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GS
0
PA
31 tháng 7 2017
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
NV
30 tháng 7 2015
a. \(\Rightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\Rightarrow\sqrt{x+5}\left(2-3+4\right)=6\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)
b.\(\Rightarrow5\sqrt{x-1}-\frac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\Rightarrow\sqrt{x-1}\left(5-\frac{5}{2}-1\right)=6\Rightarrow\sqrt{x-1}=4\Rightarrow x-1=16\Rightarrow x=17\)
1 tháng 7 2019
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
1 tháng 7 2019
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
19 tháng 11 2016
\(P=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\left(1-\sqrt{x}\right)\left(5\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{2-5\sqrt{x}}{3+\sqrt{x}}\)
19 tháng 11 2016
\(N=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}-3}{2-\sqrt{x}}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}+3\right)}-1\right):\left(\frac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=-3:\left(\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)}\right)\)
\(=3.\left(\frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)}\right)\)
T
24 tháng 11 2019
b)\(\frac{2}{3}.\sqrt{4x^2-20}+2\sqrt{\frac{x^2-5}{9}}-3\sqrt{x^2-5}=2\)
\(< =>\frac{2}{3}.\sqrt{4\left(x^2-5\right)}+2\cdot\frac{\sqrt{x^2-5}}{3}-3\sqrt{x^2-5}=2\)
\(< =>\frac{2}{3}.2\sqrt{\left(x^2-5\right)}+2\cdot\frac{\sqrt{x^2-5}}{3}-3\sqrt{x^2-5}=2\)
\(< =>\frac{4}{3}\sqrt{\left(x^2-5\right)}+\frac{2}{3}.\sqrt{x^2-5}-3\sqrt{x^2-5}=2\)
\(< =>-\sqrt{\left(x^2-5\right)}=2\)
\(< =>\sqrt{\left(x^2-5\right)}=-2\)(vô nghiệm)
T
24 tháng 11 2019
a)\(\sqrt{25x-25}-\frac{15}{2}\sqrt{\frac{x-1}{9}}=6+\frac{3}{2}\sqrt{x-1}\)
\(< =>\sqrt{25\left(x-1\right)}-\frac{15}{2}.\frac{\sqrt{x-1}}{3}-\frac{3}{2}\sqrt{x-1}=6\)
\(< =>5\sqrt{x-1}-\frac{5}{2}.\sqrt{x-1}-\frac{3}{2}\sqrt{x-1}=6\)
\(< =>\sqrt{x-1}=6\)
\(< =>x-1=36\)
\(< =>x=37\)
vậy ...
NV
24 tháng 6 2016
a/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
c/ đề \(=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 6 2019
ĐKXĐ:...
\(\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)
\(=\left(\frac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{\left(9-x\right)}\)
\(=\frac{5\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{5}{\sqrt{x}+3}\)
