![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4x}{x^2-1}\) (1)
điều kiện xác định: \(x\ne\pm1\)
(1) => \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{4x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1+x-1\right)\left(x+1-x+1\right)-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x.2-4x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{0x}{\left(x-1\right)\left(x+1\right)}=0\)
Vậy phương trình có nghiệm với mọi x \(\ne\pm1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\left(x\ne\pm5\right)\)
\(\Leftrightarrow\frac{x+5}{x-5}+\frac{x-5}{x+5}-\frac{2\left(x^2+25\right)}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{\left(x-5\right)\left(x+5\right)}+\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}-\frac{2x^2+50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{x^2+10x+25+x^2-10x+25-2x^2-50}{\left(x-5\right)\left(x+5\right)}=0\)
\(\Rightarrow\frac{0}{\left(x-5\right)\left(x+5\right)}=0\)
=> PT đúng với mọi x khác \(\pm5\)
Refund QB nhìn logic :V
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{x^2-25}\)
\(\frac{x+5}{x-5}+\frac{x-5}{x+5}=\frac{2\left(x^2+25\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\left(x+5\right)^2-\left(x-5\right)^2=2\left(x^2+25\right)\)
\(20x=2x^2+50\)
\(20x-2x^2-50=0\)
\(2\left(10x-x^2-25\right)=0\)
\(-x^2+10x+25=0\)
\(x^2-10x+25=0\)
\(x^2-2\left(x\right)\left(5\right)+5^2=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
2+4+...+2016+2018/1019090 = -3x² - 4x
Ta có: số số hạng tử của phân số 2+4+...+2016+2018/1019090 là:( 2018 - 2 ) : 2 + 1 = 1009 ( số hạng)
Tổng của tử đó là: ( 2018 + 2 ) . 1009 : 2 = 1019090
=> Ta được: 1019090/1019090 = -3x² - 4x
<=> -3x² - 4x = 1
<=> -3x² - 4x - 1 = 0
<=> -3x² - 3x - x - 1 = 0
<=> -3x( x + 1 ) -( x + 1 ) = 0
<=> ( x + 1 )( -3x - 1 ) = 0
<=> x + 1 = 0 hoặc -3x - 1 = 0
<=> x = -1 hoặc x = 1/-3
Vậy nghiệm phương trình là: S = { -1; -1/3 }
![](https://rs.olm.vn/images/avt/0.png?1311)
<=> \(x^2-25=10x+35-2x^2-7x\)
<=> \(3x^2-3x-60=0\)
<=> \(x^2-x-20=0\)
<=> \(\left(x-5\right)\left(x+4\right)=0\)
<=> \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
Vay \(x\in\left\{-4;5\right\}\)
Chuc ban hoc tot
![](https://rs.olm.vn/images/avt/0.png?1311)
(x2+5)(x-1)(2x+3)=0
<=> x2+5=0 hoặc x-1=0 hoặc 2x+3=0
<=> x2=-5(loại) hoặc x=1 hoặc 2x=-3
<=> x=1 hoặc x=-3/2
Vậy x=1; x=-3/2
Trả lời:
\(\left(x^2+5\right)\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\)\(x^2+5=0\)hoặc\(x-1=0\)hoặc\(2x+3=0\)
\(\Leftrightarrow\)\(x^2=-5\)hoặc \(x=1\)hoặc \(2x=-3\)
\(\Leftrightarrow\)\(x\in\varnothing\)(Vì\(x^2\ge0\)với \(\forall x\)) hoặc \(x=1\)hoặc \(x=\frac{-3}{2}\)
Vậy\(x=1\)hoặc \(x=\frac{-3}{2}\)
Hok tốt!
Bad boy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-2x=-x^2+2\)
\(\Leftrightarrow x^3+x^2-2x-2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-2\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có: \(x^3-2x=-x^2+2\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(2x+2\right)=0\)
\(\Leftrightarrow x^2.\left(x+1\right)-2.\left(x+1\right)=0\)
5\(\Leftrightarrow\left(x+1\right).\left(x^2-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\pm\sqrt{2}\end{cases}}\)
Vậy \(S=\left\{-\sqrt{2};-1;\sqrt{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\frac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)\(\Leftrightarrow x^2+x=0\)\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
So sánh với ĐKXĐ ta thấy: \(x=0\)không thoả mãn
Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)
Ta có: \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x.\left(x+2\right)-\left(x-2\right)}{\left(x-2\right).x}=\frac{2}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-x+2}{x^2-2x}=\frac{2}{x^2-2x}\)
\(\Rightarrow x^2+x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy \(S=\left\{-1;0\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đkxđ: \(\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\)
\(\frac{x+3}{x-2}+\frac{x+2}{x}=2\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x-2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)x}=\frac{2x\left(x-2\right)}{x\left(x-2\right)}\)
\(\Rightarrow x\left(x+3\right)+\left(x-2\right)\left(x+2\right)=2x\left(x-2\right)\)
\(\Leftrightarrow x^2+3x+x^2-4=2x^2-4x\)
\(\Leftrightarrow x^2+3x+x^2-2x^2+4x=4\)
\(\Leftrightarrow7x=4\)
\(\Leftrightarrow x=\frac{4}{7}\)
x2+4x-5=0
<=> x2-5x+x-5=0
<=> x(x-5)+(x-5)=0
<=> (x-5)(x+1)=0
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
lollolol là j