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

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

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

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

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

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

<=> x - 89 = 0   <=> x = 89

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xin lỗi mk mới hok lớp 5

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

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

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

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)

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

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

\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)

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

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

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

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

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

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

\(\Leftrightarrow x-89=0\)

\(\Leftrightarrow x=89\)

Vậy x = 89

gfffrgfrff

mik ko biết

mà nhìn như toán nâng cao ý

Lời giải:

a)

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

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

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

Dễ thấy \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}<0\) nên $x+100=0$

$\Rightarrow x=-100$

b)

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

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

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

Dễ thấy \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}< 0\)

\(\Rightarrow x-89=0\Rightarrow x=89\)

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

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

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

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

=> \(\left(x-89\right).\left[\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)\right]=0\Rightarrow x-89=0\left(\text{vì }\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0\right)\)

=> x = 89

Vậy x = 89

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

\(\Rightarrow\dfrac{x-12}{77}+\dfrac{x-11}{78}-\dfrac{x-74}{15}-\dfrac{x-73}{16}=0\)

\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1-\dfrac{x-74}{15}+1-\dfrac{x-73}{16}+1=0+1+1-1-1\)

\(\Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)-\left(\dfrac{x-74}{15}-1\right)-\left(\dfrac{x-73}{16}-1\right)=0\)

\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)

\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-89=0\\\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\end{matrix}\right.\)

\(x-89=0\\ \Rightarrow x=89\)

\(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\)(vô lí)

Vậy \(x=89\)

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

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

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

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

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

\(\Rightarrow x-89=0\)

\(\Rightarrow x=89\)

Vậy x = 89