![](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)
Ta có : x2 + 8x - 7 = 0
=> x2 + 8x + 16 - 9 = 0
=> (x + 4)2 - 9 = 0
=> (x + 4)2 = 9
=> \(\orbr{\begin{cases}x+4=3\\x+4=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-7\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
\(S=\left\{1;\frac{1}{3}\right\}\)
b)\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(S=\left\{2;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 4x(x - 5) - (x - 1)(4x - 3) = 5
4x2 - 20x - (4x2 - 3x - 4x + 3) = 5
4x2 - 20x - 4x2 + 3x + 4x - 3 = 5
-13x - 3 = 5
\(\Rightarrow\) -13x = 8
\(\Rightarrow\) x = \(\dfrac{-8}{13}\)
b) (3x - 4)(x - 2) = 3x(x - 9) - 3
3x2 - 6x - 4x + 8 = 3x2 - 27x - 3
3x2 - 10x + 8 - 3x2 + 27x + 3 = 0
17x + 11 = 0
\(\Rightarrow\) 17x = -11
\(\Rightarrow\) x = \(\dfrac{-11}{17}\)
c) x2 - 81 = 0
\(\Rightarrow\) x2 = 81
\(\Rightarrow\) x = \(\pm\) 9
d) 3x2 - 75 = 0
3(x2 - 25) = 0
\(\Rightarrow\) x2 - 25 = 0
\(\Rightarrow\) x2 = 25
\(\Rightarrow\) x = \(\pm\)5
e) x2 - 4x + 3 = 0
x2 - x - 3x + 3 = 0
(x2 - x) - (3x - 3) = 0
x(x - 1) - 3(x - 1) = 0
(x - 3)(x - 1) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
xin lỗi vì chữa đề
nhầm tí mjk sửa lại nè
(x+2)(x+3)-(x-2)(x+5)=0
<=>x2+5x+6-(x2+3x-10)=0
<=>x2+5x+6-x2-3x+10=0
<=>2x+16=0
<=>2x=-16
<=>x=-8
nhầm tí
(x+2)(x+3)-(x-2)(x+5)=0
<=>x2+5x+6-(x2+3x-10)=0
<=>x2+5x+6-x2-3x+10=0
<=>2x+16=0
<=>2x=-16
<=>x=-8