3/x-2/+/4x-8/=/-2/-/1/3/
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)

`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`

https://olm.vn/hoi-dap/detail/227952918582.html vào link này xem câu a nha Lê Phương Nhung
b)Q = (x - 1)3 - 4x(x + 1)(x - 1) + 3(x - 1)(x2 + x + 1)
Q = (x - 1)3 - 4x(x2 - 1) + 3(x3 - 1)
Thay x = -2 vào Q ta dc :
(-3)3 - 4 . (-2) . 3 + 3 . (-9) = -27 + 24 - 27 = -30

1) 2(3x + 5) - 6 = 0
2(3x + 5) = 6 + 0
2(3x + 5) = 6
3x + 5 = 6 : 2
3x + 5 = 3
3x = 5 - 3
3x = 2
x = 2 : 3
x = 2/3
2) 5x + 3(4 + 2x) = 25
11x + 12 = 25
11x = 25 - 12
11x = 13
x = 13 : 11
x = 13/11
3) 3(4x + 1) + 2(x - 1) = 105
14x + 1 = 105
14x = 105 - 1
14x = 104
x = 104 : 14
x = 104/14 = 52/7
4) 30 - [2(x - 3) - 2] = 14
[2(x - 3) - 2] = 30 - 14
[2(x - 3) - 2] = 16
2x - 8 = 16
2x = 16 + 8
2x = 24
x = 24 : 2
x = 12
5) 3(x - 8)(4x + 5) - 8(x - 8) = 0
12x2 - 89x - 56 = 0
x = 8 hoặc -7/12
6) 4x = 8
Vì: 41 = 4
42 = 16
=> x thuộc tập hợp rỗng

a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1