Ta có\(8< 16\Rightarrow\sqrt{8}< \sqrt{16}=4\)

và \(5< 9\Rightarrow\sqrt{5}< \sqrt{9}=3\)

\(\Rightarrow\sqrt{8}-\sqrt{5}< \sqrt{16}-\sqrt{9}=4-3=1\)

Vậy \(\sqrt{8}-\sqrt{5}< 1\)

Ta có \(\sqrt{63-27}=\sqrt{36}=6\)

lại có\(63< 64\Rightarrow\sqrt{63}< \sqrt{64}=8\)và \(27>4\Rightarrow\sqrt{27}>\sqrt{4}=2\)

\(\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{64}-\sqrt{4}=8-2=6\)

mà\(\sqrt{63-27}=6\Rightarrow\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)

Vậy\(\sqrt{63}-\sqrt{27}< \sqrt{63-27}\)

√8_√5< 1

√63−27 > √63_√27

bạn có thể giải kĩ hơn đc ko?

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

\(\sqrt{27}\) +  \(\sqrt{6}\)>  \(\sqrt{35}\)

căn 27 + căn 6 = 7,196156423

căn 35 = 5,916079783

=>căn 27 + căn 6 > căn 35

Ta có: \(\sqrt{27}+\sqrt{6}=\sqrt{6}+3\sqrt{3}=7,64...\)

Và: \(\sqrt{35}=5,91\)

Vì \(7,64>5,91\)

Vậy \(\sqrt{27}+3\sqrt{3}>\sqrt{35}\)

Ủng hộ mik nha!

a, ta có:
\(\sqrt{24}=4,89\\ \sqrt{3}=1,73\)

\(\Rightarrow\sqrt{24}+\sqrt{3}=4,89+1,73=6,62\)
vì 7>6,62 nên 7>\(\sqrt{24}+\sqrt{3}\)

\(7=5+2=\sqrt{25}+\sqrt{4}>\sqrt{24}+\sqrt{3}\)

\(\left(\sqrt{50+2}\right)^2=\left|50+2\right|=50+2\\ \left(\sqrt{50}+\sqrt{2}\right)^2=50+20+2>50+2=\left(\sqrt{50+2}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{50+2}\right)^2}< \sqrt{\left(\sqrt{50}+\sqrt{2}\right)^2}\\ \Leftrightarrow\left|\sqrt{50+2}\right|< \left|\sqrt{50}+\sqrt{2}\right|\\ \Leftrightarrow\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)

\(\left(\sqrt{63-27}\right)^2=\left|63-27\right|=63-27\\ \left(\sqrt{63}-\sqrt{27}\right)^2=63-2\sqrt{90}+27>63-20+27=63-7>63-27=\left(\sqrt{63-27}\right)^2\\ \Rightarrow\sqrt{\left(\sqrt{63-27}\right)^2}< \sqrt{\left(\sqrt{63}-\sqrt{27}\right)^2}\\ \Leftrightarrow\left|\sqrt{63-27}\right|< \left|\sqrt{63}-\sqrt{27}\right|\\ \Leftrightarrow\sqrt{63-27}< \sqrt{63}-\sqrt{27}\)

b, Ta có:
\(\sqrt{50+2}=\sqrt{52}=7,21\\ \sqrt{50}+\sqrt{2}=7,07+1,41=8,48 \)

vì 7,21<8,48 nên \(\sqrt{50+2}< \sqrt{50}+\sqrt{2}\)