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\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4}{\left(x-2\right)\left(x+3\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{-3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)
<=> 4x-16=-3x+6
<=> 4x-16+3x-6=0
<=> 7x-22=0
<=> 7x=22
<=> \(x=\frac{22}{7}\)(TMĐK)
ĐẶt x+1/x = m
suy ra x2+1/x2=m2-2
Vậy m2-2+9/2m+7=0
2m2+9m+10=0
(2m2+4m) +(5m+10)=0
2m(m+2)+5(m+2)=0
\(\Leftrightarrow\orbr{\begin{cases}m+2=0\\2m+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=-2\\m=\frac{-5}{2}\end{cases}}\)
Với m=-2
x+1/x=-2 hay x2+2x+1=0
x=-1
Với m=-5/2 làm tương tự
ĐKXĐ: x\(x\ne\)1,-1
a) pt <=> \(\left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{x^2-1}=\frac{10}{9}\)
<=> \(\frac{4x^4}{\left(x^2-1\right)^2}-\frac{2x^2}{x^2-1}=\frac{10}{9}\)
Đặt: t=\(\frac{2x^2}{x^2-1}\)
Pt trở thành: \(t^2-t-\frac{10}{9}=0\)\(\Leftrightarrow9t^2-9t-10=0\)<=> \(\orbr{\begin{cases}t=-\frac{1}{3}\\t=\frac{5}{6}\end{cases}}\)
Nếu: \(\frac{2x^2}{x^2-1}=-\frac{1}{3}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{1}{7}}\\x=-\sqrt{\frac{1}{7}}\end{cases}\left(tm\right)}\)
Nếu: \(\frac{2x^2}{x^2-1}=\frac{5}{6}\)(vô nghiệm)
Vậy nghiệm là ...
http://vchat.vn/pictures/service/2017/02/iit1486637364.PNG
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)
<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
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
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
\(ĐKXĐ:\) \(x\ne0\)
Đặt \(x+\frac{1}{x}=y\) \(\left(\text{*}\right)\), thì khi đó \(x^2+\frac{1}{x^2}=y^2-2\)
Do đó, \(y^2-2-\frac{9}{2}y+7=0\)
\(\Leftrightarrow\) \(y^2-\frac{9}{2}y+5=0\)
\(\Leftrightarrow\) \(2y^2-9y+10=0\)
\(\Leftrightarrow\) \(2y^2-4y-5y+10=0\)
\(\Leftrightarrow\) \(2y\left(y-2\right)-5\left(y-2\right)=0\)
\(\Leftrightarrow\) \(\left(y-2\right)\left(2y-5\right)=0\)
\(\Leftrightarrow\) \(^{y-2=0}_{2y-5=0}\) \(\Leftrightarrow\) \(^{y=2}_{y=\frac{5}{2}}\)
\(\text{*)}\) Với trường hợp \(y=2\) thì khi đó, \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=2\) \(\left(1\right)\)
Vì \(x\ne0\) nên từ \(\left(1\right)\) suy ra \(x^2+1=2x\) \(\Leftrightarrow\) \(x^2-2x+1=0\) \(\Leftrightarrow\) \(\left(x-1\right)^2=0\) \(\Leftrightarrow\) \(x-1=0\) \(\Leftrightarrow\) \(x=1\) ( thỏa mãn điều kiện xác định)
\(\text{*)}\) Với \(y=\frac{5}{2}\) thì \(\left(\text{*}\right)\) \(\Rightarrow\) \(x+\frac{1}{x}=\frac{5}{2}\) \(\left(2\right)\)
Từ \(\left(2\right)\) \(\Rightarrow\) \(2x^2+2=5x\) (do \(x\ne0\) )
\(\Leftrightarrow\) \(2x^2-5x+2=0\)
\(\Leftrightarrow\) \(2x^2-4x-x+2=0\)
\(\Leftrightarrow\) \(2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\) \(\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\) \(^{x-2=0}_{2x-1=0}\) \(\Leftrightarrow\) \(^{x=2}_{x=\frac{1}{2}}\) (t/mãn điều kiện xác định)
Vậy, \(S=\left\{1;2;\frac{1}{2}\right\}\)