Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
=>x-99=0
hay x=99
7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)
=>x+100=0
hay x=-100
8:
Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)
=>200-x=0
hay x=200
Bài làm
\(2x.\left(x-3\right)=x-3\)
\(2x.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(2x-1\right).\left(x-3\right)=0\)
\(\orbr{\begin{cases}2x-1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=3\end{cases}}}\)
Vậy phương trình có 2 nghiêm \(x\in\left\{\frac{1}{2};3\right\}\)
Lời giải:
a. $9x^2-16-(3x-4)(2x+5)=0$
$\Leftrightarrow [(3x)^2-4^2]-(3x-4)(2x+5)=0$
$\Leftrightarrow (3x-4)(3x+4)-(3x-4)(2x+5)=0$
$\Leftrightarrow (3x-4)(3x+4-2x-5)=0$
$\Leftrightarrow (3x-4)(x-1)=0$
$\Leftrightarrow 3x-4=0$ hoặc $x-1=0$
$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=1$.
b.
$x^2+4x=12$
$\Leftrightarrow x^2+4x-12=0$
$\Leftrightarrow (x^2-2x)+(6x-12)=0$
$\Leftrightarrow x(x-2)+6(x-2)=0$
$\Leftrightarrow (x-2)(x+6)=0$
$\Leftrightarrow x-2=0$ hoặc $x+6=0$
$\Leftrightarrow x=2$ hoặc $x=-6$
c.
$x^2-2x=35$
$\Leftrightarrow x^2-2x-35=0$
$\Leftrightarrow (x^2+5x)-(7x+35)=0$
$\Leftrightarrow x(x+5)-7(x+5)=0$
$\Leftrightarrow (x+5)(x-7)=0$
$\Leftrightarrow x+5=0$ hoặc $x-7=0$
$\Leftrightarrow x=-5$ hoặc $x=7$
có sai đầu bài ko (x+36)/70 í
Ta có:\(\frac{x+25}{75}+\frac{x+36}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
⟺\(\frac{x+25}{75}+1+\frac{x+36}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
⟺\(\frac{x+100}{75}+\frac{x+100}{70}-\frac{x+100}{65}-\frac{x+100}{60}=0\)
⟺(x+100)(....)=0
⟺x=-100(vì cái trng ngoặc luôn khác 0)