Bài 3:

b: ĐKXĐ: \(\begin{cases}1-x^2\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}x^2\le1\\ x\ge-1\end{cases}\Rightarrow\begin{cases}x=-1\\ x\ge1\end{cases}\)

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

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

=>\(\sqrt{1+x}=0\)

=>x+1=0

=>x=-1(nhận)

c: Sửa đề: \(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

ĐKXĐ: x>=2; y>=3; z>=5

\(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

=>\(x+y+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\)

=>\(x-2-2\sqrt{x-2}+1+y-3-4\sqrt{y-3}+4+z-5-6\sqrt{z-5}+9=0\)

=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

=>\(\begin{cases}x-2=1\\ y-3=4\\ z-5=9\end{cases}\Rightarrow\begin{cases}x=3\\ y=7\\ z=14\end{cases}\) (nhận)

d: \(x^2+2x-\sqrt{x^2+2x+1}-5=0\)

=>\(x^2+2x+1-\sqrt{x^2+2x+1}-6=0\)

=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|-6=0\)

=>(|x+1|-3)(|x+1|+2)=0

=>|x+1|-3=0

=>|x+1|=3

=>\(\left[\begin{array}{l}x+1=3\\ x+1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-4\end{array}\right.\)

Bài 2:

a: DKXĐ: x>=0

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

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

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

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

=>|x+2|+|x-1|=4(1)

TH1: x<-2

=>x+2<0; x-1<0

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>\(x=-\frac52\) (nhận)

TH2: -2<=x<1

=>x+2>=0; x-1<0

(1) sẽ trở thành: x+2+1-x=4

=>3=4(loại)

TH3: x>=1

=>x+2>0; x-1>=0

(1) sẽ trở thành: x+2+x-1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

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

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

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

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

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>x-1<=1

=>x<=2

=>1<=x<=2

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)

\(\Leftrightarrow x=6\)

1: =>3x^2+5x-7=3x+14

=>2x=21

=>x=21/2

2;=>x+4=4

=>x=0

3: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-20x+25-4x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{2}\\4x^2-24x+32=0\end{matrix}\right.\)

=>x>=5/2 và x^2-6x+8=0

=>x=4

4: \(\Leftrightarrow\left\{{}\begin{matrix}x>=1\\x^2+2x-1=x^2-2x+1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

5: \(\Leftrightarrow\sqrt{2x+16}=x-4\)

=>x>=4 và x^2-8x+16=2x+16

=>x>=4 và x^2-10x=0

=>x=10

Câu 6:

ĐK: $x\geq 1$

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

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

$\Leftrightarrow |\sqrt{x-1}-1|=\sqrt{x-1}+1$

Nếu $\sqrt{x-1}-1\geq 0$ thì PT trở thành:

$\sqrt{x-1}-1=\sqrt{x-1}+1\Leftrightarrow 2=0$ (vô lý)

Nếu $\sqrt{x-1}-1< 0$ (tương đương với $1\leq x< 2$ thì PT trở thành:

$1-\sqrt{x-1}=\sqrt{x-1}+1$

$\Leftrightarrow \sqrt{x-1}=0\Rightarrow x=1$ (thỏa mãn)

Vậy PT có nghiệm $x=1$

Câu 5:

ĐK: $x\geq 1$

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

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

$\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$

Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:

$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=|\sqrt{x-1}-2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}-2+3-\sqrt{x-1}|=1$

Dấu "=" xảy ra khi $(\sqrt{x-1}-2)(3-\sqrt{x-1})\geq 0$

$\Leftrightarrow 3\geq \sqrt{x-1}\geq 2$

$\Leftrightarrow 10\geq x\geq 5$. Kết hợp ĐKXĐ ta thấy những giá trị $x$ thỏa mãn $10\geq x\geq 5$ là nghiệm của pt.

\(DK:x\ge\frac{5}{2}\)

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

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

\(\Leftrightarrow2\sqrt{2x-5}=0\)

\(\Leftrightarrow x=\frac{5}{2}\left(n\right)\)

Vay PT co nghiem la \(x=\frac{5}{2}\)

\(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4\)

⇔ \(\sqrt{2x-5+2.3\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\)

⇔ \(\text{ |}\sqrt{2x-5}+3\text{ |}+\text{ |}\sqrt{2x-5}-1\text{ |}=4\)

⇔ \(\sqrt{2x-5}+3+\text{ |}\sqrt{2x-5}-1\text{ |}=4\) ( x ≥ \(\dfrac{5}{2}\) ) ( 1)

+) Với : \(\sqrt{2x-5}\text{≥}1\) ⇔ x ≥ 3 , ta có :

\(\left(1\right)\text{⇔}\sqrt{2x-5}+3+\sqrt{2x-5}-1=4\)

\(\text{⇔}2\sqrt{2x-5}=2\)

\(\text{⇔}x=3\left(TM\right)\)

+) Với : \(\sqrt{2x-5}< 1\) ⇔ x < 3 , ta có :

\(\left(1\right)\text{⇔}\sqrt{2x-5}+3+1-\sqrt{2x-5}=4\)

\(\text{⇔}4=4\) ( luôn đúng với : \(3>x\text{≥}\dfrac{5}{2}\) )

KL...............

\(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}-3+\sqrt{2x-5}+1=4\\\sqrt{2x-5}-3+\sqrt{2x-5}+1=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}-2=4\\2\sqrt{2x-5}-2=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{2x-5}=6\\2\sqrt{2x-5}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-5}=3\\\sqrt{2x-5}=-1\left(L\right)\end{cases}}\)

\(\Leftrightarrow2x-5=9\)

\(\Leftrightarrow x=7\)