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\(\dfrac{41}{36}-\dfrac{11}{12}=\dfrac{41}{36}-\dfrac{33}{36}=\dfrac{8}{36}=\dfrac{2}{9}\)
Chúc bạn học tốt
\(\dfrac{11}{12}=\dfrac{33}{36}\)
Trừ 2 phân số cùng mẫu số ta lấy tử trừ tử, mẫu số là mẫu số chung
A= \(\dfrac{8}{36}=\dfrac{2}{9}\)
\(=\sqrt{\dfrac{2\left(3-\sqrt{5}\right)}{2\left(2-\sqrt{3}\right)}}=\sqrt{\dfrac{6-2\sqrt{5}}{4-2\sqrt{3}}}=\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{3}-1\right)^2}}=\dfrac{\sqrt{5}-1}{\sqrt{3}-1}\)
\(\left(\dfrac{y\sqrt{y}-x\sqrt{x}}{\sqrt{y}-\sqrt{x}}+\sqrt{xy}\right)\left(\dfrac{\sqrt{y}-\sqrt{x}}{y-x}\right)^2\)
\(=\left(\sqrt{xy}+\sqrt{xy}\right)\left(\sqrt{y}+\sqrt{x}\right)^2=2\sqrt{xy}\left(x+2\sqrt{xy}+y\right)\)
\(=2x\sqrt{xy}+4xy+2y\sqrt{xy}\)
sửa bài : \(\left(\dfrac{y\sqrt{y}-x\sqrt{x}}{\sqrt{y}-\sqrt{x}}+\sqrt{xy}\right)\left(\dfrac{\sqrt{y}-\sqrt{x}}{y-x}\right)^2\)ĐK : \(x\ne y;x;y>0\)
\(=2\sqrt{xy}\left(\dfrac{1}{\sqrt{y}+\sqrt{x}}\right)^2=\dfrac{2\sqrt{xy}}{x+2\sqrt{xy}+y}\)
\(a,\dfrac{2}{8}=\dfrac{1}{4}\\ b,\dfrac{6}{18}=\dfrac{1}{3}\\ c,\dfrac{17}{24}-\dfrac{8}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)
\(a,\dfrac{5}{8}-\dfrac{3}{8}=\dfrac{5-3}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
\(b,\dfrac{23}{18}-\dfrac{17}{18}=\dfrac{23-17}{18}=\dfrac{6}{18}=\dfrac{1}{3}\)
\(c,\dfrac{17}{24}-\dfrac{1}{3}=\dfrac{17}{24}-\dfrac{8}{24}=\dfrac{17-8}{24}=\dfrac{9}{24}=\dfrac{3}{8}\)
13) \(\sqrt{x^2}=\left|x\right|\)
14) \(\sqrt{x^4}=x^2\)
15) \(\sqrt{\left(3x-2\right)^2}=\left|3x-2\right|\)
16) \(\sqrt{\left(2x+1\right)^2}=\left|2x+1\right|\)
17) \(\sqrt{a^2-2a+1}=\left|a-1\right|\)
18) \(\sqrt{4a^2-4a+1}=\left|2a-1\right|\)
19) \(\sqrt{x^2+6x+9}=\left|x+3\right|\)
20) \(\sqrt{a^2-a+\dfrac{1}{4}}=\left|a-\dfrac{1}{2}\right|\)
21) \(\sqrt{4x^2-4xy+y^2}=\left|2x-y\right|\)
\(=\dfrac{x^2+2x-x^2+4x-4+6-5x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)
`sqrt{8-4sqrt3}-sqrt{14+8sqrt3}`
`=sqrt{2(4-2sqrt3)}-sqrt{2(7+4sqrt3)}`
`=sqrt{2(3-2sqrt3+1)}-sqrt{2(4+2.2.sqrt3+3)}`
`=sqrt{2(sqrt3-1)^2}-sqrt{2(2+sqrt3)^2}`
`=sqrt2(sqrt3-1)-sqrt2(2+sqrt3)`
`=sqrt6-sqrt2-2sqrt2-sqrt6`
`=-3sqrt2`
Sửa đề: a = b => x = y
\(P=\left(1-\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)\left(1+\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)\) (ĐK: \(x,y>0\))
\(=1-\left(\dfrac{\sqrt{2xy}}{\sqrt{x^2+y^2}}\right)^2\)
\(=1-\dfrac{\left(\sqrt{2xy}\right)^2}{\left(\sqrt{x^2+y^2}\right)^2}\)
\(=1-\dfrac{2xy}{x^2+y^2}\)
\(=\dfrac{x^2+y^2-2xy}{x^2+y^2}\)
\(=\dfrac{\left(x-y\right)^2}{x^2+y^2}\)
Khi x = y, ta được: \(P=\dfrac{\left(x-y\right)^2}{x^2+y^2}=\dfrac{\left(x-x\right)^2}{x^2+y^2}=0\)
#Urushi