leftrightarrow x-2020=1\)

hay x=2021

Vậy: x=2021

x=2021