Question

GIẢI PHƯƠNG TRÌNH

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

Answer 1

Bài làm:

Ta có: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

\(\Leftrightarrow\sqrt{\left(x-1\right)-4\sqrt{x-1}+4}+\sqrt{\left(x-1\right)-6\sqrt{x-1}+9}=1\)

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

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

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Answer 2

Ta có: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) (ĐKXĐ: x \(\ge\)1)

<=> \(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

<=> \(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)

<=> \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|=1\)

Do \(\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=\left|1\right|=1\)

Dấu "=" xảy ra <=> \(\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\) 

TH1: \(\hept{\begin{cases}\sqrt{x-1}-2\ge0\\3-\sqrt{x-1}\ge0\end{cases}}\) <=> \(\hept{\begin{cases}\sqrt{x-1}\ge2\\0\le\sqrt{x-1}\le3\end{cases}}\) <=> \(\hept{\begin{cases}x\ge5\\1\le x\le10\end{cases}}\)=> \(5\le x\le10\)

TH2: \(\hept{\begin{cases}\sqrt{x-1}-2\le0\\3-\sqrt{x-1}\le0\end{cases}}\) <=> \(\hept{\begin{cases}0\le\sqrt{x-1}\le2\\\sqrt{x-1}\ge3\end{cases}}\) <=> \(\hept{\begin{cases}1\le x\le5\\x\ge10\end{cases}}\)(loại)

Vậy S = \(\left\{x\left|5\le x\le10\right|\right\}\)

Answer 3

Mik cũng lớp 8

Answer 4

Trả lời;

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)\(\left(ĐK:x\ge1\right)\)

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

\(\Leftrightarrow2\sqrt{x-1}=6\)

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

\(\Leftrightarrow x-1=9\)

\(\Leftrightarrow x=10\)

Vậy \(x=10\)

Answer 5

Bạn Conan cho mik hỏi

Ở Th1 tại sao \(0\le\sqrt{x-1}\le\) 

Answer 6

PHẠM PHƯƠNG DUYÊN

ở TH1: \(\sqrt{x-1}\ge0\forall x\); \(3-\sqrt{x-1}\ge0\) <=> \(3\ge\sqrt{x-1}\)

=> \(0\le\sqrt{x-1}\le3\)

Answer 7

Cảm ơn bạn Conan