giúp mình câu 2 và câu 3 gấp nhé
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\(2.\)
\(a,1=\sqrt{1}\)
\(1< 2< =>\sqrt{1}< \sqrt{2}< =>1< \sqrt{2}\)
\(b,2-\sqrt{2}-1\)
\(1-\sqrt{2}\)
mà \(1< \sqrt{2}< =>1-\sqrt{2}< 0\)
\(2< \sqrt{2}+1\)
c, 7 và \(5\sqrt{2}\)
\(7=\sqrt{49}\)
\(5\sqrt{2}=\sqrt{50}< =>\sqrt{49}< \sqrt{50}\)
\(< =>7< 5\sqrt{2}\)
d, 7 và \(\sqrt{47}\)
\(7=\sqrt{49}< =>\sqrt{49}>\sqrt{47}\)
\(7>\sqrt{47}\)
e, 1 và \(\sqrt{3}-1\)
\(1-\sqrt{3}+1=2-\sqrt{3}=\sqrt{4}-\sqrt{3}>0\)
\(1>\sqrt{3}-1\)
f,\(2\sqrt{31}\)và \(10\)
\(2\sqrt{31}=\sqrt{124}\)
\(10=\sqrt{100}< =>\sqrt{124}>\sqrt{100}\)
\(2\sqrt{31}>10\)
\(3.A=\sqrt{1-4a+4a^2}-2a\)
\(A=\sqrt{\left(1-2a\right)^2}-2a\)
\(A=\left|1-2a\right|-2a\)
kết hợp với ĐKXĐ \(x\ge0,5\)
\(A=2a-1-2a=-1\)
\(b,B=\sqrt{x-2+2\sqrt{x-3}}\)
\(B=\sqrt{x-3+2\sqrt{x-3}+1}\)
\(B=\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(B=\left|\sqrt{x-1}+1\right|\)
kết hợp đkxđ
\(B=\sqrt{x-1}+1\)
\(C=\sqrt{x-2\sqrt{x}+1}+\sqrt{x+2\sqrt{x}+1}\)
\(C=\sqrt{\left(\sqrt{x}-1\right)^2}+\sqrt{\left(\sqrt{x}+1\right)^2}\)
\(C=\left|\sqrt{x}-1\right|+\left|\sqrt{x}+1\right|\)
\(TH1:0\le x\le1\)
\(C=1-\sqrt{x}+\sqrt{x}+1=2\)
\(TH2:x>1\)
\(C=\sqrt{x}-1+\sqrt{x}+1=2\sqrt{x}\)
\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
\(D=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
kết hợp với đkxđ
\(TH1:1\le x\le4\)
\(D=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(TH2:x>4\)
\(D=\sqrt{x-1}+1+\sqrt{x-1}+1=2\sqrt{x-1}\)
Ta có:\(Q=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\frac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
\(\left(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{x+\sqrt{x}-2}:\frac{1}{x-1}\)
\(\frac{3x+5\sqrt{x}-2}{x+\sqrt{x}-2}.x-1\)
\(\frac{3x-\sqrt{x}+6\sqrt{x}-2}{\sqrt{x}+2}\)
\(\frac{\sqrt{x}\left(3\sqrt{x}-1\right)+2\left(3\sqrt{x}-1\right)}{\sqrt{x}+2}\)
\(M=3\sqrt{x}-1\)
Ta có:\(M=\left(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(=\left(\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\frac{1}{x-1}\)
\(=\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\frac{1}{x-1}\)
\(=\frac{3x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\frac{x-1}{1}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(3\sqrt{x}-1\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\left(3\sqrt{x}-1\right)\left(x-1\right)}{\left(\sqrt{x}-1\right)}\)
\(a,\left(3-\frac{\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\right)\left(3+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\)
\(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(3^2-\sqrt{5}^2\)
\(=4\)
\(b,\frac{6\sqrt{2}-2\sqrt{6}}{\sqrt{6}-\sqrt{2}}+\frac{3}{3-\sqrt{2}}\)
\(\frac{\sqrt{6}\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{6}-\sqrt{2}}+\frac{3}{3-\sqrt{2}}\)
\(\sqrt{12}+\frac{3}{3-\sqrt{2}}\)
\(\frac{6\sqrt{3}-2\sqrt{6}-3}{3-\sqrt{2}}\)
\(c,\frac{4+2\sqrt{3}}{4-\sqrt{15}}\frac{\left(\sqrt{5}-\sqrt{3}\right)}{2}-2\sqrt{3}\)
\(\frac{2+\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{4-\sqrt{15}}-2\sqrt{3}\)
\(\frac{2\sqrt{5}+\sqrt{15}-2\sqrt{3}-3}{4-\sqrt{15}}-2\sqrt{3}\)
\(\frac{2\sqrt{5}+\sqrt{15}-2\sqrt{3}-3+6\sqrt{5}-8\sqrt{3}}{4-\sqrt{15}}\)
\(\frac{8\sqrt{5}+\sqrt{15}-10\sqrt{3}-3}{4-\sqrt{15}}\)
\(\sqrt{49a^2}+3a\)
\(\sqrt{49}\sqrt{a^2}+3a\)
\(7a+3a\)
\(=10a\)
Đk: \(a^2>b^2\); b \(\ne\)0
Q = \(\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\)
Q = \(\frac{a}{\sqrt{a^2-b^2}}-\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\cdot\frac{a-\sqrt{a^2-b^2}}{b}\)
Q = \(\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
Q = \(\frac{ab-b^2}{b\sqrt{a^2-b^2}}=\frac{b\left(a-b\right)}{b\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(1=\sqrt{1}\)
\(\sqrt{1}< \sqrt{2}\)
\(\Rightarrow1< \sqrt{2}\)
\(2=1+1\)
\(1+1< \sqrt{2}+1\)
\(\Rightarrow2< \sqrt{2}+1\)
\(7=\sqrt{49}\)
\(5\sqrt{2}=\sqrt{50}\)
\(\sqrt{49}< \sqrt{50}\Rightarrow7< 5\sqrt{2}\)
\(7=\sqrt{49}\)
\(\sqrt{49}>\sqrt{47}\)
\(\Rightarrow7>\sqrt{47}\)
\(1=2-1=\sqrt{4}-1\)
\(\sqrt{4}-1>\sqrt{3}-1\)
\(\Rightarrow1>\sqrt{3}-1\)
\(2\sqrt{31}=\sqrt{124}\)
\(10=\sqrt{100}\)
\(\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)