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26 tháng 1 2018
\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\) <=>\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\) <=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\) <=>\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\) <=> (x+95)(\(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\)) vì \(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\) khác 0 =>x+95=0=>x=-95 chức bạn học tốt ^ ^
16 tháng 3 2017

Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)

Quy đồng mẫu số :

\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)

\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)

Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)

16 tháng 3 2017

Cảm ơn!

6 tháng 1 2018

\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\)

\(\Rightarrow\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)

\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\)

\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)

\(\Rightarrow\left(x+95\right)\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)=0\)

\(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\ne0\) nên \(x+95=0\Leftrightarrow x=-95\)

6 tháng 1 2018

Mk làm luôn nhé , không chép lại đề đâu !!! Ahihi

\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)

\(\left(x+95\right)\)\(\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)\) = 0

\(x+95=0\)

\(x=-95\)

Vậy , ......

31 tháng 12 2022

a: \(\Leftrightarrow4x+4+9\left(2x+1\right)=2\left(5x+3\right)+12x+7\)

=>4x+4+18x+9=10x+6+12x+7

=>22x+13=22x+13(luôn đúng)

b: \(\Leftrightarrow\left(\dfrac{x+1}{94}+1\right)+\left(\dfrac{x+2}{93}+1\right)+\left(\dfrac{x+3}{92}+1\right)=\left(\dfrac{x+4}{91}+1\right)+\left(\dfrac{x+5}{90}+1\right)+\left(\dfrac{x+6}{89}+1\right)\)

=>x+95=0

=>x=-95

=>360+57<10x<58x4+59x5

=>417<10x<527

\(\Leftrightarrow10x\in\left\{420;430;440;...;510;520\right\}\)

hay \(x\in\left\{42;43;44;...;51;52\right\}\)

a: Ta có: \(-\left(x+5\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x+5\right)^2+2021\le2021\forall x\)

Dấu '=' xảy ra khi x=-5

17 tháng 8 2021

Xĩn lỗi nha nhma mình chx hiểu lắm bạn trình bày rõ ra hơn đc ko huhu:((

 

21 tháng 7 2023

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(100-10\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=1.2=2\)

\(\Rightarrow\left(x+\dfrac{206}{100}\right)=\dfrac{5}{2}:2=\dfrac{5}{2}.\dfrac{1}{2}=\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{5}{4}-\dfrac{206}{100}=\dfrac{125}{100}-\dfrac{206}{100}\)

\(\Rightarrow x=-\dfrac{81}{100}\)