a, \(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)(ĐK: \(0\le x\le1\))

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

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

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

\(\Leftrightarrow\sqrt{1-x}=1-2\sqrt{x}\)(ĐK: \(0\le x\le\frac{1}{4}\))

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

\(\Leftrightarrow1-x=1-4\sqrt{x}+4x\)

\(\Leftrightarrow5x-4\sqrt{x}=0\)

\(\Leftrightarrow5x=4\sqrt{x}\)

\(\Leftrightarrow25x^2=16x\)

\(\Leftrightarrow25x^2-16x=0\)

\(\Leftrightarrow x\left(25x-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\25x-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=\frac{16}{25}\left(L\right)\end{cases}}\)

Vậy PT có nghiệm là x = 0

a) Ta có pt \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=\sqrt{\left(\sqrt{3}+1\right)^2}\Leftrightarrow\left|x-3\right|=\sqrt{3}+1...\)

b) Ta có pt \(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=1\Leftrightarrow\left|x-1\right|+\left|x+2\right|=1\)

đến đây tự phá dấu trị tuyệt đối !

^_^

a) dat x-1=a

x=a+1

\(a+1+\sqrt{5+\sqrt{a}}=6\)

\(5-a=\sqrt{5+\sqrt{a}}\)

\(25-10a+a^2=5+\sqrt{a}\)

\(20-10a+a^2-\sqrt{a}=0\)

(a - \sqrt{5} - 5) (a + \sqrt{a} - 4) = 0

đúng nhưng b,c,d đâu

d/ \(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{\left(x-1\right)^2}+\sqrt[3]{x^2-1}=1\)

Đặt \(\hept{\begin{cases}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{cases}\Rightarrow a^3-b^3=2}\)

\(\Rightarrow\hept{\begin{cases}a^3-b^3=2\\a^2+b^2+ab=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a^2+b^2+ab\right)=2\\a^2+b^2+ab=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a-b=2\\a^2+b^2+ab=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a-b=2\\b^2+2b+1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=1\\b=-1\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt[3]{x+1}=1\\\sqrt[3]{x-1}=-1\end{cases}\Leftrightarrow}x=0}\)

bài b , lập phương lên 

bài c , đặt cái căn đưa về hệ 

mới nhìn dc làm dc liền thế thui

a)Đk:\(0\le x\le1\)

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

\(pt\Leftrightarrow\sqrt{x}+\sqrt{1-x}-1+\sqrt{x+1}-1=0\)

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

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

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

\(\Rightarrow x=0\)

b)\(\frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^2-x+1}}\)

\(pt\Leftrightarrow\frac{3x+3}{\sqrt{x}}-6=\frac{x+1}{\sqrt{x^2-x+1}}-2\)

\(\Leftrightarrow\frac{3x+3-6\sqrt{x}}{\sqrt{x}}=\frac{x+1-2\sqrt{x^2-x+1}}{\sqrt{x^2-x+1}}\)

\(\Leftrightarrow\frac{\frac{\left(3x+3\right)^2-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{\left(x+1\right)^2-4\left(x^2-x+1\right)}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)

\(\Leftrightarrow\frac{\frac{9x^2+18x+9-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{x^2+2x+1-4x^2+4x-4}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)

\(\Leftrightarrow\frac{\frac{9x^2-18x+9}{3x+3+6\sqrt{x}}}{\sqrt{x}}-\frac{\frac{-3x^2+6x-3}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)

\(\Leftrightarrow\frac{\frac{9\left(x-1\right)^2}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{3\left(x-1\right)^2}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)

\(\Leftrightarrow3\left(x-1\right)^2\left(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\right)=0\)

Dêx thấy: \(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}>0\forall....\)

\(\Rightarrow3\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

a ) x = 0 

b ) x = 1

k tui nha

thanks

aを見つける= 175度はどれくらい尋ねる