Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha
.
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.
Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)
<=>1/x+2 -1/x+6 +1/x+6 -1/x+13
<=>1/x+2-1/x+13
=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)
<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)
<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)
<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)
<=> 2x+1=14
<=> 2x=14-1
<=> 2x=13
<=> x=13:2
<=> x=13/2
Vậy x=13/2
Chúc bạn học tốt
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt t = x2 + x
<=> t(t - 2) - 24 = 0
<=> t2 - 2t - 24 = 0
<=> t2 - 6t + 4t - 24 = 0
<=> (t + 4)(t - 6) = 0
<=> \(\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\\x^2+3x-2x-6=0\end{cases}}\)
<=> \(\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x-2\right)\left(x+3\right)=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy S = {2; -3}
(lưu ý: thay "ktm" thành vô lý và giải thích thêm)
\(\left(x+3\right)^4+\left(x+5\right)^4=2\)
<=> (x + 4 - 1)4 + (x + 4 + 1)4 - 2 = 0
Đặt y = x + 4
<=> (y - 1)4 + (y + 1)4 - 2 = 0
<=> y4 - 4y3 + 6y2 - 4y + 1 + y4 + 4y3 + 6y2 + 4y + 1 - 2 = 0
<=> 2y4 + 12y2 = 0
<=> 2y2(y2 + 6) = 0
<=> \(\orbr{\begin{cases}y^2=0\\y^2+6=0\left(ktm\right)\end{cases}}\)
<=> y = 0
<=> x + 4 = 0
<=> x = -4
Vậy S = {-4}
\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)
<=> \(\frac{x^2+x+4}{2}-3+\frac{x^2+x+7}{3}-3=\frac{x^2+x+13}{5}-3+\frac{x^2+x+16}{6}-3\)
<=> \(\frac{x^2+x+4-6}{2}+\frac{x^2+x+7-9}{3}=\frac{x^2+x+13-15}{5}+\frac{x^2+x+16-18}{6}\)
<=> \(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)
<=> \(\left(x^2+2x-x-2\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=0\)
<=> (x + 2)(x - 1) = 0 (do \(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\ne0\))
<=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
Vậy S = {-2; 1}
câu cuối: + 3 vào sau các phân số của pt như trên