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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
10 tháng 11 2021
\(a,ĐK:x\ne\pm1;x\ne0\\ M=\dfrac{1-x+2x}{\left(1+x\right)\left(1-x\right)}:\dfrac{1-x}{x}\\ M=\dfrac{x+1}{\left(x+1\right)\left(1-x\right)}\cdot\dfrac{x}{1-x}=\dfrac{x}{\left(1-x\right)^2}\\ b,ĐK:x\ge0;x\ne4\\ N=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ N=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
Tất cả đều phải tìm điều kiện
1 tháng 12 2021
\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)
23 tháng 4 2021
\(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}-\frac{3-\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{3+\sqrt{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)
23 tháng 4 2021
a, \(\frac{1}{3-\sqrt{7}}-\frac{1}{3+\sqrt{7}}=\frac{3+\sqrt[]{7}-3+\sqrt{7}}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\frac{2\sqrt{7}}{9-7}=\sqrt{7}\)
AT
27 tháng 7 2021
\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{x+\sqrt{x}-2}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)-3\left(\sqrt{x}-1\right)+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
27 tháng 7 2021
Ta có: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+2}+\dfrac{3}{x+\sqrt{x}-2}\)
\(=\dfrac{x-4-3\left(\sqrt{x}-1\right)+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-1-3\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 4 2021
\(=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}\right)\left(\dfrac{x\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x-3\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}-1\right)}-\dfrac{x+3\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\right)\)
\(=\left(\dfrac{-6\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\right)=-6\)
9 tháng 4 2021
`N=(x-2sqrtx+1)/(sqrtx-1)-(x+sqrtx)/sqrtx(x>0,x ne 1)`
`=(sqrtx-1)^2/(sqrtx-1)+(sqrtx(sqrtx+1))/sqrtx`
`=sqrtx-1-(sqrtx+1)`
`=-2`
e) Ta có: \(E=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}-\sqrt{6}\)
\(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-2\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1-2\sqrt{3}}{\sqrt{2}}=0\)
f) Ta có: \(F=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)\)
\(=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2\)
=0
g) Ta có: \(G=4\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)\)
\(=2\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}+2\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}-2\sqrt{10}\)
\(=2\sqrt{2}\left(\sqrt{3}+1\right)+2\sqrt{2}\left(\sqrt{5}-1\right)-2\sqrt{10}\)
\(=2\sqrt{6}+2\sqrt{2}+2\sqrt{10}-2\sqrt{2}-2\sqrt{10}\)
\(=2\sqrt{6}\)
\(E=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}-\sqrt{6}\)
\(=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}-\sqrt{6}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}{2}}-\sqrt{6}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}-\sqrt{6}\)
\(=\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}+\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}}-\sqrt{6}\)
\(=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}-\sqrt{6}=\dfrac{2\sqrt{3}}{\sqrt{2}}-\sqrt{6}=\sqrt{6}-\sqrt{6}=0\)
\(F=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)\)
\(=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}-\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}-2\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}-2=\left|\sqrt{7}+1\right|-\left|\sqrt{7}-1\right|-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2=0\)
\(G=4\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)\)
\(=2\sqrt{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{2.\dfrac{5}{2}}\right)\)
\(=2\sqrt{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)\)
\(=2\sqrt{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)=2\sqrt{6}\)
\(H=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\sqrt{2}}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{2}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)