dk:....

đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)

=> \(\sqrt[5]{\frac{x-1}{16x}}=\frac{1}{a}\)

ta duoc: a+1/a=5/2

tự giải tiếp nhé

\(pt\Leftrightarrow2\left(x+4\right)-2\sqrt{1+16x}+x^2-3x-10=0\)

\(\Leftrightarrow\frac{2\left(x^2+8x+16-1-16x\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)\)

\(\Leftrightarrow\frac{2\left(x^2-8x+15\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\frac{2\left(x-5\right)\left(x-3\right)}{x+4+\sqrt{1+16x}}+\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2x-6}{x+4+\sqrt{1+16x}}+x+2\right)=0\)

Biết phải lm j tiếp theo r` chứ ?

Cái ngoài x=5

cái trong phức tạp quá khong biết thế nào 

<=>\(4x^2+12x+9=1+16x+8\sqrt{1+16x}+16\)

<=>\(\left(2x+3\right)^2=\left(\sqrt{1+16x}+4\right)^2\)

Như vậy bạn tiếp tục làm nha

a, \(16x^2-\left(1+\sqrt{3}\right)^2=0\\ \Rightarrow\left(4x-1-\sqrt{3}\right)\left(4x+1+\sqrt{3}\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x-1-\sqrt{3}=0\\4x+1+\sqrt{3}=0\end{matrix}\right.\)

    \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{3}}{4}\\x=\dfrac{-1-\sqrt{3}}{4}\end{matrix}\right.\)

b, \(x-2\sqrt{2x}+2=8\\ \Rightarrow x-\sqrt{8x}-6=0\\ \Rightarrow x-6=\sqrt{8x}\\ \Rightarrow\left(x-6\right)^2=\sqrt{8x}^2\\ \Rightarrow x^2-12x+36=8x\\ \Rightarrow x^2-20x+36=0\\ \Rightarrow\left(x^2-2x\right)-\left(18x-36\right)=0\)

    \(\Rightarrow x\left(x-2\right)-18\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-18\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-18=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=18\end{matrix}\right.\)

1: Ta có: \(16x^2-\left(\sqrt{3}+1\right)^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{3}+1}{4}\\x=\dfrac{-\sqrt{3}-1}{4}\end{matrix}\right.\)

2: Ta có: \(x-2\sqrt{2x}+2=8\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=8\)

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

\(\Leftrightarrow x=12+8\sqrt{2}\)

a) \(16x^2-\left(1+\sqrt{3}\right)^2=0\Leftrightarrow\left(4x-1-\sqrt{3}\right)\left(4x+1+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1-\sqrt{3}=0\\4x+1+\sqrt{3}=0\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\dfrac{1+\sqrt{3}}{4}\)

b) \(x-2\sqrt{2x}+2=8\Leftrightarrow\left(\sqrt{x}-\sqrt{2}\right)^2=8\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\sqrt{2}\\\sqrt{x}=-\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow x=18\)(do \(\sqrt{x}\ge0\ne-\sqrt{2}\))

em thử làm phát nhá chị

kkkk oke le tai bao chau

em ns ý cx đc

mà làm đc càng tốt

Đk : x\(\ge\frac{-1}{16}\)

Ta có x² - x -2.\(\sqrt{1+16x}=2\) 

          \(\Rightarrow\)4x2−4x−8\(\sqrt{1+16x}\)=8

     <=> (2x−1)²+8(2x−1)−(1+16x)−8\(\sqrt{1+16x}\)=0                        (1)

   Đặt  \(\hept{\begin{cases}2x-1=a\\\sqrt{1+16x}=b\left(b\ge0\right)\end{cases}}\)

Khi đó (1) trở thành :   a²+8a−b²−8b=0

                            <=>  (a−b)(a+b+8)=0

             Làm típ ( cả bài trên em lừa đảo đáy )

\(1,\dfrac{x-1}{3}=x+1\\ \Leftrightarrow x-1=3x+3\\ \Leftrightarrow3x-x=3+1\\ \Leftrightarrow x=2\)

PT có tập nghiệm S = {2}

\(2,\sqrt{16x^2+8x+1}-2=x\\ \Leftrightarrow\sqrt{\left(4x+1\right)^2}-2=x\\\Leftrightarrow 4x+1-2=x\\ \Leftrightarrow4x-x=2-1\\ \Leftrightarrow x=\dfrac{1}{3}\)

PT có tập nghiệm S = {1/3}

\(3,\left\{{}\begin{matrix}2x+y=17\\x-2y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=17\\2x-4y=2\end{matrix}\right.\\ \Leftrightarrow\left(2x+y\right)-\left(2x-4y\right)=17-2\\ \Leftrightarrow5y=15\\ \Leftrightarrow y=3\\ \Leftrightarrow2x+3=17\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\)

PTHH có tập nghiệm (x; y) là (7; 3)