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29 tháng 6 2021

\(1=\sqrt{1}\)   

\(\sqrt{1}< \sqrt{2}\)   

\(\Rightarrow1< \sqrt{2}\) 

\(2=1+1\)   

\(1+1< \sqrt{2}+1\)   

\(\Rightarrow2< \sqrt{2}+1\)   

\(7=\sqrt{49}\)   

\(5\sqrt{2}=\sqrt{50}\)   

\(\sqrt{49}< \sqrt{50}\Rightarrow7< 5\sqrt{2}\)

29 tháng 6 2021

\(7=\sqrt{49}\)   

\(\sqrt{49}>\sqrt{47}\)   

\(\Rightarrow7>\sqrt{47}\)   

\(1=2-1=\sqrt{4}-1\)   

\(\sqrt{4}-1>\sqrt{3}-1\)   

\(\Rightarrow1>\sqrt{3}-1\)   

\(2\sqrt{31}=\sqrt{124}\)   

\(10=\sqrt{100}\)   

\(\sqrt{124}>\sqrt{100}\Rightarrow2\sqrt{31}>10\)

29 tháng 6 2021

\(2.\)

\(a,1=\sqrt{1}\)

\(1< 2< =>\sqrt{1}< \sqrt{2}< =>1< \sqrt{2}\)

\(b,2-\sqrt{2}-1\)

\(1-\sqrt{2}\)

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

\(TH1:1\le x\le4\)

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

\(TH2:x>4\)

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

29 tháng 6 2021

lên mạng mà tìm

11 tháng 9 2021
Đề đâu bn ưi
29 tháng 6 2021

Ta có:\(Q=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)

             \(=\left(\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

             \(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)

             \(=\frac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)

29 tháng 6 2021

\(\left(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)

\(\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{x+\sqrt{x}-2}:\frac{1}{x-1}\)

\(\frac{3x+5\sqrt{x}-2}{x+\sqrt{x}-2}.x-1\)

\(\frac{3x-\sqrt{x}+6\sqrt{x}-2}{\sqrt{x}+2}\)

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

\(M=3\sqrt{x}-1\)

29 tháng 6 2021

Ta có:\(M=\left(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)

              \(=\left(\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\frac{1}{x-1}\)

              \(=\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\frac{1}{x-1}\)

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

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

29 tháng 6 2021

\(a,\left(3-\frac{\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\right)\left(3+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\)

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

\(3^2-\sqrt{5}^2\)

\(=4\)

\(b,\frac{6\sqrt{2}-2\sqrt{6}}{\sqrt{6}-\sqrt{2}}+\frac{3}{3-\sqrt{2}}\)

\(\frac{\sqrt{6}\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{6}-\sqrt{2}}+\frac{3}{3-\sqrt{2}}\)

\(\sqrt{12}+\frac{3}{3-\sqrt{2}}\)

\(\frac{6\sqrt{3}-2\sqrt{6}-3}{3-\sqrt{2}}\)

\(c,\frac{4+2\sqrt{3}}{4-\sqrt{15}}\frac{\left(\sqrt{5}-\sqrt{3}\right)}{2}-2\sqrt{3}\)

\(\frac{2+\sqrt{3}\left(\sqrt{5}-\sqrt{3}\right)}{4-\sqrt{15}}-2\sqrt{3}\)

\(\frac{2\sqrt{5}+\sqrt{15}-2\sqrt{3}-3}{4-\sqrt{15}}-2\sqrt{3}\)

\(\frac{2\sqrt{5}+\sqrt{15}-2\sqrt{3}-3+6\sqrt{5}-8\sqrt{3}}{4-\sqrt{15}}\)

\(\frac{8\sqrt{5}+\sqrt{15}-10\sqrt{3}-3}{4-\sqrt{15}}\)

29 tháng 6 2021

\(\sqrt{49a^2}+3a\)

\(\sqrt{49}\sqrt{a^2}+3a\)

\(7a+3a\)

\(=10a\)

29 tháng 6 2021
Câu trả lời bằng hình

Bài tập Tất cả

29 tháng 6 2021

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 = \(\frac{ab-b^2}{b\sqrt{a^2-b^2}}=\frac{b\left(a-b\right)}{b\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)