\(\sqrt{x-4\sqrt{x-1}+3}+\sqrt{x-6\sqrt{x-1}+8}=1\\ < =>\sqrt{x-1-2\sqrt{x-1}.2+4}+\sqrt{x-1-2\sqrt{x-1}.3+9}=1\\ < =>\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)ĐK: x>=1

\(< =>|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1\\ < =>\left(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\right)^2=1\\ < =>\sqrt{x-1}-2+2\left|\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-3\right)\right|+\sqrt{x-1}-3=1\\ < =>2\sqrt{x-1}-5+2\left|x+5-5\sqrt{x-1}\right|=1\\ < =>2\left|x+5-5\sqrt{x-1}\right|=6-2\sqrt{x-1}\\ < =>\left|x+5-5\sqrt{x-1}\right|=3-\sqrt{x-1}\)

\(< =>\left[{}\begin{matrix}x+5-5\sqrt{x-1}=3-\sqrt{x-1}\left(1\right)\\x+5-5\sqrt{x-1}=\sqrt{x-1}-3\left(2\right)\end{matrix}\right.\)

Giải (1): \(x+5-5\sqrt{x-1}=3-\sqrt{x-1}\\ < =>x+2-4\sqrt{x-1}=0\\ < =>x-1-2\sqrt{x-1}.2+4=1\\ < =>\left(\sqrt{x-1}-2\right)^2=1\\ < =>\left[{}\begin{matrix}\sqrt{x-1}-2=1\\\sqrt{x-1}-2=-1\end{matrix}\right.< =>\left[{}\begin{matrix}x=8\\x=0\left(loại\right)\end{matrix}\right.\)

Giải (2) cũng ra x=8

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

Và \(\sqrt{x+8+6\sqrt{x-1}}=\sqrt{x-1+6\sqrt{x-1}+9}=\left(\sqrt{x-1}-3\right)^2\)

Ok dễ nhé

ĐKXĐ: \(x\ge1\)

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

\(=\sqrt{4-2.2.\sqrt{x-1}+x-1}+\sqrt{x-1+2.\sqrt{x-1}.3+9}\)

\(=\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}\)\(=|2-\sqrt{x-1}|+|\sqrt{x-1}+3|\ge|2-\sqrt{x-1}+\sqrt{x-1}+3|=5\)

Dấu bằng xảy ra khi \(2-\sqrt{x-1}\ge0\Leftrightarrow\sqrt{x-1}\le2\Leftrightarrow x\le3\)

Vậy \(1\le x\le3\)

Nếu đúng cho nhé bạn.

đk: x >= 1

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}+3\right|=5\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)vì \(\sqrt{x-1}+3>0\forall x\ge1\)

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

Mà |A| = -A khi và chỉ khi A <=0 

(1)\(\Rightarrow\sqrt{x-1}-2\le0\Rightarrow0\le\sqrt{x-1}\le2\Rightarrow0\le x-1\le4\)

\(\Rightarrow1\le x\le5\)

Vậy, PT có nghiệm với mọi x thuộc [1;5].

Điều kiện \(x\ge1.\)  Phương trình trở thành \(\sqrt{\left(\sqrt{x-1}+2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=5\Leftrightarrow\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|=5.\)

Theo bất đẳng thức trị tuyệt đối \(\left|a+b\right|\le\left|a\right|+\left|b\right|.\)  Dấu bằng xảy ra khi \(ab\ge0.\) Do đó ta có phương trình tương đương với \(\left(3-\sqrt{x-1}\right)\left(2+\sqrt{x-1}\right)\ge0\Leftrightarrow3\ge\sqrt{x-1}\Leftrightarrow x\le10.\)  Kết hợp với điều kiện ta có nghiệm của phương trình là \(1\le x\le10.\)

cậu dùng hằng đẳng thưc là ra

đk: x>=1

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1+6\sqrt{x-1}+9}=5\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}=5\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\sqrt{x-1}+3=5\)

th1: x>=5 <=> \(\sqrt{x-1}-2+\sqrt{x-1}+3=5\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow x=17\)(t/m đk)

th2: x<5 <=> \(2-\sqrt{x-1}+\sqrt{x-1}+3=5\Leftrightarrow5=5\)=> pt có vô số nghiệm

=> x=17 hoặc x<5

( Nhớ tìm ĐK)

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

Thay vào ta có 

 \(\sqrt{y^2+1+3-4y}+\sqrt{y^2+1+8-6y}=5\)

\(\Leftrightarrow\sqrt{\left(y-2\right)^2}+\sqrt{\left(y-3\right)^2}=5\)

=> l y- 2 l + l y - 3 l = 5 

(+) Với 2 <= y ta có pt

       2-y + 3-y = 5

           5 - 2y  = 5

            => 2y  = 0 => y = 0

    (-) y = 0 => \(\sqrt{x-1}=0\Leftrightarrow x=1\)

(+) Còn 2 trường hợp nua twowg tụ

1.

ĐKXĐ: \(x< 5\)

\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)

\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)

\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)

\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

b.

ĐKXĐ: \(x\ge2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

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

\(\Rightarrow x=2\)

3.

ĐKXĐ: \(x\ge-1\)

\(x^2+x-12+12\left(\sqrt{x+1}-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\dfrac{12\left(x-3\right)}{\sqrt{x+1}+2}=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+4+\dfrac{12}{\sqrt{x+1}+2}\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

a. \(\sqrt[3]{1-2x}+3=0\left(ĐK:x\le\dfrac{1}{2}\right)\)

<=> \(\sqrt[3]{1-2x}=-3\)

<=> \(1-2x=\left(-3\right)^3\)

<=> \(1-2x=-27\)

<=> \(-2x=-28\)

<=> \(x=14\left(TM\right)\)