![](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)
\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(\left(x+1\right)^2=25\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)
Vậy...
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3\left(x^2-2x+1\right)-\left(3x^2-15x\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy...
\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(x^2-2x-24=0\)
\(\Leftrightarrow\)\(\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
b)x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12
<=> 3x^2 +2x +x^2+2x+1 - 4x^2 +25 +12=0
<=> 4x+38=0
=>4x= -38
=>x= -38/4= -19/2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Rightarrow3x^2-3^2-3x^2+15x=1\)
\(\Rightarrow3x^2-9-3x^2+15x=1\)
\(\Rightarrow-9+15x=1\)
\(\Rightarrow15x=-8\)
\(\Rightarrow x=\frac{-8}{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(5-2x\right)^2-16=0\)
\(\Leftrightarrow\left(5-2x\right)^2-4^2=0\)
\(\Rightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x=4-5\\-2x=-4-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}-2x=-1\\-2x=-9\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{-2}=\frac{1}{2}\\x=\frac{-9}{-2}=\frac{9}{2}\end{cases}}\)
Vậy ................................
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=x+1-\left(x-9\right)\)
\(\Rightarrow6x^2+27x+4x+18-6x^2-x-12x-2=10\)
\(\Rightarrow18x+16=10\)
\(\Rightarrow18x=-6\)
\(\Rightarrow x=-\frac{6}{18}=-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=25\)
\(\Leftrightarrow\)\(\left(x-1\right)^2=25\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
Vậy....
b) \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)
\(\Leftrightarrow\)\(9x=-2\)
\(\Leftrightarrow\)\(x=-\frac{2}{9}\)
Vậy...
Cam on ban <3