Giải các phương trình sau:
a) \(\sqrt{2x-1}+\sqrt{x-1}=5\)
b) \(x^2+2x+7=3\sqrt{\left(x^2+1\right).\left(x+3\right)}\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
giải phương trình :
a, \(\left(x+1\right)\sqrt{x+8}=x^2+x+4\)
b, \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
c, \(\left(3x+1\right)\sqrt{x^2+3}=3x^2+2x+3\)
c.
\(\Leftrightarrow x^2+3-\left(3x+1\right)\sqrt{x^2+3}+2x^2+2x=0\)
Đặt \(\sqrt{x^2+3}=t>0\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
a.
Đề bài ko chính xác, pt này ko giải được
b.
ĐKXĐ: \(x\ge-\dfrac{7}{2}\)
\(2x+7-\left(2x+7\right)\sqrt{2x+7}+x^2+7x=0\)
Đặt \(\sqrt{2x+7}=t\ge0\)
\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)
\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=1+2\sqrt{2}\)
giải các phương trình sau a)\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
a: ĐKXĐ: y<=1/2
\(\left\{{}\begin{matrix}3\left(x-1\right)-\sqrt{1-2y}=1\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6\left(x-1\right)-2\sqrt{1-2y}=2\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x-1\right)=7\\\left(x-1\right)+2\sqrt{1-2y}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=1\\2\sqrt{1-2y}=5-1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\\sqrt{1-2y}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\1-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
b:
ĐKXĐ: \(x\in R\)
\(\left\{{}\begin{matrix}\sqrt{x^2-2x+1}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{\left(x-1\right)^2}-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|x-1\right|-3y=7\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-6y=14\\2\left|x-1\right|-8y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2y=13\\\left|x-1\right|-3y=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\\left|x-1\right|=3y+7=3\cdot\dfrac{13}{2}+7=\dfrac{39}{2}+7=\dfrac{53}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x-1\in\left\{\dfrac{53}{2};-\dfrac{53}{2}\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{13}{2}\\x\in\left\{\dfrac{55}{2};-\dfrac{51}{2}\right\}\end{matrix}\right.\)
c: ĐKXĐ: y>=4
\(\left\{{}\begin{matrix}2\left(x^2-x\right)+\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-x\right)+2\sqrt{y-4}=0\\3\left(x^2-x\right)-2\sqrt{y-4}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-x\right)=-7\\2\left(x^2-x\right)+\sqrt{y-4}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x=-1\\\sqrt{y-4}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-x+1=0\\y-4=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vôlý\right)\\y=8\end{matrix}\right.\)
=>\(\left(x,y\right)\in\varnothing\)
1. Giải các phương trình sau:
a)\(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt[]{x+\sqrt{x^2-1}}=2\)
b)\(x^2-x-\sqrt{x^2-x+13}=7\)
c)\(x^2+2\sqrt{x^2-3x+1}=3x+4\)
d)\(2x^2+5\sqrt{x^2+3x+5}=23-6x\)
e)\(\sqrt{x^2+2x}=-2x^2-4x+3\)
f)\(\sqrt{\left(x+1\right)\left(x+2\right)}=x^2+3x+4\)
2. Giải các bất phương trình sau:
1)\(\sqrt{x^2-4x+5}\ge2x^2-8x\)
2)\(2x^2+4x+3\sqrt{3-2x-x^2}>1\)
3)\(\dfrac{\sqrt{-3x+16x-5}}{x-1}\le2\)
4)\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
5)\(\dfrac{9x^2-4}{\sqrt{5x^2-1}}\le3x+2\)
Giải phương trình 1, \(x^2+9x+7=\left(2x+1\right)\sqrt{2x^2+4x+5}\)
2, GPT \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
3. GHPT \(\left\{{}\begin{matrix}x^2-2y-1=2\sqrt{5y+8}+\sqrt{7x-1}\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
ĐKXĐ: ...
Từ pt dưới:
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow y=x-2\)
Thế vào pt trên:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
Giải phương trình:
\(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\)
\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\)
Tham khảo:
1) Giải phương trình : \(11\sqrt{5-x}+8\sqrt{2x-1}=24+3\sqrt{\left(5-x\right)\left(2x-1\right)}\) - Hoc24
\(\sqrt{x+3}+2\sqrt{x}=2+\sqrt{x\left(x+3\right)}\left(đk:x\ge0\right)\)
\(\Leftrightarrow x+3+4x+4\sqrt{x\left(x+3\right)}=4+x\left(x+3\right)+4\sqrt{x\left(x+3\right)}\)
\(\Leftrightarrow5x+3=4+x^2+3x\)
\(\Leftrightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\left(tm\right)\)
Giải các bất phương trình sau:
\(a,\left(x+1\right)\left(x+4\right)< 5\sqrt{x^2+5x+28}\)
\(b,4\sqrt{x}+\dfrac{2}{\sqrt{x}}< 2x+\dfrac{1}{2x}+2\)
a, ĐKXĐ : \(D=R\)
BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)
Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)
BPTTT : \(5\sqrt{a+24}>a\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)
\(\Leftrightarrow-24\le a< 40\)
- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)
\(\Leftrightarrow-9< x< 4\)
Vậy ....
b, ĐKXĐ : \(x>0\)
BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)
- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)
\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)
BPTTT : \(2a\le a^2\)
\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)
\(\Leftrightarrow a\ge2\)
\(\Leftrightarrow a^2\ge4\)
- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)
\(\Leftrightarrow4x^2-12x+1\ge0\)
\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)
Vậy ...
Giải các bất phương trình sau :
a) \(\left(x+1\right)\left(2x-1\right)+x\le3+2x^2\)
b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
c) \(x+\sqrt{x}>\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\)
d) \(\left(\sqrt{1-x}+3\right)\left(2\sqrt{1-x}-5\right)>\sqrt{1-x}-3\)
lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x\ge0
x+\sqrt{x}>\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)
\Leftrightarrow x+\sqrt{x}>2x+\sqrt{x}-3
\Leftrightarrow x-3>0
\Leftrightarrow x>3. (tmđk).
d) Đkxđ: \(1-x\ge0\)\(\Leftrightarrow x\le1\).
\(\left(\sqrt{1-x}+3\right)\left(2\sqrt{1-x}-5\right)>\sqrt{1-x}-3\)
Đặt \(\sqrt{1-x}=t\left(t\ge0\right)\) bpt trở thành:
\(\left(t+3\right)\left(2t-5\right)>t-3\)\(\Leftrightarrow2t^2+t-15>t-3\)
\(\Leftrightarrow2t^2>12\)\(\Leftrightarrow t^2>6\)\(\Leftrightarrow t>\sqrt{6}\) ( do \(t\ge0\) ).
Trở lại phép đặt: \(\sqrt{1-x}>\sqrt{6}\)\(\Leftrightarrow1-x>6\)\(\Leftrightarrow x< -5\).