\(\frac{148-x}{13}-1+\frac{169-x}{17}-2+\frac{186-x}{17}-3+\frac{199-x}{16}-4=0\)\(\frac{135-x}{13}+\frac{135-x}{17}+\frac{135-x}{17}+\frac{135-x}{16}=0\)

(135-x)(\(\frac{1}{13}+\frac{1}{17}+\frac{1}{17}+\frac{1}{16}\))=0

135-x=0

x=135

Có : \(\frac{148-x}{13}+\frac{169-x}{17}+\frac{186-x}{17}+\frac{199-x}{16}=10\)

\(\Leftrightarrow\)\(\left(\frac{148-x}{13}-1\right)+\)\(\left(\frac{169-x}{17}-2\right)+\)\(\left(\frac{186-x}{17}-3\right)\) + \(\left(\frac{199-x}{16}-4\right)=10\)

\(\Leftrightarrow\) \(\frac{135-x}{13}+\frac{135-x}{17}+\frac{135-x}{17}+\frac{135-x}{16}\)= 10

\(\Leftrightarrow\) \(\left(135-x\right)\left(\frac{1}{13}+\frac{1}{17}+\frac{1}{17}+\frac{1}{16}\right)=0\)

\(\Leftrightarrow\) \(135-x=0\) \(\left(\frac{1}{13}+\frac{1}{17}+\frac{1}{17}+\frac{1}{16}\right)\ne0\)

\(\Leftrightarrow\) \(x=135\)

Vậy \(x=135\)

x=135

cộng thêm 1 vào mỗi vế là ra ấy mà. bạn động não chút đi

Bạn xem lại có sai đề ko,mk thấy sao sao ý

Sửa đề:

\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\\\Leftrightarrow \frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=0\\ \Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\\ \Leftrightarrow\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\\ \Leftrightarrow123-x=0\left(Vi\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\right)\\ \Leftrightarrow x=123\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{123\right\}\)

=10 chứ ko phải bằng 0 nha bạn

Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!

CÁM ƠN bn tuy chưa có lời jai n cx đk

\(\Leftrightarrow\frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=0\)

\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)

\(\Rightarrow123-x=0\Rightarrow x=123\)

Vậy Tập nghiệm của phương trình là \(S=\left\{123\right\}\)

<=> 148-×/25 -1 + 169-x/23 -2 + 186-x/21 - 3 + 199-×/19 - 4=0  

<=>  (123-x)(1/25+1/23+1/21+1/19)=0

<=> x=123

Chúc bạn học tốt

chị lớp 9 chứ ko phải lớp 8 ok

\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)

\(\Leftrightarrow\frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=0\)

\(\Leftrightarrow\frac{148-x}{25}-\frac{25}{25}+\frac{169-x}{23}-\frac{46}{23}+\frac{186-x}{21}-\frac{63}{21}+\frac{199-x}{19}-\frac{76}{19}=0\)

\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right).\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

\(\Leftrightarrow123-x=0\left(\text{vì }\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\right)\)

<=>x=123

Vậy S={123}

a) \(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)

\(\Leftrightarrow\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

\(\Leftrightarrow x=123\)

c) \(x^4-10.2^x+16=0\)

\(\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Ta có: 

\(2^x=t\)

\(\Rightarrow t^2-10t+16=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)