Đặt \(N=\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\)

\(\Rightarrow N^2=x-1+2\sqrt{x-2}+x-1-2\sqrt{x-2}+2\sqrt{\left(x-1+2\sqrt{x-2}\right)\left(x-1-2\sqrt{x-2}\right)}\)

\(\Leftrightarrow N^2=2x-2+2\sqrt{\left(x-1\right)^2-\left(2\sqrt{x-2}\right)^2}\)

\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-2x+1-4\left(x-2\right)}\)

\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-2x+1-4x+8}\)

\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-6x+9}\)

\(\Leftrightarrow N^2=2x-2+2\sqrt{\left(x-3\right)^2}\)

\(\Leftrightarrow N^2=2x-2+2\left|x-3\right|\)

* Với \(x\ge3\)thì \(N^2=2x-2+2\left(x-3\right)=4x-8=4\left(x-2\right)\Rightarrow N=2\sqrt{x-2}\)

* Với \(2\le x\le4\)thì \(N^2=2x-2-2\left(x-3\right)=4\Rightarrow N=\sqrt{4}=2\)

(Bạn xem thử coi đúng hông nha, và 1 cái k nhá!)

cam on ban

ban la vi cuu' tinh cua~ to' 

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a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

liên hợ thôi !

\(DK:x\ge0\)

\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)

\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)

\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)

\(\Leftrightarrow x=1\)

Vay nghiem cua PT la \(x=1\)

Em làm bừa thôi, mới học dạng này .

ĐK: \(1\le x\le7\)

Đặt \(\sqrt{6}\ge a=\sqrt{7-x}\ge0;\sqrt{6}\ge b=\sqrt{x-1}\ge0\)

PT<=>\(b^2+2a=2b+ab\left(1\right)\)

(1) \(\Leftrightarrow\left(a-b\right)\left(2-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\\b=2\end{cases}}\). Nếu a = b thì \(\sqrt{7-x}=\sqrt{x-1}\Leftrightarrow7-x=x-1\Leftrightarrow x=4\) (TM)

Nếu b = 2 thì \(\sqrt{x-1}=2\Leftrightarrow x=5\left(TM\right)\)

Vậy...

ĐK: \(0< x\le4\)

Đặt \(\sqrt{2+\sqrt{x}}=a\left(a>0\right)\) ; \(\sqrt{2-\sqrt{x}}=b\left(b\ge0\right)\)

=> \(a^2+b^2=2+\sqrt{x}+2-\sqrt{x}=4\) (1)

Ta có: \(\dfrac{a^2}{\sqrt{2}+a}+\dfrac{b^2}{\sqrt{2}-b}=\sqrt{2}\)

<=> \(\dfrac{a^2.\sqrt{2}-a^2b+b^2.\sqrt{2}+ab^2}{2+\sqrt{2}\left(a-b\right)-ab}=\sqrt{2}\)

<=> \(\left(a^2+b^2\right)\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\)

<=> \(4\sqrt{2}+ab\left(b-a\right)=2\sqrt{2}+2\left(a-b\right)-ab.\sqrt{2}\) ( Theo 1)

<=> \(\left(a-b\right)\left(2+ab\right)=2\sqrt{2}+ab.\sqrt{2}\)

<=> \(\left(a-b-\sqrt{2}\right)\left(ab+2\right)=0\)

<=> \(\left[{}\begin{matrix}ab+2=0\\a-b-\sqrt{2}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}ab=-2\\a-b=\sqrt{2}\end{matrix}\right.\) mà a² + b² = 4

Xét \(\left\{{}\begin{matrix}ab=-2\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left(a-b\right)^2=8\\\left(a+b\right)^2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a-b=\pm\sqrt{8}\\a+b=0\end{matrix}\right.\) ( Loại vì \(a>0;b\ge0\) )

Xét \(\left\{{}\begin{matrix}a-b=\sqrt{2}\\a^2+b^2=4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left(b+\sqrt{2}\right)^2+b^2=4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\2b^2+2b.\sqrt{2}-2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\b^2+b.\sqrt{2}-1=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left[{}\begin{matrix}b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\b=\dfrac{-\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

#Lề: Bn lấy cái đề ở đâu hay v?

v cac bac e giai xong lau roi cac bac a voi lai co cach giai ko can dai nhu the dau nhe

ĐK: x>= -1/3

Ta có: \(pt\Leftrightarrow2x\sqrt{x^2-x+1}+4\sqrt{3x+1}=2x^2+2x+6\)

<=> \(x^2-2x\sqrt{x^2-x+1}+\left(x^2-x+1\right)+\left(3x+1\right)-2.\sqrt{3x+1}.2+4=0\)

\(\Leftrightarrow\left(x-\sqrt{x^2-x+1}\right)^2+\left(\sqrt{3x+1}-2\right)^2=0\)

Mà : \(\left(x-\sqrt{x^2-x+1}\right)^2\ge0;\left(\sqrt{3x+1}-2\right)^2\ge0\)

Khi đó: \(\left(x-\sqrt{x^2-x+1}\right)^2+\left(\sqrt{3x+1}-2\right)^2\ge0\)

Dấu "=" xảy ra khi và chỉ khi: 

\(\hept{\begin{cases}\left(x-\sqrt{x^2-x+1}\right)^2=0\\\left(\sqrt{3x+1}-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2=x^2-x+1,x\ge0\\3x+1=4\end{cases}}\Leftrightarrow x=1\)tm đk

Vậy x=1

Ta có thể dùng cô si chăng?

ĐK: \(x\ge-\frac{1}{3}\)

\(VT=\sqrt{x^2\left(x^2-x+1\right)}+\sqrt{4\left(3x+1\right)}\)

\(\le\frac{x^2+x^2-x+1}{2}+\frac{4+3x+1}{2}=\frac{2x^2+2x+6}{2}=x^2+x+3=VP\)

Để đẳng thức xảy ra, tức là xảy ra đẳng thức ở phương trình thì:

\(\hept{\begin{cases}x^2=x^2-x+1\\4=3x+1\end{cases}}\Leftrightarrow x=1\)

Vậy...

Is it true??

tth_new nếu thế thì em phải xét 2 TH \(x\ge0\) ( là trường hợp em làm ) và \(\frac{1}{3}\le x< 0\)

TH: \(\frac{1}{3}\le x< 0\)

\(VT< 0+2=2\)

\(VP=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}>\frac{1}{36}+\frac{11}{4}=\frac{25}{9}>\frac{18}{9}=2>VT\) => loại TH này