Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)

\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)

\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)

\(=7-4\sqrt{3}+4\sqrt{3}+2\)

\(=9\)

Làm luôn phần y :D

y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)

y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)

y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)

y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)

y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)

y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)| 

y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\)   (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)

y = 9

Vậy y = 9

Chúc bn học tốt!

d)

ĐK: $x\leq \frac{16}{7}$

PT $\Rightarrow 16-7x=11^2=121$

$\Rightarrow 7x=16-121=-105$

$\Leftrightarrow x=-15$ (thỏa mãn)

e) ĐK: $x\geq 3$

PT $\Rightarrow 10(x-3)=30$ (bình phương 2 vế)

$\Leftrightarrow x-3=3\Leftrightarrow x=6$

(thỏa mãn)

f)

ĐK: $x\geq 2$

PT $\Leftrightarrow \sqrt{x-2}=6$

$\Rightarrow x-2=6^2=36\Leftrightarrow x=38$ (thỏa mãn)

a)

ĐK: $x\geq \frac{-5}{2}$

PT $\Rightarrow 2x+5=25$ (bình phương 2 vế)

$\Leftrightarrow 2x=10\Leftrightarrow x=5$ (thỏa mãn)

b) ĐK: $x\geq \frac{-1}{3}$

PT $\Rightarrow 3x+1=10$ (bình phương 2 vế)

$\Leftrightarrow 3x=9\Leftrightarrow x=3$ (thỏa mãn)

c)

ĐK: $x\geq 7$

PT $\Leftrightarrow \sqrt{x-7}=3+0=3$

$\Rightarrow x-7=3^2$

$\Leftrightarrow x=16$ (thỏa mãn)

a,ĐK: x\(\ge\)1

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

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

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

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha

a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)

=>x+1=5

=>x=4

b: =>x^2/10=1,1

=>x^2=11

=>x=căn 11 hoặc x=-căn 11

c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0

=>4x+3=9x+9

=>-5x=6

=>x=-6/5

d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0

=>2x-3=4x-4 và x>=3/2

=->-2x=-1 và x>=3/2

=>x=1/2 và x>=3/2

=>Ko có x thỏa mãn

e: Đặt căn x=a(a>=0)

PT sẽ là a^2-a-5=0

=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)

=>x=(1+căn 21)^2/4=(11+căn 21)/2

\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)

Áp dụng BDT: Cô-si:

\(\Rightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)\ge2\sqrt{\dfrac{9}{\sqrt{x-19}}\cdot\sqrt{x-19}}+2\sqrt{\dfrac{16}{\sqrt{y-5}}\cdot\sqrt{y-5}}+2\sqrt{\dfrac{25}{\sqrt{z-91}}\cdot\sqrt{z-91}}\\ =2\cdot3+2\cdot4+2\cdot5=24\)Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{x-19}}=\sqrt{x-19}\\\dfrac{16}{\sqrt{y-5}}=\sqrt{y-5}\\\dfrac{25}{\sqrt{z-91}}=\sqrt{z-91}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-19=9\\y-5=16\\z-91=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=21\\z=116\end{matrix}\right.\)

Vậy các số \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)

1/Liên hợp đi cho nó nhẹ:D

ĐKXĐ: \(x\ge16\)

PT \(\Leftrightarrow\sqrt{x+24}-7+\sqrt{x-16}-3=0\)

\(\Leftrightarrow\frac{x-25}{\sqrt{x+24}+7}+\frac{x-25}{\sqrt{x-16}+3}=0\)

\(\Leftrightarrow\left(x-25\right)\left(\frac{1}{\sqrt{x+24}+7}+\frac{1}{\sqrt{x-16}+3}\right)=0\)

\(\Leftrightarrow x=25\)

a) ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

b) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

c) ĐKXĐ: \(x>-\dfrac{5}{3}\)

d) ĐKXĐ: \(3\le x\le10\)

e) ĐKXĐ: \(\left\{{}\begin{matrix}x>-4\\x\ne4\end{matrix}\right.\)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

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

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

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

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)