Thêm 2 vào pt có :

\(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)                (1)

\(\Leftrightarrow\frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1\)

\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\) (2)

\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\ne0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

k mk nha!

thanks

thanks

Lời giải:

PT $\Leftrightarrow \frac{x+16}{49}+1+\frac{x+18}{47}+1=\frac{x+20}{45}+1$

$\Leftrightarrow \frac{x+65}{49}+\frac{x+65}{47}=\frac{x+65}{45}$

$\Leftrightarrow (x+65)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0$

Thấy rằng $\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\neq 0$

Do đó $x+65=0\Rightarrow x=-65$

thank ạ

đúng là toán 8 khó thật nhìn mà hoa cả mắt *_*    T_T 

duyệt đi

chẳng hoa j cả

áp dụng tỉ lệ thức ta có :

\(\Leftrightarrow\frac{96x+1634}{2303}=\frac{x-25}{45}\Rightarrow\left(96x+1634\right)45=2303\left(x-25\right)\)

tự giải tiếp ra

=>x=-65

\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)

\(\Rightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\frac{51-x}{49}+1\)\(=-5+5\)

\(\Rightarrow\frac{59-x+49}{41}+\frac{57-x+43}{43}+\frac{55-x+45}{45}+\frac{53-x+47}{47}\)\(+\frac{51-x+49}{49}=0\)

\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)

\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)

Vì \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)

\(\Rightarrow100-x=0\)

\(\Rightarrow x=100\)

\(=\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\)

\(\frac{51-x}{49}+1=-5+5\)

đoạn này có 5 là do mik mượn 5 con 1 khi đó nha

\(=\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\)

\(\frac{100-x}{49}=0\)

\(=\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)

mà \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}< 0\)

nên 100-x=0 

còn lại bn từ lm

a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)

\(-18x^3+51x^2+9x-60=0\)

\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)

Bạn hỏi j dzậy!

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