Ta có:

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

\(=1+2\sqrt{27\sqrt{2}-38}\)

Áp dụng vào bài toán t được

\(\dfrac{\sqrt{1+2\sqrt{27\sqrt{2}-38}}-\sqrt{5-3\sqrt{2}}}{\sqrt{3\sqrt{2}-4}}\)

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

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

\(A=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{16.5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)

\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)

\(B=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)

\(=2\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}+3\sqrt{22}=22\)

Đk: tự tìm

\(pt\Leftrightarrow\sqrt{\left(x-4\right)\left(x+4\right)}+\sqrt{x-4}=0\)

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

Dễ thấy: \(\sqrt{x+4}\ge0\forall x\)

\(\Rightarrow\sqrt{x+4}+1\ge1>0\forall x\) (vô nghiệm)

\(\Rightarrow\sqrt{x-4}=0\Rightarrow x-4=0\Rightarrow x=4\)

??????????????????????????????????????

??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

a, ĐK: \(x\ge11\)

\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)

\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)

Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{​​}8\)

\(\Rightarrow\) phương trình vô nghiệm.

\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)

\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)

b, ĐK: \(x\ge\dfrac{5}{2}\)

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

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

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

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

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

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

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

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

\(=4\)

Đẳng thức xảy ra khi: 

\(\left(\sqrt{2x-5}+3\right)\left(1-\sqrt{2x-5}\right)\ge0\)

\(\Leftrightarrow1-\sqrt{2x-5}\ge0\)

\(\Leftrightarrow\sqrt{2x-5}\le1\)

\(\Leftrightarrow0\le2x-5\le1\)

\(\Leftrightarrow\dfrac{5}{2}\le x\le3\)

Điều kiện xác định của phương trình : \(1\le x\le5\)

Xét vế trái của phương trình , áp dụng bđt Bunhiacopxki , ta có: 

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

\(\Leftrightarrow\left(2\sqrt{x-1}+3\sqrt{5-x}\right)^2\le52\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{5-x}\le2\sqrt{13}\)

Dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}1\le x\le5\\\frac{\sqrt{x-1}}{3}=\frac{\sqrt{5-x}}{2}\end{cases}}\)\(\Leftrightarrow x=\frac{29}{13}\)

Vậy pt có nghiệm \(x=\frac{29}{13}\)

tks bạn nhìu

help

11/18

THAM KHẢO: