![](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)
bạn đăng vừa thôi nhé chứ đăng nhiều thế này ít người khiên trì giải hết lắm bạn nên đăng từng bài cho đỡ dài
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 5 kq =0
câu 6: góc C=90 độ (tam giác vuông tại C)(Định lý Pytago)
câu 7: 0 giá trị
câu 8:x=1
câu 10: x=3;y=1
x+y=4
bye
nếu đúng tích cho mik nha
Mik cảm ơn trc
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+2\sqrt{2x^2}+2x^3=0\)
\(\Leftrightarrow x+2x\sqrt{2}+2x^3=0\)
\(\Leftrightarrow x\left(1+2\sqrt{2}+2x^2\right)=0\)
\(\Leftrightarrow x=0\) ( Vì \(1+2\sqrt{2}+2x^2>0\) )
Tìm x biết :
\(x+2\sqrt{2}x^2+2x^3=0\)
\(x\left(1+2\sqrt{2}x+2x^2\right)=0\)
\(x\left(1+\sqrt{2}x\right)^2=0\)
TH1 : x=0
TH2 : \(\left(1+\sqrt{2}x\right)^2=0\)
\(1+\sqrt{2}x=0\)
\(x=\frac{-1}{\sqrt{2}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(3x-2)(2x-1)=(2-3x)(x+3)
(3x-2)(2x-1)-(2-3x)(x+3)=0
(3x-2)(2x-1)+(3x-2)(x+3)=0
(3x-2)(2x-1+x+3)=0
(3x-2)(3x+2)=0
\(\orbr{\begin{cases}3x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\3x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-2}{3}\end{cases}}}\)
Vậy........
(=) 6x2 - 3x - 4x - 2 = 2x + 6 - 3x2 -9x
(=) 6x2 +3x2 - 7x + 7x + 2 -6 = 0
(=) 9x2 - 4 = 0
(=) 9x2 = 4
(=) x2 = \(\frac{9}{4}\)
(=) x = +- \(\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(6x^2+6\)
\(=6\left(x^2+1\right)\)
b) \(2x^2-18\)
\(=2\left(x^2-9\right)\)
\(=2\left(x-3\right)\left(x+3\right)\)
c) \(3x^2-3xy+4x-4y\)
\(=\left(3x^2-3xy\right)+\left(4x-4y\right)\)
\(=3x\left(x-y\right)+4\left(x-y\right)\)
\(=\left(3x-4\right)\left(x-y\right)\)
a) \(\left(x^3-9x^2+27x-27\right)\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^3\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^2\)
c) \(\frac{x^2-4}{2x}:\frac{3x-6}{6}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{2x}.\frac{6}{3\left(x-2\right)}\)
\(=\frac{\left(x+2\right)}{x}\)