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

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

\(\Leftrightarrow2x+2\sqrt{x^2-2x+1}=2\Leftrightarrow2\left(x+\sqrt{\left(x-1\right)^2}\right)=2\Leftrightarrow x+|^{ }_{ }x-1|=1\)

\(\Leftrightarrow|^{ }_{ }x-1|^{ }_{ }=1-x\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)

vậy x<1

mik hỏi bạn mik bảo x<10

a) \(\sqrt{39-12\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{36-12\sqrt{3}+3}+\sqrt{9-12\sqrt{3}+12}\)

\(=\sqrt{\left(6-\sqrt{3}\right)^2}+\sqrt{\left(3-\sqrt{12}\right)^2}\)

\(=6-\sqrt{3}+\sqrt{12}-3=3+\sqrt{3}\)

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\frac{\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

Đặt a=2013

\(\Rightarrow M=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{a^2+2a+1+a^4+2a^3+a^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\frac{\left(a^4+2a^3+a^2\right)+2\left(a^2+a\right)+1}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(\Rightarrow M=\sqrt{\left(\frac{a^2+a+1}{a+1}\right)^2}+\frac{a}{a+1}\)

\(\Rightarrow M=\frac{a^2+a+1+a}{a+1}\)(Bỏ trị tuyệt đối vì a=2013)

\(\Rightarrow M=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=1013+1=1014\)

đặt \(\sqrt{7-x}=a\) , \(\sqrt{x-1}=b\)

rồi thay vào và ptđttnt

ĐK: \(1\le x\le7\)

\(x+2\sqrt{7-x}=2\sqrt{x-1}+\sqrt{-x^2+8x-7}+1\)

\(x-1+2\sqrt{7-x}-2\sqrt{x-1}-\sqrt{-x^2+8x-7}=0\)

Đặt \(\sqrt{x-1}=a;\sqrt{7-x}=b\left(a,b\ge0\right)\)

\(pt\Rightarrow a^2+2b-2a-ab=0\Leftrightarrow\left(a^2-ab\right)-\left(2a-2b\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a-b\right)=0\Leftrightarrow\orbr{\begin{cases}a-2=0\\a=b\end{cases}}\)

TH1: \(a-2=0\Rightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(tm\right)\)

TH2: \(a=b\Rightarrow\sqrt{x-1}=\sqrt{7-x}\Rightarrow x=4\left(tm\right)\)

Vậy pt có 2 nghiệm x = 4 hoặc x = 5.

dat a=\(\sqrt{x^2+x+1}\)    b=\(\sqrt{x^2-x+1}\) dk \(a,b\ge0\)

t a co he phuong trinh \(\hept{\begin{cases}a^2-b^2=2x\\a-b=2x\end{cases}}\) \(\Rightarrow a^2-b^2=a-b\Leftrightarrow\left(a-b\right)\left(a+b-1\right)=0\)

voi a=b \(\sqrt{x^2+x+1}=\sqrt{x^2-x+1}\Rightarrow x^2+x+1=x^2-x+1\)

\(\Rightarrow x=0\)

vs a+b=1  ket hop vs a-b=2x \(\Rightarrow a=\frac{2x+1}{2}\) \(b=\frac{-2x+1}{2}\)

do \(a\ge0,b\ge0\Rightarrow\frac{-1}{2}\le x\le\frac{1}{2}\)

tu \(\sqrt{x^2+x+1}=\frac{2x+1}{2}\Rightarrow x^2+x+1=\frac{4x^2+4x+1}{4}\)

\(\Rightarrow4\left(x^2+x+1\right)=4x^2+4x+1\)

\(\Rightarrow\) ko co no nao tm

kl x=0 la no cua pt da cho

Bạn làm được chưa giải bài này giúp mình với

pt(1)\(\Leftrightarrow\left(\sqrt{2x^2+x+1}-2x\right)+\left(\sqrt{x^2-x+1}-x\right)=0\left(đk;x\ge0\right)\)

\(\Leftrightarrow\frac{-2x^2+x+1}{\sqrt{2x^2+x+1}+2x}+\frac{-x+1}{\sqrt{x^2-x+1}+x}=0\)

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

\(\Leftrightarrow x=1\)