`x=root{3}{14sqrt2+20}+sqrt{-14sqrt2+20}`

`<=>x^3=14sqrt2+20-14sqrt2+20+3root{3}{(14sqrt2+20)(20-14sqrt2)}(root{3}{14sqrt2+20}+sqrt{-14sqrt2+20})`

`<=>x^3=40+3root{3}{400-392}.x`

`<=>x^3=40+6x`

`<=>x^3-6x=40`

\(VT=\left(\dfrac{\sqrt{14}.\sqrt{14}}{\sqrt{14}}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right)\sqrt{5-\sqrt{21}}\\ =\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\\ =\sqrt{14}.\sqrt{5-\sqrt{2}1}+\sqrt{6}.\sqrt{5-\sqrt{21}}\\ =\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\\ =\sqrt{49-2.7.\sqrt{21}+21}+\sqrt{9-2.3.\sqrt{21}+21}\\ =\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(3-\sqrt{21}\right)^2}\\ =\left|7-\sqrt{21}\right|+\left|3-\sqrt{21}\right|\\ =7-\sqrt{21}+\sqrt{21}-3\\ =7-3=4=VP\)

help me now

\(VT=\left(\sqrt{14}+\sqrt{6}\right)\cdot\sqrt{5-\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\cdot\sqrt{10-2\sqrt{21}}\)

=(căn 7+căn 3)(căn 7-căn 3)

=7-3

=4=VP

a/ ĐKXĐ: \(x\ge1\)

Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm

b/ \(x\ge1\)

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

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

Đặt \(\sqrt{x-1}=a\ge0\) ta được:

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

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

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

- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)

- Với \(0\le a\le1\) ta được:

\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)

- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)

c/ ĐKXĐ: \(x\ge\frac{49}{14}\)

\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)

Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(7-\sqrt{14x-49}\ge0\)

\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)

Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)

d/ ĐKXĐ: \(x\ge\frac{1}{2}\)

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

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

TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)

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

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

\(\Leftrightarrow x=13\)

TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)

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

\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)

TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)

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

\(\Leftrightarrow4=4\) (luôn đúng)

TH4: \(\frac{1}{2}\le x< 1\)

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

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

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)

ta có

\(2A=\left(\sqrt{x^2-5x+14}-\sqrt{x^2-5x+10}\right)\left(\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}\right)\)

⇔ 2A=x²-5x+14-x²+5x-10

⇔2A= 4

⇔ A=2

Đặt  \(\sqrt{x^2-5x+14}=a\) và \(\sqrt{x^2-5x+10}=b\) \(\left(a,b>0\right)\)

\(\Rightarrow a-b=2\)

\(\Rightarrow a^2-b^2=x^2-5x+14-x^2+5x-10=4\)

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=4\)

\(\Leftrightarrow a-b=2\)

\(\Leftrightarrow\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}=2\left(đpcm\right)\)

Xem lại đề là - 5x hay 5x nha bạn

\(x=\sqrt[3]{30+14\sqrt{2}}-\sqrt[3]{20+14\sqrt{2}}\)

\(=\sqrt[3]{\left[2^3+3.2^2.\sqrt{2}+3.2+\sqrt{2^2}+\left(\sqrt{2}\right)^3\right]}+\sqrt[3]{\left[2^3-3.2.\sqrt{2}+3.2.\sqrt{2^2}-\left(\sqrt{2}\right)^3\right]}\)

\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)

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

\(=4\)

Vậy x = 4.