148-x/25-169-x/23+186-x/4+199-x/19=10.tìm x
Tìm x:
(148-x)/25+(169-x/23+(186-x)/21+(199-x)/10=10
Tìm x: \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
148-x/25-1 + 169-x/23-2 + 186-x/21-3 + 199-x/19-4
123-x/25 + 123-x/23 + 123-x/21 + 123-x/19 =0
123-x=0 => x=123
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
\(\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)
=> \(\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)
=> \(\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
=> 123 - x = 0
=> x = 123
duongtiendung vế bên trái có thêm -1,-2,-3,-3 thì bên vế phải ,phải là 0+(-1)+(-2)+(-3)+(-4)
=-10 chứ = 0 sao đc
148-x/29+169-x/23+186-x/21+199-x/19=10
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!
GIÚP MK VS Ạ!!!
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}-\frac{199-x}{19}=10\)
Câu cuối r , giúp mk vs nha
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}-\frac{199-x}{19}=10\)
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=0\) giải phương trình
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
\(\dfrac{148-x}{25}\) + \(\dfrac{169-x}{23}\) + \(\dfrac{186-x}{21}\) +\(\dfrac{199-x}{19}\) =10
\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)
\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)
\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)
\(\Leftrightarrow123-x=0\Leftrightarrow x=123\)
Vậy x = 123
Tìm x biết
\(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}\)=0