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 $x+65=0\Rightarrow x=-65$

c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)

\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)

\(\Rightarrow42x-6=60+18-10x\)

\(\Leftrightarrow52x-84=0\)

\(\Leftrightarrow x=\dfrac{21}{13}\)

Vậy....

d) tương tự

a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)

\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )

Vậy....

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

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

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

\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)

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

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

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

Vậy....

bạn chỉ mình cách gõ phần như vậy đi mình mới chỉ bạn lời giải được

<=> (x+16/49 +2)+(x+18/47 +2)= (x+20/45 +2)-1

<=>x+65/49 + x+65/47 - x+65/45 +1=0

<=>(x+65)(1/49 + 1/47 - 1/45 + 1)=0

<=>x+65=0 (vì 1/49 + 1/47 - 1/45 + 1 khác 0)

<=>x= -65

Vậy S={-65}

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

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

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

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

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

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

\(\Rightarrow x+65=0\)

\(\Rightarrow x=-65\)

Bài này dễ mà bạn để ý một chút là giải ra ngay.

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

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

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

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

a, <=> (59-x/41 + 1) + (57-x/43 + 1) + (55-x/45 + 1) + (53-x/47 + 1) + (51-x/49 + 1) = 0

<=> 100-x/41 + 100-x/43 + 100-x/45 + 100-x/47 + 100-x/49 = 0

<=> (100-x).(1/41+1/43+1/45+1/47+1/49) = 0

<=> 100-x=0 ( vì 1/41+1/43+1/45+1/47+1/49 > 0 )

<=> x=100

Vậy x = 100

b, <=> 2-x/2016 + 1 = (1-x/2017 + 1) + (1 - x/2018)

<=> 2018-x/2016 = 2018-x/2017 + 2018-x/2018

<=> 2018-x/2016 - 2018-x/2017 - 2018-x/2018 = 0

<=> (2018-x).(1/2016-1/2017-1/2018) = 0

<=> 2018-x=0 ( vì 1/2016-1/2017-1/2018 khác 0 )

<=> x=2018

Vậy x=2018

Tk mk nha