Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)
\(\Leftrightarrow x=-10\)
Vậy x = -10 là nghiệm của phương trình.
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)
\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)
\(\Leftrightarrow96x+744=-6x+48\)
\(\Leftrightarrow102x=-696\)
\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)
Vậy .....
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)
\(\Leftrightarrow x=-5\) (nhận)
Vậy ....
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
1.Tìm điều kiện xác định của phương trình:
a) 1x2+1 -4xx =0 (1)
b) 1x2−1 -2020 (2)
c) x2020x−2019 +
a) Dễ thấy: x2 + 1 ≠ 0 \(\forall\) x
Vậy điều kiện để phương trình (1) xác định là x ≠ 0.
b) Để phương trình (2) xác định thì x2 - 1 ≠ 0 ⇔ (x + 1)(x - 1) ≠ 0
⇔ \(\left[{}\begin{matrix}x+1\ne0\\x-1\ne0\end{matrix}\right.\) ⇔ x ≠ \(\pm\) 1
Vậy điều kiện để phương trình (2) xác định là x ≠ \(\pm\) 1.
c) Dễ thấy: x2 + 1 ≠ 0 \(\forall\) x
Vậy điều kiện để phương trình (3) xác định là x ≠ 2019.
`Answer:`
a. \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\left(ĐKXĐ:x\ne-1;x\ne-8\right)\)
\(\Leftrightarrow14\left(x+8\right)-14\left(x+1\right)=\left(x+1\right)\left(x+8\right)\)
\(\Leftrightarrow x^2+9x+8=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-15\end{cases}}}\)
b. \(\frac{2x}{x-2}+\frac{7}{x+2}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{7\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=2\)
\(\Leftrightarrow\frac{2x^2+4x+7x-14}{x^2-4}=2\)
\(\Leftrightarrow2x^2+11x-14=2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2+11x-12=2x^2-8\)
\(\Leftrightarrow11x-14=-8\)
\(\Leftrightarrow11x=6\)
\(\Leftrightarrow x=\frac{6}{11}\)
c. \(\frac{x+1}{2022}+\frac{x+2}{2021}=\frac{x+3}{2020}+\frac{x+4}{2019}\) (Câu này mình sửa lại đề nhé. Vì đề bạn cho sai hoặc thiếu.)
\(\Leftrightarrow\left(\frac{x+1}{2022}+1\right)+\left(\frac{x+2}{2021}+1\right)=\left(\frac{x+3}{2020}+1\right)+\left(\frac{x+4}{2019}+1\right)\)
\(\Leftrightarrow\frac{x+1+2022}{2022}+\frac{x+2+2021}{2021}=\frac{x+3+2020}{2020}+\frac{x+4+2019}{2019}\)
\(\Leftrightarrow\frac{x+2023}{2022}+\frac{x+2023}{2021}-\frac{x+2023}{2020}-\frac{x+2023}{2019}\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2022}+\frac{1}{2021}-\frac{1}{2020}-\frac{1}{2019}\right)=0\)
Do \(\frac{1}{2022}+\frac{1}{2021}-\frac{1}{2020}-\frac{1}{2019}\ne0\)
\(\Rightarrow x+2023=0\Leftrightarrow x=-2023\)