\(\dfrac{x+5}{97}+\dfrac{x+5}{98}+\dfrac{x+5}{99}=0\\ \Rightarrow\left(x+5\right).\left(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\right)=0\)
Vì \(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\ne0\) nên:
\(x+5=0\\ \Rightarrow x=-5\)
Vậy...

\(\dfrac{x+5}{97}+\dfrac{x+5}{98}+\dfrac{x+5}{99}=0\)

=>\(\left(x+5\right)\left(\dfrac{1}{97}+\dfrac{1}{98}+\dfrac{1}{99}\right)=0\)

=>x+5=0

=>x=-5

\(2\left(x-3\right)-\left(4x-1\right)=0\)

\(2x-6-4x+1=0\)

\(-2x-5=0\)

\(2x=-5\)

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

\(2\cdot\left(x-3\right)-\left(4\cdot x-1\right)=0\\ \Rightarrow2x-6-4x+1=0\\ \Rightarrow\left(2x-4x\right)+\left(-6+1\right)=0\\ \Rightarrow-2x-5=0\\ \Rightarrow-2x=5\\ \Rightarrow x=-\dfrac{5}{2}\)

Ý bạn là sao nhỉ???

Bn viết sai hoặc là viết thiếu câu hỏi r.

Chớ mik đọc hổng hỉu j hớt á

Sửa đề: \(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\)

\(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

\(=\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\)

\(=\dfrac{2}{7}-1:\dfrac{7}{2}=\dfrac{2}{7}-\dfrac{2}{7}=0\)

a: ΔDEF đều

=>DE=DF=EF và \(\widehat{DEF}=\widehat{EDF}=\widehat{DFE}=60^0\)

EM là phân giác của góc DEF

=>\(\widehat{DEM}=\widehat{FEM}=\dfrac{\widehat{DEF}}{2}=30^0\)

Ta có: ΔDEP vuông tại D

=>\(\widehat{DEP}+\widehat{DPE}=90^0\)

=>\(\widehat{DPE}=90^0-60^0=30^0\)

Xét ΔNEP có \(\widehat{NEP}=\widehat{NPE}\left(=30^0\right)\)

nên ΔNEP cân tại N

b: Xét ΔDEN và ΔFEN có

DE=FE

\(\widehat{DEN}=\widehat{FEN}\)

EN chung

Do đó: ΔDEN=ΔFEN

=>\(\widehat{EDN}=\widehat{EFN}\)

=>\(\widehat{EFN}=90^0\)

=>NF\(\perp\)EP

c: ΔNEP cân tại N

mà NF là đường cao

nên F là trung điểm của EP

Ox là phân giác của góc BOA

=>\(\widehat{xOA}=\dfrac{\widehat{BOA}}{2}\)

Oy là phân giác của góc COA

=>\(\widehat{yOA}=\dfrac{\widehat{COA}}{2}\)

\(\widehat{xOy}=\widehat{xOA}+\widehat{yOA}=\dfrac{1}{2}\left(\widehat{BOA}+\widehat{COA}\right)\)

\(=\dfrac{1}{2}\cdot180^0=90^0\)

=>Ox\(\perp\)Oy

a.

Để A là phân số

\(\Rightarrow x+7\ne0\)

\(\Rightarrow x\ne7\)

b.

Để P nguyên \(\Rightarrow-\dfrac{3}{x+7}\) là số nguyên

\(\Rightarrow3\) chia hết `x+7`

\(\Rightarrow x+7\) là ước của 3

\(\Rightarrow x+7=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow x=\left\{-10;-8;-6;-4\right\}\)

c.

\(P=-\dfrac{2}{3}\Rightarrow-\dfrac{3}{x+7}=-\dfrac{2}{3}\)

\(\Rightarrow\left(-3\right).\left(-3\right)=2.\left(x+7\right)\)

\(\Rightarrow9=2x+14\)

\(\Rightarrow2x=-5\)

\(\Rightarrow x=-\dfrac{5}{2}\)

Cảm ơn bạn nghen !

\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(2x=0\) hoặc \(x-\dfrac{1}{7}=0\)

\(x=0\) hoặc \(x=\dfrac{1}{7}\)

Ta có:
+)
 \(\dfrac{2023.2024-1}{2023.2024}\\ =\dfrac{2023.2024}{2023.2024}-\dfrac{1}{2023.2024}\\ =1-\dfrac{1}{2023.2024}\)

+)
​ \(\dfrac{2022.2023-1}{2022.2023}\\ =\dfrac{2022.2023}{2022.2023}-\dfrac{1}{2022.2023}\\ =1-\dfrac{1}{2022.2023}\)

Nhận xét:
Vì \(2023.2024>2022.2023\) nên:
\(\dfrac{1}{2023.2024}< \dfrac{1}{2022.2023}\\\Rightarrow1-\dfrac{1}{2023.2024}>1-\dfrac{1}{2022.2023}\)
hay \(\dfrac{2023.2024-1}{2023.2024}>\dfrac{2022.2023-1}{2022.2023}\)
Vậy...

\(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{32}\right)^{16}\\ =>\left[\left(\dfrac{1}{2}\right)^4\right]^x=\left[\left(\dfrac{1}{2}\right)^5\right]^{16}\\ =>\left(\dfrac{1}{2}\right)^{4\cdot x}=\left(\dfrac{1}{2}\right)^{5\cdot16}\\ =>\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{80}\\ =>4x=80\\ =>x=\dfrac{80}{4}\\ =>x=20\)

Vậy: ..