a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

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

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

\(\Leftrightarrow a=b\)

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

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

Đk:\(x\ge-1\)

Đặt \(\left(a,b,c\right)=\left(x;\sqrt{x+1};\sqrt{2}\right)\)

Pt tt: \(a^3+b^3+c^3=\left(a+b+c\right)^3\)

\(\Leftrightarrow a^3+b^3+c^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(\Leftrightarrow0=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

\(\Leftrightarrow3\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)

\(\Leftrightarrow3\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\a+c=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{x+1}=0\\\sqrt{x+1}+\sqrt{2}=0\left(vn\right)\\x+\sqrt{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=-x\\x=-\sqrt{2}\left(ktm\right)\end{matrix}\right.\)\(\Rightarrow\)\(\sqrt{x+1}=-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le0\\x+1=x^2\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{5}}{2}\) (tm)

Vậy...

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

\(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)

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

TH1: \(x\ge-1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

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

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

\(\sqrt[3]{x^2}+\sqrt[3]{x+1}=\sqrt[3]{x}+\sqrt[3]{x^2+x}\)

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

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

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

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

Suy ra x=1. pt kia chịu :v nghiệm lẻ quá

Thắng Nguyễn đúng là thánh troll

đặt \(\sqrt[3]{x}=a;\sqrt[3]{x+1}=b\)

pt trở thành:

a²+b=a+ab

<=>a(a-1)-b(a-1)=0

<=>(a-b)(a-1)=0

từ đó thay vào rồi giải tìm x

Sửa lại đề bài cho mk là: \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)