\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy x=0 hoặc x=4 là giá trị cần tìm

\(x-2\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

vậy phương trình có tập nghiệm là S={0;4}

\(x-2\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\sqrt{x}=2\end{matrix}\right.\Rightarrow}}\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Bài này chỉ yêu cầu tìm x thôi đúng ko bạn .

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\sqrt{2}=0\\y-\sqrt{2}=0\\x+y+z=0\end{cases}\Rightarrow x=\sqrt{2}}\)

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)\(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(ĐK:x\ge3\right)\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\left(tm\right)\\x=6\left(tm\right)\end{array}\right.\)

\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(x-\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\sqrt{2}=0\\x+y+z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x+y=-z\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=\sqrt{2}\\x=-z-y\end{cases}}\)

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

=> ( 2x + 2x + .... + 2x + 2x ) + ( 1 + 2 + 3 + .... + 2015 ) = 0

=> ( 2x.2015 ) + { [ 2015.( 2015 + 1 ] : 2 } = 0

=> 4030x + [ ( 2015 . 2016 ) : 2 ] = 0

=> 4030x + ( 4062240 : 2 ) = 0

=> 4030x + 2031120 = 0

=> 4030x = - 2031120

=> x = - 2031120 : 4030 = - 504

Vậy x = - 504

\(y=\frac{1}{x^2+\sqrt{x}}\)

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

TH1: 2x-6=0                TH2: x-2=0

        2x=6                           x =2

        x=3

(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}