Nghiệm của phương trình 148-x/13+169-x/17+186-x/17+199-x/16=10. vậy x=
Tìm nghiệm của phương trình sau:
\(\frac{148-x}{13}+\frac{169-x}{17}+\frac{186-x}{17}+\frac{199-x}{16}=10\)
\(\frac{148-x}{13}-1+\frac{169-x}{17}-2+\frac{186-x}{17}-3+\frac{199-x}{16}-4=0\)\(\frac{135-x}{13}+\frac{135-x}{17}+\frac{135-x}{17}+\frac{135-x}{16}=0\)
(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 \(x=135\)
Tìm nghiệm của phương trình:
\(\frac{148-x}{13}+\frac{169-x}{17}+\frac{186-x}{17}+\frac{199-x}{16}=10\)
\(\left(\frac{148-x}{13}-1\right)+\left(\frac{169-x}{17}-2\right)+\left(\frac{186-x}{17}-3\right)+\left(\frac{199-x}{16}-4\right)=10-1-2-3-4\)
VT có tử =(135-x) VP=0
Vậy: x=135
Câu 1. Tìm các số x, y thỏa mãn đẳng thức: 3x2+ 3y2 + 4xy + 2x - 2y + 2 = 0
Câu 2. Tìm x để: \(\dfrac{148-x}{13}\)+\(\dfrac{169-x}{17}\)+\(\dfrac{186-x}{17}\)+\(\dfrac{199-x}{16}\)=10
148-x/29+169-x/23+186-x/21+199-x/19=10
Tìm x:
(148-x)/25+(169-x/23+(186-x)/21+(199-x)/10=10
148-x/25-169-x/23+186-x/4+199-x/19=10.tìm x
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
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=0\) 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
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
Mk chi p bang 123 vi bam may tinh, con ck jai thi hk p!