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7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
ĐKXĐ: x>=0; y>=1 ; z>=2.
câu 1:Từ giả thiết ta có:
\(2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\sqrt{y-1}+1+\left(z-2\right)-2\sqrt{z-2}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1;\sqrt{y-1}=1;\sqrt{z-2}=1\)
Vậy x=1;y=2;z=3.
Có gì ko hiểu bạn cứ bình luận phía dưới :)
a)\(pt\Leftrightarrow\sqrt{3x^2-6x+4}+\sqrt{2x^2-4x+6}+x^2-2x-2=0\)
\(\Leftrightarrow\sqrt{3x^2-6x+4}-1+\sqrt{2x^2-4x+6}-2+x^2-2x+1=0\)
\(\Leftrightarrow\dfrac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}+\dfrac{2x^2-4x+6-4}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}+\dfrac{2\left(x-1\right)^2}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1\right)=0\)
Dễ thấy: \(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1>0\)
\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=3-4x-2x^2\)
\(pt\Leftrightarrow\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}+2x^2+4x-3=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x-8=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x+2=0\)
\(\Leftrightarrow\dfrac{3x^2+6x+12-9}{\sqrt{3x^2+6x+12}+3}+\dfrac{5x^4-10x^2+9-4}{\sqrt{5x^4-10x^2+9}+2}+2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\right)=0\)
Dễ thấy: \(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2>0\)
\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
f/
ĐKXĐ: ...
Đặt \(\sqrt{2-x}+\sqrt{x+2}=a>0\)
\(\Rightarrow a^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}\)
Phương trình trở thành:
\(a+\frac{a^2-4}{2}=2\)
\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}=0\)
\(\Rightarrow4-x^2=0\Rightarrow x=\pm2\)
e/ ĐKXĐ: ...
Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\)
\(\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\)
Pt trở thành:
\(a+\frac{a^2-5}{2}=5\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}+\sqrt{4-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=2\)
\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)
\(\Leftrightarrow-x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)