rút gọn: x3+2x2+ 2\(\sqrt{2}\)x+ 2\(\sqrt{2}\)
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Những câu hỏi liên quan
5 tháng 5 2021
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
10 tháng 11 2021
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
19 tháng 8 2023
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
H9
HT.Phong (9A5)
CTVHS
16 tháng 8 2023
\(\left(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right):\left(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\right)\) (ĐK: \(x>0;x\ne4\))
\(=\left[\dfrac{2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\sqrt{x}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\right]\)
\(=\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left(\sqrt{x}-\sqrt{x}+2\right)\)
\(=-\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}:2\)
\(=-\dfrac{1}{\sqrt{x}}:2\)
\(=-\dfrac{1}{\sqrt{x}}\cdot\dfrac{1}{2}\)
\(=-\dfrac{1}{2\sqrt{x}}\)
HL
1
H9
HT.Phong (9A5)
CTVHS
8 tháng 11 2023
\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-2}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne4\right)\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{5\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2x-5\sqrt{x}+2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
HV
2
TA
28 tháng 4 2023
\(A=\dfrac{-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{-2\left(\sqrt{x}+2\right)-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-2\sqrt{x}-4-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-4-x}{x-4}\)
28 tháng 4 2023
\(=\dfrac{-2\sqrt{x}-4-x+2\sqrt{x}}{x-4}=\dfrac{-x-4}{x-4}\)
T
1
26 tháng 6 2023
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{x-2\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)