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a,dk x>0
\(\Leftrightarrow\)\(\dfrac{\left(\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}\right)\left(\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}\right)}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}=3x\)
\(\Leftrightarrow x\left(\dfrac{x+2}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}-3\right)=0\)
\(\Rightarrow\dfrac{x+2}{\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}}=3\)
\(\Rightarrow\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\)
kh vs dé bài ta có hệ \(\left\{{}\begin{matrix}\sqrt{2x^2+x+1}+\sqrt{x^2-x+1}=3x\\\sqrt{2x^2+x+1}-\sqrt{x^2-x+1}=\dfrac{x+2}{3}\end{matrix}\right.\)
cộng vs nhau ta có
\(2\sqrt{2x^2+x+1}=3x+\dfrac{x+2}{2}\)
\(\Leftrightarrow3\sqrt{2x^2+x+1}=5x+1\)
giải ra ta có x=1(tm) x=-8/7 (l)
b, dk tu xd nhé 
\(\Leftrightarrow\dfrac{\left(\sqrt{x^2+x+1}-\sqrt{x^2-x+1}\right)\left(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\right)}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-2x=0\)
\(\Leftrightarrow2x\left(\dfrac{1}{\sqrt{x^2+x+1}+\sqrt{x^2-x+1}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+x+1}+\sqrt{x^2-x+1}=1\left(l\right)\end{matrix}\right.\)
ns \(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}>1\)
\(\Rightarrow x=0\left(tm\right)\)
6.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+3}\\\sqrt{x-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\left(vn\right)\\x=2\end{matrix}\right.\)
4.
ĐKXĐ: \(x\ge4\)
Đặt \(\sqrt{x-4}=t\ge0\Rightarrow x=t^2+4\)
\(\Rightarrow3\left(t^2+4\right)+7t=14t-20\)
\(\Leftrightarrow3t^2-7t+34=0\)
Phương trình vô nghiệm
5.
ĐKXĐ: ...
- Với \(x=0\) ko phải nghiệm
- Với \(x\ne0\Rightarrow\sqrt{x+1}-1\ne0\) , nhân 2 vế của pt cho \(\sqrt{x+1}-1\) và rút gọn ta được:
\(\sqrt{x+1}+2x-5=\sqrt{x+1}-1\)
\(\Leftrightarrow2x=4\Rightarrow x=2\)
1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)
<=> \(\sqrt{\left(x-10\right)^2}=10\)
<=> \(\left|x-10\right|=10\)
=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)
Vậy S = \(\left\{20;0\right\}\)
2) \(\sqrt{x +2\sqrt{x}+1}=6\)
<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)
<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)
<=> \(\left|\sqrt{x}+1\right|=6\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)
Vậy S = \(\left\{25\right\}\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)
<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)
<=> \(\left|x-3\right|=\sqrt{3}+1\)
=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)
Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)
<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)
<=> \(\left|\sqrt{3x}+1\right|=5\)
=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)
Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)
<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)
<=> \(\left|\sqrt{6x}+2\right|=7\)
=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)
=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)
Điều kiện:\(-2\le x\le2\)
Ta có: \(10-3x=\left(2+x\right)+4\left(2-x\right)\)
Đặt \(a=\sqrt{2+x}\ge0\)
\(b=\sqrt{2-x}\ge0\)
Pt trở thành:\(3a-6b+4ab=a^2+4b^2\)
Chuyển vế cùng 1 vế sau đó nhóm lại và đặt nhân tử chung
\(\left(a^2-2ab\right)-\left(2ab-4b^2\right)-\left(3a-6b\right)=0\)
\(a\left(a-2b\right)-2b\left(a-2b\right)-3\left(a-2b\right)=0\)
\(\left(a-2b\right)\left(a-2b-3\right)=0\)
\(\Rightarrow\sqrt{2+x}-2\sqrt{2-x}=0\)
\(\Rightarrow x=\frac{6}{5}\left(tm\right)\)
\(\Rightarrow\sqrt{2+x}-2\sqrt{2-x}-3=0\)
=> vô nghiệm
Vậy pt trên có nghiệm là \(x=\frac{6}{5}\)
Câu 1:
Ta có 2 vế luôn dương nên bình phương 2 vế được:
\(2x^2+4=5x^3+5\)
\(5x^3-2x^2-1=0\)
<=> x = 0,7528596306