\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
K
Khách
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)
VT
2
![](https://rs.olm.vn/images/avt/0.png?1311)
2 tháng 7 2017
x\(\sqrt{4x^2-4x+1}+\sqrt{x^2-2x+1}\)
=\(\sqrt{\left(2x-1\right)^2}+\sqrt{\left(x-1\right)^2}\)
=I2x-1I+Ix-1I=0
=-2x+1-x+1=0
=-3x+2=0
=>x=2/3
H
4
![](https://rs.olm.vn/images/avt/0.png?1311)
2 tháng 7 2017
Câu a mk sửa đề tí nha:
\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
2 tháng 7 2017
\(b,\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(=\sqrt{x-4}+2+\sqrt{x-4}-2\)
\(=2\sqrt{x-4}\)
NT
1
NN
0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
\(pt\Leftrightarrow\sqrt[3]{x-2}-1+\sqrt{x+1}-2=0\)
\(\Leftrightarrow\frac{x-2-1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x+1-4}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{x-3}{\sqrt{x+1}+2}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}\right)=0\)
Dễ thấy: \(\frac{1}{\sqrt[3]{\left(x-2\right)^2}+\sqrt[3]{x-2}+1}+\frac{1}{\sqrt{x+1}+2}>0\) (loại)
\(\Rightarrow x-3=0\Rightarrow x=3\)