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12 tháng 10 2019
ĐK: \(x\ge-1\)
pt <=> \(\left(14\sqrt{x+35}-84\right)+\left(6\sqrt{x+1}-\sqrt{x^2+36x+35}\right)=0\)
<=> \(14\left(\sqrt{x+35}-6\right)+\sqrt{x+1}\left(6-\sqrt{x+35}\right)=0\)
<=> \(\left(\sqrt{x+35}-6\right)\left(11-\sqrt{x+1}\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x+35}-6=0\\11-\sqrt{x+1}=0\end{cases}}\)Em làm tiếp nhé!
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
1/ \(\dfrac{5}{3}\le x\le\dfrac{7}{3}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{3x-5}=a>0\\\sqrt{7-3x}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\17-6x=2b^2+3\\6x-7=2a^2+3\end{matrix}\right.\)
Mặt khác theo BĐT Bunhiacốpxki:
\(a+b=\sqrt{3x-5}+\sqrt{7-3x}\le\sqrt{\left(1+1\right)\left(3x-5+7-3x\right)}=2\)
\(\Rightarrow0< a+b\le2\)
Ta được hệ pt:
\(\left\{{}\begin{matrix}a^2+b^2=2\\\left(2b^2+3\right).a+\left(2a^2+3\right)b=2+8ab\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2ab^2+3a+2a^2b+3b-8ab-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-2\\2ab\left(a+b\right)+3\left(a+b\right)-8ab-2=0\end{matrix}\right.\)
\(\Rightarrow\left(\left(a+b\right)^2-2\right)\left(a+b\right)+3\left(a+b\right)-4\left(a+b\right)^2+6=0\)
\(\Leftrightarrow\left(a+b\right)^3-4\left(a+b\right)^2+\left(a+b\right)+6=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=-1< 0\left(l\right)\\a+b=2\\a+b=3>2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow a+b=2\) , dấu "=" xảy ra khi và chỉ khi:
\(3x-5=7-3x\Rightarrow x=2\)
Vậy pt có nghiệm duy nhất \(x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
2/ ĐKXĐ: \(x\ne\pm2\)
\(\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-\left(\dfrac{15}{x^2-4}+5\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-5.\left(\dfrac{x^2-1}{x^2-4}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{x^2-1}{x^2-4}\right)-4\left[\left(\dfrac{x^2-1}{x^2-4}\right)-\left(\dfrac{x+1}{x-2}\right)^2\right]=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)-4\left(\dfrac{x+1}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{x+2}-\dfrac{4\left(x+1\right)}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{4\left(x+1\right)}{x-2}\\\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=4\left(x^2+3x+2\right)\\x^2-3x+2=x^2+3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2+15x+6=0\\6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5+\sqrt{17}}{2}\\x=\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)
DH
20 tháng 6 2017
a, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\)
\(=\dfrac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{7}}\)
b, \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
\(=\dfrac{2.\sqrt{5}.\sqrt{3}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{2.\sqrt{5}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}.\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}.\left(1-\sqrt{2}\right)-\sqrt{3}.\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right).\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
c, \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}.\sqrt{x}+\sqrt{x}.\sqrt{y}}{\sqrt{y}.\sqrt{y}+\sqrt{x}.\sqrt{y}}\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)
Chúc bạn học tốt!!!
20 tháng 6 2017
d) \(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\) = \(-\dfrac{\sqrt{a}\left(1+\sqrt{ab}\right)-\sqrt{b}\left(1+\sqrt{ab}\right)}{1-ab}\)
= \(-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(1+\sqrt{ab}\right)}{\left(1+\sqrt{ab}\right)\left(1-\sqrt{ab}\right)}\) = \(-\dfrac{\sqrt{a}-\sqrt{b}}{1-\sqrt{ab}}\) = \(\dfrac{\sqrt{b}-\sqrt{a}}{1-\sqrt{ab}}\)
KT
6 tháng 7 2018
a) \(1+\sqrt{3}+\sqrt{5}+\sqrt{15}\)
\(=\left(1+\sqrt{3}\right)+\sqrt{5}\left(1+\sqrt{3}\right)\)
\(=\left(1+\sqrt{3}\right)\left(1+\sqrt{5}\right)\)
b) \(\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}\)
\(=\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{7}\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{7}\right)\)
c) \(\sqrt{35}-\sqrt{15}+\sqrt{14}-\sqrt{6}\)
\(=\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{2}\right)\)
KT
6 tháng 7 2018
e) \(xy+y\sqrt{x}+\sqrt{x}+1\)
\(=y\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)\left(y\sqrt{x}+1\right)\)
g) \(3+\sqrt{x}+9-x\)
\(=\left(3+\sqrt{x}\right)+\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)
\(=\left(3+\sqrt{x}\right)\left(4-\sqrt{x}\right)\)
30 tháng 6 2019
c) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
\(\Leftrightarrow\sqrt{x^2+6}+2\sqrt{x^2-1}=x\)
\(\Leftrightarrow x^2+6+4\left(x^2-1\right)+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=x^2\)
\(\Leftrightarrow6+4x^2-4+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)
\(\Leftrightarrow4x^2+2+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)
\(\Leftrightarrow2x^2+2\sqrt{\left(x^2+6\right)\left(x^2-1\right)}+1=0\)
Dễ thấy \(VT>0\forall x\)
Do đó pt vô nghiệm
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
Lời giải:
a)
ĐK: \(0\leq x\leq 1\)
PT \(\Leftrightarrow \sqrt{x+\sqrt{1-x}}=1-\sqrt{x}\)
\(\Rightarrow x+\sqrt{1-x}=1+x-2\sqrt{x}\) (bình phương 2 vế)
\(\Leftrightarrow \sqrt{1-x}-1+2\sqrt{x}=0\)
\(\Leftrightarrow \frac{-x}{\sqrt{1-x}+1}+2\sqrt{x}=0\)
\(\Leftrightarrow \sqrt{x}(2-\frac{\sqrt{x}}{\sqrt{1-x}+1})=0\)
Ta thấy \(\sqrt{1-x}+1\geq 1\Rightarrow \frac{\sqrt{x}}{\sqrt{1-x}+1}\leq \sqrt{x}\leq 1< 2\) với mọi $0\leq x\leq 1$
\(\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}>0\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}\neq 0\)
Do đó $\sqrt{x}=0\Leftrightarrow x=0$ là nghiệm duy nhất
b)
ĐK: \(1 \leq x\leq \frac{1+\sqrt{5}}{2}\) hoặc \(0\geq x\geq \frac{1-\sqrt{5}}{2}\)
PT \(\Rightarrow \left\{\begin{matrix} \sqrt{x}-1\geq 0\\ 1-\sqrt{x^2-x}=x-2\sqrt{x}+1\end{matrix}\right.\) (bình phương 2 vế)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1(1)\\ x+\sqrt{x^2-x}-2\sqrt{x}=0(2)\end{matrix}\right.\)
(1) kết hợp với ĐKXĐ suy ra \(1\leq x\leq \frac{1+\sqrt{5}}{2}(*)\)
(2) \(\Leftrightarrow \sqrt{x}(\sqrt{x}+\sqrt{x-1}-2)=0\)
Từ $(*)$ suy ra $x\neq 0$. Do đó \(\sqrt{x}+\sqrt{x-1}-2=0\)
\(\Leftrightarrow \sqrt{x-1}=2-\sqrt{x}\)
\(\Rightarrow x-1=4+x-4\sqrt{x}\) (bình phương)
\(\Leftrightarrow 4\sqrt{x}=5\Rightarrow x=\frac{25}{16}\) (thỏa mãn $(*)$)
Vậy......
\(\Leftrightarrow14\sqrt{x+35}+6\sqrt{x+1}-84-\sqrt{\left(x+35\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\left(\sqrt{x+35}-6\right)\left(14-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x=195;1\)
tick nha