g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=0-100\)

\(\Leftrightarrow x=-100.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)

h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)

\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)

\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)

\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)

Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)

\(\Leftrightarrow x-89=0\)

\(\Leftrightarrow x=0+89\)

\(\Leftrightarrow x=89.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)

Chúc bạn học tốt!

Câu g) bạn cộng 1 vào mỗi hạng tử của 2 vế

Câu h) bạn trừ một vào mỗi hạng tử ở hai vế

Quy đồng mẫu thì được tử giống nhau sau đó đặt nhân tử chung là xong

x=8.

y=12.

\(x=\frac{903}{391}\)

Bài này sử dụng MTCT đó bạn!

903/391 mình nghĩ vậy

x = 903/391

\(\frac{x-2}{99}+\frac{x-3}{98}=\frac{x-4}{97}+\frac{x-5}{96}\Leftrightarrow\frac{x-2}{99}-1+\frac{x-3}{98}-1=\frac{x-4}{97}-1+\frac{x-5}{96}-1\)

<=>\(\frac{x-101}{99}+\frac{x-101}{98}=\frac{x-101}{97}+\frac{x-101}{96}\)

<=>\(\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

<=>x-101=0 \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\right)\)

<=>x=101

101

Ta có: \(x+\frac{x}{3}=96-12.5\)

\(\Rightarrow\frac{2x}{3}=96-60\)

\(\Rightarrow2x:3=36\)

\(\Rightarrow2x=36.3=108\)

\(\Rightarrow x=108:2=54\)

Vậy x = 54

\(x+\frac{x}{3}=96-60=36\)

\(x+x\times\frac{1}{3}=36\)

\(x\times\frac{4}{3}=36\)

\(x=36:\frac{4}{3}=27\)

\(x+\frac{x}{3}=96-60\)

\(\Leftrightarrow x.1+x.\frac{1}{3}=36\)\(\Leftrightarrow x.\left(1+\frac{1}{3}\right)=36\)

\(\Leftrightarrow x.\frac{4}{3}=36\Rightarrow x=36:\frac{4}{3}\)\(\Rightarrow x=27\)

a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)

(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)

\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)

\(-x^3-x^2+9x+9=0\)

\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)

\(\left(x+1\right)\left(9-x^2\right)\)=0

(x+1)(3-x)(3+x)=0

*x+1=0 =>x=-1

*3-x=0=>x=3

*3+x=0=>x=-3

       \(\frac{x-144}{10}+\frac{x-130}{12}+\frac{x-112}{14}+\frac{x-106}{16}+\frac{x-96}{17}=15\)

\(\Leftrightarrow\)\(\frac{x-144}{10}-1+\frac{x-130}{12}-2+\frac{x-112}{14}-3+\frac{x-106}{16}-4+\frac{x-96}{17}-5=0\)

\(\Leftrightarrow\)\(\frac{x-154}{10}+\frac{x-154}{12}+\frac{x-154}{14}+\frac{x-154}{16}+\frac{x-154}{17}=0\)

\(\Leftrightarrow\)\(\left(x-154\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

\(\Leftrightarrow\)\(x-154=0\)  (do   1/10 + 1/12 + 1/14 + 1/16 + 1/17  khác   0)

\(\Leftrightarrow\)\(x=154\)

Vậy...

Hình như đề có gì sai nha bạn.

\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+4}{96}+\frac{x+1}{99}=-4\)

\(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+1}{99}+1\right)=-4+4\)

\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{96}+\frac{x+100}{99}=0\)

\(\left(x+100\right).\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\right)=0\)

=> \(\orbr{\begin{cases}x+100=0\\\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}=0\end{cases}}\)

Mà \(\frac{1}{97}+\frac{1}{95}+\frac{1}{96}+\frac{1}{99}\ne0\)

=> x + 100 = 0

=> x = -100

Vậy x = -100

Câu b trừ mỗi số đi 1 tức là trừ cả cụm đó cho 3 rùi lm tương tự câu a

a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)

\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)

\(\Rightarrow\)\(x+102=0\)

\(\Rightarrow x=-102\)

c) \(\left(x+2\right)-\left(x+3\right)>0\)

\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)

\(\Rightarrow x\in\varnothing\)

d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)

TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\ge\dfrac{-7}{3}\)

TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\le5\)

TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)