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 1:

ĐK: $x\geq 4$

PT $\Leftrightarrow \sqrt{2x-7}=\sqrt{x-4}+1$

$\Rightarrow 2x-7=x-4+1+2\sqrt{x-4}$ (bình phương 2 vế)

$\Leftrightarrow x-4=2\sqrt{x-4}$

$\Leftrightarrow \sqrt{x-4}(\sqrt{x-4}-2)=0$

\(\Rightarrow \left[\begin{matrix} \sqrt{x-4}=0\\ \sqrt{x-4}=2\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=8\end{matrix}\right.\) (đều thỏa mãn)

Vậy........

Câu 2:

ĐK: $\frac{3}{2}\leq x\leq \frac{5}{2}$

Áp dụng BĐT AM-GM ta có:

\(3x^2-12x+14=\sqrt{2x-3}+\sqrt{5-2x}\leq \frac{(2x-3)+1}{2}+\frac{(5-2x)+1}{2}\)

\(\Leftrightarrow 3x^2-12x+14\leq 2\Leftrightarrow 3x^2-12x+12\leq 0\)

\(\Leftrightarrow x^2-4x+4\leq 0\Leftrightarrow (x-2)^2\leq 0(*)\)

Mà $(x-2)^2\geq 0, \forall x$ nên $(x-2)^2=0$

$\Rightarrow x=2$

Thử lại thấy thỏa mãn nên $x=2$ là nghiệm của PT.

Câu 3:

ĐK: $x\geq \frac{2}{3}$

PT $\Leftrightarrow x^2-x+6-4\sqrt{3x-2}=0$

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

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

$\Leftrightarrow (x-2)^2+(\sqrt{3x-2}-2)^2=0$

Ta thấy:

$(x-2)^2\geq 0$ với mọi $x\geq \frac{2}{3}$

$(\sqrt{3x-2}-2)^2\geq 0$ với mọi $x\geq \frac{2}{3}$

Do đó để tổng của chúng bằng $0$ thì $(x-2)^2=(\sqrt{3x-2}-2)^2=0$

$\Leftrightarrow x=2$ (thỏa mãn)

Vậy...........

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

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

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

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

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

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...............

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

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

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

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

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

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

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

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

\(\Leftrightarrow2x-5=9\)

\(\Leftrightarrow x=7\)