rút gọn biểu thức \(\sqrt{x}+\sqrt{\left(x+1\right)^2=1+x}\)X
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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: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
6 tháng 4 2022
\(P=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
AT
25 tháng 7 2021
\(A=\left(\dfrac{\sqrt{x}}{x-\sqrt{x}}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}}{x+1}\right)\left(x>0,x\ne1\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\dfrac{x-\sqrt{x}+1}{x+1}\)
\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{x+1}{x-\sqrt{x}+1}\)
\(=\dfrac{x+1-2}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+1\right)}.\dfrac{x+1}{x-\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
AH
Akai Haruma
Giáo viên
25 tháng 7 2021
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(A=\left[\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)}-\frac{2}{(\sqrt{x}-1)(x+1)}\right]:\frac{x-\sqrt{x}+1}{x+1}\)
\(=\left[\frac{1}{\sqrt{x}-1}-\frac{2}{(\sqrt{x}-1)(x+1)}\right].\frac{x+1}{x-\sqrt{x}+1}=\frac{x+1-2}{(\sqrt{x}-1)(x+1)}.\frac{x+1}{x-\sqrt{x}+1}=\frac{x-1}{(\sqrt{x}-1)(x-\sqrt{x}+1)}=\frac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
29 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\);\(ĐK:x\ge0;x\ne1\)
\(A=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{x-\sqrt{x}+2\sqrt{x}-2-2\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
29 tháng 5 2022
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right).\left(\sqrt{x-1}\right)\left(đk:x\ne1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-2\right).\left(\sqrt{x}-1\right)\)
\(A=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\left(\dfrac{\sqrt{x}+2-2\sqrt{x}-2}{\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\\ A=\dfrac{-\sqrt{x}}{\sqrt{x}+1}.\left(\sqrt{x}-1\right)\\ A=\dfrac{-x+\sqrt{x}}{\sqrt{x}+1}\)
25 tháng 7 2021
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
TA
21 tháng 7 2021
ĐK: `x>0`
`A=((\sqrtx)/(\sqrtx+2) - 4/(x+2\sqrtx)):(1+1/(\sqrtx))`
`=((\sqrtx)/(\sqrtx+2)-4/(\sqrtx(\sqrtx+2))):((\sqrtx+1)/(\sqrtx))`
`=(x -4)/(\sqrtx(\sqrtx+2)) . (\sqrtx)/(\sqrtx+1)`
`=((\sqrtx+2)(\sqrtx-2))/(\sqrtx+2) . 1/(\sqrtx+1)`
`=(\sqrt-2)/(\sqrtx+1)`
21 tháng 7 2021
Ta có:\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{x-4}{x\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2}{x}.\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
18 tháng 7 2021
`P=((xsqrtx)/(sqrtx-1)-x^2/(xsqrtx-x))(1/sqrtx-2)(x>0,x ne 1,x ne 1/4)`
`=((xsqrtx)/(sqrtx-1)-x^2/(x(sqrtx-1))((1-2sqrtx)/sqrtx)`
`=((xsqrtx)/(sqrtx-1)-x/(sqrtx-1))((1-2sqrtx)/sqrtx)`
`=(xsqrtx-x)/(sqrtx-1)((1-2sqrtx)/sqrtx)`
`=(x(sqrtx-1))/(sqrtx-1)((1-2sqrtx)/sqrtx)`
`=x*((1-2sqrtx)/sqrtx)`
`=sqrtx(1-2sqrtx)`
`=sqrtx-2x`
H9
HT.Phong (9A5)
CTVHS
4 tháng 9 2023
\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\) (ĐK: \(x\ne1;x\ge0\))
\(A=\left[\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(A=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(A=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
\(A=\dfrac{1}{\sqrt{x}+2}\)
GH
2 tháng 7 2023
Với `x>0` có:
\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}+1\right)\\ =(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)})\left(\dfrac{2}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}\right)\)
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}\right)\\ =\dfrac{-2\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right).\sqrt{x}}=-\dfrac{2}{x}\)
2 tháng 7 2023
\(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2+\sqrt{x}}{\sqrt{x}}=\dfrac{-2}{x}\)