![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)
\(\Leftrightarrow-2x^2+6x-4=0\)
a=-2; b=6; c=-4
Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:
\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình làm mẫu câu a nha
a, pt <=> ( x-2/7 - 1 ) + ( x-1/8 - 1 ) = ( x-4/5 - 1 ) + ( x-3/6 - 1 )
<=> x-9/7 + x-9/8 = x-9/5 + x-9/6
<=> x-9/5 + x-9/6 - x-9/7 - x-9/8 = 0
<=> (x-9).(1/5+1/6-1/9-1/8) = 0
<=> x-9 = 0 ( vì 1/5+1/6-1/9-1/8 > 0 )
<=> x = 9
Vậy x = 9
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m^2\left(x-1\right)+\left(m-2\right)x=1-2m\)
\(\Rightarrow\left(m^2+m-2\right)x-m^2=1-2m\)
\(\Rightarrow\left(m^2+m-2\right)x-m^2-\left(-2m\right)-1=0\)
\(\Rightarrow\left(\left(m+2\right)x-m+1\right)\left(m-1\right)=0\)
\(\Rightarrow\left(m+2\right)x-m+1=0\)
\(\Rightarrow m-1=0\)
\(\Rightarrow m+2=0\)
\(\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x-2=2\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Mình liệt kê nghiệm ra để bạn không nhầm :
\(\orbr{\begin{cases}m=-5\\x=2\end{cases}}\)
\(\orbr{\begin{cases}m=-3\\x=4\end{cases}}\)
\(\orbr{\begin{cases}m=-1\\x=-2\end{cases}}\)
\(\orbr{\begin{cases}m=1\\x=0\end{cases}}\)
\(\orbr{\begin{cases}m=-2\\x=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)đặt x^2-5x=y
<=> y^2+10y+24=0
<=>(y^2+2.5y+25)=1
<=>(y+5)^2=1
\(\left[\begin{matrix}y+5=1\\y+5=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}y=-4\\y=-6\end{matrix}\right.\)
với y=-4 <=> x^2-5x=-4<=> x(x-4)-(x-4)=0
<=> (x-4)(x-1)=0=>\(\left[\begin{matrix}x=1\\x=4\end{matrix}\right.\)
với y=-6<=> x^2-5x=-6<=> x(x-2)-3(x-2)=(x-2)(x-3)=>\(\left[\begin{matrix}x=2\\x=3\end{matrix}\right.\)
d) trôi hết đề bạn đăng quá nhiều
(x+2)(x+3)(x+4)(x+5)-24=0
<=>[(x+2)(x+5)][(x+3)(x+4)]-24=0
<=>(x^2+7x+10)(x^2+7x+12)-24=0
đặt x^2+7x+11=t
<=> (t-1)(t+1)-24=0
<=>t^2-1-25=0
<=>t^2=25=> t=+-5
với t=5
x^2+7x+11=5<=> x^2+7x+6=0
{a-b+c=0}=> x=-1 hoặc -6
với t=-5
x^2+7x+11=-5<=> x^2+7x+17=0=> vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
=( x\(^2\)+x)(x\(^2\)+x -2)=24
đặt x\(^2\)+ x= a\(\Rightarrow\)a(a-2)=24
chuển vế sang rồi tìm a, thay x vào rồi tìm x. tương tự mấy cau trên thui
Ta có : \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)(1)
Đặt \(t=x^2+x-1\Rightarrow\hept{\begin{cases}x^2+x=t+1\\x^2+x-2=t-1\end{cases}}\)
Suy ra pt \(\left(1\right)\Leftrightarrow\left(t+1\right)\left(t-1\right)=24\Leftrightarrow t^2-1=24\)
\(\Leftrightarrow t^2=25\Leftrightarrow\left(x^2+x-1\right)=25\)
\(\Leftrightarrow\hept{\begin{cases}x^2+x-1=5\\x^2+x-1=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+x-6=0\\x^2+x+4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\hept{\begin{cases}x+5=y\\x-4=z\end{cases}}\)
\(\Leftrightarrow2x+1=y+z\)
=> PT có dạng
\(y^4+z^4=\left(y+z\right)^4\)
\(\Rightarrow y^4+z^4=y^4+4y^3z+6y^2z^2+4yz^3+z^4\)
\(\Leftrightarrow2yz\left(2y^2+3yz+2z^2\right)=0\)
\(\Leftrightarrow2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)
Dễ thấy \(7x^2+7x+22=7\left(x+\frac{1}{2}\right)^2+\frac{81}{4}>0\)
Từ \(2\left(x+5\right)\left(x-4\right)\left(7x^2+7x+22\right)=0\)(1)
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=4\end{cases}}\)
Vậy .....
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,4x^2+(x-1)^2-(2x+1)^2=0`
`<=>4x^2+3x(-x-2)=0`
`<=>x(4x-3x-6)=0`
`<=>x(x-6)=0`
`<=>` $\left[ \begin{array}{l}x=0\\x=6\end{array} \right.$
`b)(x^2-3x)^2+5(x^2-3x)+6=0`
Đặt `x^2-3x=a(a>=-9/4)`
`pt<=>a^2+5a+6=0`
`<=>(a+2)(a+3)=0`
`<=>` $\left[ \begin{array}{l}a=-2\\a=-3(l)\end{array} \right.$
`<=>x^2-3x=-2`
`<=>x^2-3x+2=0`
`<=>(x-1)(x-2)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=1\end{array} \right.$
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
Đặt \(x^2+x-1=a\)
Ta có : \(x^2+x-1=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
\(\Rightarrow a\ge-\frac{5}{4}\)
Ta có pt : \(\left(a+1\right)\left(a-1\right)=24\)
\(\Leftrightarrow a^2-1=24\)
\(\Leftrightarrow a^2=25\)
\(\Leftrightarrow a=5\left(Do\text{ }a\ge-\frac{5}{4}\right)\)
\(\Leftrightarrow x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)