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1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có pt
<=>\(\sqrt{\left(x+3\right)\left(x+7\right)}=3\sqrt{x+3}+2\sqrt{x+7}=6\)
đặt \(\sqrt{x+3}=a;\sqrt{x+7}=b\)
nên pt <=>\(ab=3a+2b-6\Leftrightarrow ab-3a-2b+6=0\)
\(\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\)
đến đây thì dễ rồi
biêu thức dài dài trong căn pt thành nhân tử là \(\sqrt{\left(x+3\right)\left(x+7\right)}\)
xong rùi bn pt thành nhân tử sẽ có dạng \(\left(\sqrt{x+3}-2\right)\left(\sqrt{x+7}-3\right)=0\)
đến day bn làm tiếp nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
3x2+6x+7=3.(x2+2x+1)+4=3.(x+1)2+4 >= 4
=> căn của nó >=
..................................................... ko thích giải
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a)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)
\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)
Vậy pt có một nghiệm duy nhất là \(x=-1\)
b)
\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)
\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)
\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)
Lập bảng xét dấu ra nhé ~^o^~
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0\)
\(\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6+4\sqrt{5}\end{matrix}\right.\)
Vậy \(S=\left\{1;6+4\sqrt{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x>/ -3
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}-3\sqrt{x+3}+2\sqrt{x+7}-6=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+7}-3\right)+2\left(\sqrt{x+7}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+7}-3\right)\left(\sqrt{x+3}+2\right)=0\)
\(\Leftrightarrow\sqrt{x+7}-3=0\Rightarrow x+7=9\Rightarrow x=2\left(TM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-11=0\)
<=> \(x^2-\sqrt{11}=0\)
<=> \(\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)=0\)
<=> \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\) => x = \(\pm\sqrt{11}\) Vậy S ={ \(\pm\sqrt{11}\)}
b) \(x^2-2\sqrt{13}x+13=0\)
\(\Leftrightarrow\left(x-\sqrt{13}\right)^2=0\)
=> x = \(\sqrt{13}\)
Vậy S = {\(\sqrt{13}\) }
\(c\)) \(\sqrt{x^2-10x+25}=7-2x\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\)
\(\Leftrightarrow\left|x-5\right|=7-2x\)
=> Có 2 TH xảy ra
* Khi x - 5 \(\ge0\Leftrightarrow x\ge5\) Ta có PT :
x - 5 = 7 - 2x
<=> 3x = 12
=> x= 4 (KTM)
* Khi x - 5 < 0 => x < 5
Ta có pT
-x + 5 = 7-2x
<=> x = 2 (TM)
Vậy S = { 2 }
\(a\text{)} x^2-11=0\\ x^2=11\\ x=\pm\sqrt{11}\)
\(b\text{)}\:x^2-2\sqrt{13x}+13=0\\ \left(x-\sqrt{13}\right)^2=0\\ x-\sqrt{13}=0\\ x=\sqrt{13}\)
\(c\text{)}\:\sqrt{x^2-10x+25}=7-2x\\ \left|x-5\right|=7-2x\\ \Rightarrow\left[{}\begin{matrix}x-5=7-2x\left(với\:x\ge5\right)\\5-x=7-2x\left(với\:x< 5\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
X=2
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đặt x+5=a\(\left(a\ge0\right)\) khi đó phương trình trở thành:
\(a^2-4+\sqrt{a}+\sqrt{16-a}=0\)
lại có \(\sqrt{a}+\sqrt{16-a}\ge\sqrt{a+16-a}=4\)
nên ta có:
\(a^2-4+\sqrt{a}+\sqrt{16-a}\ge a^2\)
Suy ra \(0\ge a^2\)
\(\Rightarrow a=0\)hay x+5=0
\(\Leftrightarrow x=-5\)
Cảm ơn