Tim x biet: x+\(2\sqrt{2x^2}\) +2x3=0
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\(x-2\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy x=0 hoặc x=4 là giá trị cần tìm
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này chỉ yêu cầu tìm x thôi đúng ko bạn .
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-\sqrt{2}=0\\y-\sqrt{2}=0\\x+y+z=0\end{cases}\Rightarrow x=\sqrt{2}}\)
Tim x biet
a)\(\left(2\sqrt{x}-3\right).\left(2+\sqrt{x}\right)+6=0\)
b)\(\sqrt{x^2-9}-3\sqrt{x-3}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2\sqrt{x}-3\right)\left(2+\sqrt{x}\right)+6=0\left(ĐK:x\ge0\right)\)
\(\Leftrightarrow4\sqrt{x}+2x-6-3\sqrt{x}+6=0\)
\(\Leftrightarrow2x+\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\2\sqrt{x}+1=0\left(loại\right)\end{array}\right.\)\(\Leftrightarrow x=0\)
b)\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(ĐK:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\left(tm\right)\\x=6\left(tm\right)\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(x-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}x-\sqrt{2}=0\\x+y+z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x+y=-z\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x=-z-y\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có 2015 số hạng
Tổng là (2x+1+2x+2015).2015:2=0
4x+2016=0
4x=-2016
x=-504
Ủng hộ mk nha
( 2x + 1 ) + ( 2x + 2 ) + ... + ( 2x + 2015 ) = 0
=> ( 2x + 2x + .. + 2x ) + ( 1 + 2 + ... + 2015 ) = 0
2015 số 2x
=> 2x . 2015 + 2031120 = 0
=> 2x . 2015 = 0 - 2031120
=> 2x . 2015 = - 2031120
=> 2x = ( - 2031120 ) : 2015
=> 2x = - 1008
=> x = ( - 1008 ) : 2
=> x = - 504
![](https://rs.olm.vn/images/avt/0.png?1311)
Có tất cả số 2x là:
(2015-1):1+1=2015(số)
Ta có:(2x+1)+(2x+2)+............+(2x+2015)=0
=>2015*2x+(1+2+3+............+2015)=0
=>4030x+2031120=0
=>4030x=-2031120
=>x=-504
![](https://rs.olm.vn/images/avt/0.png?1311)
(2x-6).(x-2) = 0
=>2x - 6 = 0 hoặc x - 2 = 0
2x - 6 = 0
2x = 6
x = 3
x - 2 = 0
x = 2
Vậy x thuộc {2;3}