\(\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{4-\sqrt{7}}}{\sqrt{4+\sqrt{7}}}\)

\(=\left(4+\sqrt{7}\right)\cdot\dfrac{\sqrt{7}-1}{\sqrt{7}+1}\)

\(=\dfrac{\left(\sqrt{7}+1\right)^2\cdot\left(\sqrt{7}-1\right)}{\sqrt{7}+1}\cdot\dfrac{1}{2}\)

\(=\dfrac{6}{2}=3\)

\(=\dfrac{\left(8+2\sqrt{7}\right)\sqrt{8-2\sqrt{7}}}{2\sqrt{8+2\sqrt{7}}}=\dfrac{\left(\sqrt{7}+1\right)^2\sqrt{\left(\sqrt{7}-1\right)^2}}{2\sqrt{\left(\sqrt{7}+1\right)^2}}\)

\(=\dfrac{\left(\sqrt{7}+1\right)^2\left(\sqrt{7}-1\right)}{2\left(\sqrt{7}+1\right)}=\dfrac{\left(\sqrt{7}+1\right)\left(\sqrt{7}-1\right)}{2}\)

\(=\dfrac{7-1}{2}=3\)

√4−√7−√4+√7+√7=√2(√4−√7−√4+√7+√7)√2=√8−2√7−√8+2√7+√14√2=√7−2√7+1−√7+2√7+1+√14√2=√(√7−1)2−√(√7+1)2+√14√2=∣∣√7−1∣∣−∣∣√7+1∣∣+√14√2=√7−1−√7−1+√14√2=√14−2√2=√2(√7−√2)√2=√7−√2

Lời giải:
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\sqrt{\frac{8-2\sqrt{7}}{2}}-\sqrt{\frac{8+2\sqrt{7}}{2}}=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-\sqrt{\frac{(\sqrt{7}+1)^2}{2}}\)

\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-\frac{|\sqrt{7}+1|}{\sqrt{2}}=\frac{\sqrt{7}-1-(\sqrt{7}+1)}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

a) \(\sqrt{2}\left(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\right)\)

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

\(=\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=\left|\sqrt{7}+1\right|+\left|\sqrt{7}-1\right|\)

\(=\sqrt{7}+1+\sqrt{7}-1\)

\(=2\sqrt{7}\)

b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

\(=\dfrac{\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{2\cdot\left(2-\sqrt{3}\right)}-\sqrt{2\cdot\left(2+\sqrt{3}\right)}}{\sqrt{2}}\)

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

\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\)

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

\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{ }\)

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

\(=-\sqrt{2}\)

a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}=4\)

b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

\(=2\sqrt{3}\)

c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

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

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

A= \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\)\(1+\sqrt{7}+\sqrt{7}-1=2\sqrt{7}\)

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=0\)

:) trình bày các bước đi bạn :)) ai lại làm thế :v Bấm casio à :)

\(H=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

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

\(=8-2\sqrt{16-7}=8-6=2\)

\(\Rightarrow H=\sqrt{2}\Rightarrow\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-2=0\)

Vậy .....................

b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

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

\(=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)

\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)

\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)

\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)

\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)

B1.

Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)

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

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

Bài 1 : 

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

Bài 2 : 

a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)

b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)

Bài 3 : 

\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)

\(B=\frac{1}{\sqrt{5}+\sqrt{7}}-\frac{1}{\sqrt{5}-\sqrt{7}}=\frac{\sqrt{5}-\sqrt{7}-\sqrt{5}-\sqrt{7}}{5-7}=\frac{-2\sqrt{7}}{-2}=\sqrt{7}\)

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

\(C=\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}+2\sqrt{\frac{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}}+\frac{4-\sqrt{7}}{4+\sqrt{7}}}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}\right)^2}{16-7}+\frac{\left(4-\sqrt{7}\right)^2}{16-7}+2}\)

\(C=\sqrt{\frac{\left(4+\sqrt{7}+4-\sqrt{7}\right)^2-2\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}{16-7}+2}\)

\(C=\sqrt{\frac{16^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{238}{9}+2}=\sqrt{\frac{256}{9}}=\frac{16}{3}\)

Chúc bạn học tốt ~ 

thanks ban 

đoạn cuối sửa lại nhé -,- tính ngu 

\(C=\sqrt{\frac{8^2-2\left(16-7\right)}{9}+2}=\sqrt{\frac{46}{9}+2}=\sqrt{\frac{64}{9}}=\frac{8}{3}\)

Chúc bạn học tốt ~