\(\dfrac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\)
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30 tháng 6 2021
c) Ta có: \(C=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
d)
Sửa đề: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
Ta có: \(D=\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 9 2023
\(\left(\dfrac{2}{2-\sqrt{x}}+\dfrac{3+\sqrt{x}}{x-2\sqrt{x}}\right):\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right)\) (ĐK: \(x\ne4;x>0\))
\(=\left[\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]:\left[\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{-2\sqrt{x}+\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left[\dfrac{-\left(\sqrt{x}+2\right)^2+\left(\sqrt{x}-2\right)^2-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-x-4\sqrt{x}-4+x+4\sqrt{x}+4-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{-4x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{-4x}\)
\(=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{-4x}\)
\(=-\dfrac{3\sqrt{x}+6-x-2\sqrt{x}}{4x}\)
\(=-\dfrac{\sqrt{x}-x+6}{4x}\)
H9
HT.Phong (9A5)
CTVHS
14 tháng 9 2023
Mình sửa lại phần mẫu số của 3 dòng cuối nhé !
\(-4x\Rightarrow-4x\sqrt{x}\)
\(4x\Rightarrow4x\sqrt{x}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
H9
HT.Phong (9A5)
CTVHS
31 tháng 8 2023
\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(A=\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2021
1. ĐKXĐ:...
\(8-2x-\dfrac{2}{x}-2\sqrt{2-x^2}-2\sqrt{2-\dfrac{1}{x^2}}=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(\dfrac{1}{x^2}-\dfrac{2}{x}+1\right)+\left(2-x^2-2\sqrt{2-x^2}+1\right)+\left(2-\dfrac{1}{x^2}-2\sqrt{2-\dfrac{1}{x^2}}+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(\dfrac{1}{x}-1\right)^2+\left(\sqrt{2-x^2}-1\right)^2+\left(\sqrt{2-\dfrac{1}{x^2}}-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x}-1=0\\\sqrt{2-x^2}-1=0\\\sqrt{2-\dfrac{1}{x^2}}-1=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2021
2.
ĐKXĐ:...
Ta có:
\(VT=x\sqrt{x}+1.\sqrt{12-x}\le\sqrt{\left(x^2+1\right)\left(x+12-x\right)}=2\sqrt{3\left(x^2+1\right)}\)
Dấu "=" xảy ra khi và chỉ khi: \(x\sqrt{12-x}=\sqrt{x}\)
\(\Leftrightarrow x^3-12x^2+x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=6-\sqrt{35}\\x=6+\sqrt{35}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
AH
Akai Haruma
Giáo viên
17 tháng 7 2021
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
25 tháng 8 2021
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
26 tháng 8 2021
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
= \(\dfrac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}\)
=\(\dfrac{1-2\sqrt{x}+x}{1-\sqrt{x}}\)
=1-\(\sqrt{x}\)