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Câu a)
ĐKXĐ:.....
Đặt \(\sqrt{2+x}=a; \sqrt{2-x}=b\Rightarrow a^2+b^2=4\)
PT đã cho tương đương với:
\(3\sqrt{2+x}-6\sqrt{2-x}+4\sqrt{(2-x)(2+x)}=4+3(2-x)\)
\(\Leftrightarrow 3a-6b+4ab=a^2+b^2+3b^2\)
\(\Leftrightarrow 3(a-2b)=a^2+4b^2-4ab=(a-2b)^2\)
\(\Leftrightarrow (a-2b)^2-3(a-2b)=0\)
\(\Leftrightarrow \left[\begin{matrix} a-2b=0\\ a-2b=3\end{matrix}\right.\)
Nếu \(a-2b=0\Leftrightarrow a^2=4b^2\Leftrightarrow 2+x=4(2-x)\)
\(\Rightarrow x=\frac{6}{5}\) (t/m)
Nếu \(a-2b=3\Leftrightarrow a=2b+3\Rightarrow \sqrt{x+2}=2\sqrt{2-x}+3\geq 3\)
(vô lý vì \(x\leq 2\rightarrow \sqrt{x+2}\leq 2\))
Vậy pt có nghiệm duy nhất $x=\frac{6}{5}$
Câu b)
ĐKXĐ: \(-2\leq x\leq 2\)
Đặt \(\sqrt{4-x^2}=a\) (\(a\geq 2)\) Ta có hpt sau:
\(\left\{\begin{matrix} a^2+x^2=4\\ x+a=2+3xa\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} (a+x)^2-2ax=4\\ x+a=3xa+2\end{matrix}\right.\)
\(\Rightarrow (3ax+2)^2-2ax=4\)
\(\Leftrightarrow 9a^2x^2+10ax=0\Leftrightarrow ax=0 \) hoặc $ax=\frac{-10}{9}$
Nếu \(ax=0\Rightarrow \left[\begin{matrix} x=0\\ a=0\rightarrow x=\pm 2\end{matrix}\right.\)
Thử lại thấy $x=0; x=2$ thỏa mãn
Nếu \(ax=\frac{-10}{9}\Rightarrow x+a=\frac{-4}{3}\)
Theo định lý Vi-et đảo thì $x,a$ là nghiệm của pt:
\(X^2+\frac{4}{3}X-\frac{10}{9}=0\)
\((x,a)=(\frac{-2+\sqrt{14}}{3}; \frac{-2-\sqrt{14}}{3})\) và hoán vị. Thử lại ta thấy \(x=\frac{-2-\sqrt{14}}{3}\) thỏa mãn
Vậy........
a/ \(x\le8\)
\(\Leftrightarrow x^2+x+12=\left(8-x\right)^2\)
\(\Leftrightarrow x^2+x+12=x^2-16x+64\)
\(\Leftrightarrow17x=52\Rightarrow x=\frac{52}{17}\)
b/ \(x\le4\)
\(\Leftrightarrow x^2+3x-1=\left(4-x\right)^2\)
\(\Leftrightarrow x^2+3x-1=x^2-8x+16\)
\(\Leftrightarrow11x=17\Rightarrow x=\frac{17}{11}\)
c/ \(\left\{{}\begin{matrix}x^2-3x\ge0\\2x-1\ge0\end{matrix}\right.\) \(\Rightarrow x\ge3\)
\(x^2-3x=2x-1\)
\(\Leftrightarrow x^2-5x+1=0\Rightarrow\left[{}\begin{matrix}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
d/ \(2-x\ge0\Rightarrow x\le2\)
\(x^2+2x+4=2-x\)
\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
e/ \(2x^2-x\ge0\Rightarrow\left[{}\begin{matrix}x\le0\\x\ge\frac{1}{2}\end{matrix}\right.\)
\(x^2+2x+4=2x^2-x\)
\(\Leftrightarrow x^2-3x-4=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
f/ \(x\ge2\)
\(2x-1=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-6x+5=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=5\end{matrix}\right.\)
a) \(đkxđ:x\ge-1\)
\(\sqrt{x+1}+x=\sqrt{x+1}+2\Leftrightarrow x=2\left(tm\right)\).
b) đkxđ: \(\)\(\left\{{}\begin{matrix}3-x\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Thay x = 3 vào phương trình ta có:
\(3-\sqrt{3-3}=\sqrt{3-3}+3\Leftrightarrow3=3\left(tm\right)\)
Vậy x = 3 là nghiệm của phương trình.
c) Đkxđ \(\left\{{}\begin{matrix}2-x\ge0\\x-4\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\ge4\end{matrix}\right.\) \(\Leftrightarrow x\in\varnothing\)
Vậy phương trình vô nghiệm.
d) Đkxđ: \(-x-1\ge0\Leftrightarrow-x\ge1\) \(\Leftrightarrow x\le-1\).
Pt\(\Leftrightarrow x^2=4\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(l\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy x = -2 là nghiệm của phương trình.
a/ ĐKXĐ: \(0\le x\le1\)
Đặt \(\sqrt{x}+\sqrt{1-x}=a>0\Rightarrow\sqrt{x-x^2}=\frac{a^2-1}{2}\)
Ta được:
\(1+\frac{a^2-1}{3}=a\Leftrightarrow a^2-3a+2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x\left(1-x\right)}=0\\2\sqrt{x-x^2}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\left(1-x\right)=0\\-4x^2+4x-9=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b/ ĐKXĐ: ...
Đặt \(\sqrt{x+5}=a\ge0\Rightarrow a^2-x=5\)
\(x^2+a=a^2-x\)
\(\Leftrightarrow x^2-a^2+a+x=0\)
\(\Leftrightarrow\left(a+x\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-x\left(x\le0\right)\\\sqrt{x+5}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2\left(x\le0\right)\\x+5=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\) \(\Leftrightarrow...\)
c/ ĐKXĐ: \(2\le x\le5\)
\(\Leftrightarrow\sqrt{3x-3}=\sqrt{2x-4}+\sqrt{5-x}\)
\(\Leftrightarrow3x-3=x+1+2\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow x-2=\sqrt{\left(2x-4\right)\left(5-x\right)}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x-4\right)\left(5-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
a) \(\dfrac{3x^2+1}{\sqrt{x-1}}=\dfrac{4}{\sqrt{x-1}}\)
ĐKXĐ: \(x>1\)
\(3x^2+1=4\)
\(3x^2=3\)
\(x^2=1\)
\(x=\pm1\)
=> Pt vô nghiệm
b) ĐKXĐ: x>-4
\(x^2+3x+4=x+4\)
\(x^2+2x=0\)
\(x\left(x+2\right)=0\)
\(\left[{}\begin{matrix}x=0\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)
1) ĐK: \(x\ge\frac{3}{2}\)
pt \(\Leftrightarrow\frac{2x-2-\left(6x-9\right)}{\sqrt{2x-2}+\sqrt{6x-9}}=16x^2-28x-20x+35\)
\(\Leftrightarrow\frac{-4x+7}{\sqrt{2x-2}+\sqrt{6x-9}}=4x\left(4x-7\right)-5\left(4x-7\right)\)
\(\Leftrightarrow-\frac{4x-7}{\sqrt{2x-2}+\sqrt{6x-9}}=\left(4x-7\right)\left(4x-5\right)\)
\(\Leftrightarrow\left(4x-7\right)\left(\frac{1}{\sqrt{2x-2}+\sqrt{6x-9}}+4x-5\right)=0\)
\(\Leftrightarrow4x-7=0\Leftrightarrow x=\frac{7}{4}\) (nhận)
2) ĐK: \(2\le x\le4\)
pt \(\Leftrightarrow\sqrt{x-2}+\sqrt{a-x}=2\left(x^2-6x+9\right)+7x-19\)
\(\Leftrightarrow\sqrt{x-2}-\left(7x-20\right)+\sqrt{4-x}-1=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{x-2-\left(7x-20\right)^2}{\sqrt{x-2}+7x-20}+\frac{4-x-1}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(134-49x\right)}{\sqrt{x-2}+\left(7x-20\right)}+\frac{3-x}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\) (nhận)
a)
Pt\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=\left(x-3\right)^2\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-4=x^2-6x+9\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+13=0\\x\ge3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{9+\sqrt{29}}{2}\\x_2=\dfrac{9-\sqrt{29}}{2}\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{9+\sqrt{29}}{2}\)
Vậy \(x=\dfrac{9+\sqrt{29}}{2}\) là nghiệm của phương trình.
b) Pt \(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+3=\left(2x-1\right)^2\\2x-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x^2-2x-2=0\\x\ge\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x_1=\dfrac{1+\sqrt{7}}{3}\\x_2=\dfrac{1-\sqrt{7}}{3}\end{matrix}\right.\\x\ge\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow x=\dfrac{1+\sqrt{7}}{3}\)
Vậy phương trình có duy nhất nghiệm là: \(x=\dfrac{1+\sqrt{7}}{3}\)
a, ĐKXĐ: \(-3\le x\le6\)
\(pt\Leftrightarrow3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
b, ĐKXĐ: \(x\ge4\)
\(pt\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}+x+2+\sqrt{x-4}=8\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+x+2+\sqrt{x-4}=8\)
\(\Leftrightarrow\sqrt{x-4}+2+x+2+\sqrt{x-4}=8\)
\(\Leftrightarrow2\sqrt{x-4}=4-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\4\left(x-4\right)=\left(4-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x^2-12x+32=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)
e, Đặt \(y=x-1\) ta có
\(pt\Leftrightarrow\left(y+4\right)^4+\left(y-4\right)^4=1312\)
\(\Leftrightarrow2y^4+192y^2-800=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=4\\y^2=-100\left(l\right)\end{matrix}\right.\Leftrightarrow y=\pm2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-4\ge0\\4-x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x\le4\end{matrix}\right.\) \(\Leftrightarrow x=4\)
Thay \(x=4\) vào pt thấy thỏa mãn
Vậy pt có nghiệm duy nhất \(x=4\)