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6: \(\Leftrightarrow2x^2+3x+9+\sqrt{2x^2+3x+9}-42=0\)
Đặt \(\sqrt{2x^2+3x+9}=a\left(a>=0\right)\)
Phương trình sẽ trở thành là: a^2+a-42=0
=>(a+7)(a-6)=0
=>a=-7(loại) hoặc a=6(nhận)
=>2x^2+3x+9=36
=>2x^2+3x-27=0
=>2x^2+9x-6x-27=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
8: \(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)
=>\(\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)
=>\(\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-2=4\\z-3=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\\z=12\end{matrix}\right.\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a/ ĐXĐK: ...
\(\Leftrightarrow9x^2-1-x-8x\sqrt{x+1}=0\)
\(\Leftrightarrow x^2-x-1+8x\left(x-\sqrt{x+1}\right)=0\)
\(\Leftrightarrow x^2-x-1+\frac{8x\left(x^2-x-1\right)}{x+\sqrt{x+1}}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\Rightarrow x=...\\\frac{-8x}{x+\sqrt{x+1}}=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow-8x=x+\sqrt{x+1}\)
\(\Leftrightarrow-9x=\sqrt{x+1}\) (\(x\le0\))
\(\Leftrightarrow81x^2-x-1=0\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1-5\sqrt{13}}{162}\\x=\frac{1+5\sqrt{13}}{162}>0\left(l\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow3x^2+2\left(x^2+x+1\right)-5x\sqrt{x^2+x+1}=0\)
Đặt \(\sqrt{x^2+x+1}=a\)
\(\Leftrightarrow3x^2-5ax+2a^2=0\)
\(\Leftrightarrow\left(x-a\right)\left(3x-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a\\3x=2a\end{matrix}\right.\) (\(x\ge0\))
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=x\\2\sqrt{x^2+x+1}=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=x^2\\2\left(x^2+x+1\right)=9x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\7x^2-2x-2=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1+\sqrt{15}}{7}\)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
Dòng thứ 2 qua dòng thứ 3 anh làm chậm lại được không ạ, tại tắt quá e không hiểu
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
a/ \(\Rightarrow2x^2-3x-11=x^2-1\)
\(\Leftrightarrow x^2-3x-10=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Thay 2 nghiệm vào cả 2 căn thức thấy đều xác định
Vậy nghiệm của pt là ...
b/ \(\left\{{}\begin{matrix}x\ge-1\\2x^2+3x-5=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-6=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=2\\x=-3\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=2\)
c/
\(\Leftrightarrow x^2+4x+4=3x^2-5x+14\)
\(\Leftrightarrow2x^2-9x+10=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{5}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\left\{{}\begin{matrix}-x-9\ge0\\\left(x-1\right)\left(2x-3\right)=\left(-x-9\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\2x^2-5x+3=x^2+18x+81\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\x^2-23x-78=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=26\left(ktm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
1) \(ĐK:\orbr{\begin{cases}0\le x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\end{cases}}\)
\(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\Leftrightarrow x-5+\sqrt{x^2-4x+1}=3\sqrt{x}-6\)\(\Leftrightarrow\frac{-6\left(x-4\right)}{x-5-\sqrt{x^2-4x+1}}=\frac{9\left(x-4\right)}{3\sqrt{x}+6}\Leftrightarrow\left(x-4\right)\left(\frac{9}{3\sqrt{x}+6}+\frac{6}{x-5-\sqrt{x^2-4x+1}}\right)=0\)
Xét phương trình \(\frac{9}{3\sqrt{x}+6}+\frac{6}{x-5-\sqrt{x^2-4x+1}}=0\Leftrightarrow\left(18\sqrt{x}-9\right)+9\left(x-\sqrt{x^2-4x+1}\right)=0\)\(\Leftrightarrow\frac{81\left(4x-1\right)}{18\sqrt{x}+9}+\frac{9\left(4x-1\right)}{x+\sqrt{x^2-4x+1}}=0\Leftrightarrow\left(4x-1\right)\left(\frac{81}{18\sqrt{x}+9}+\frac{9}{x+\sqrt{x^2-4x+1}}\right)=0\)
Dễ thấy \(\frac{81}{18\sqrt{x}+9}+\frac{9}{x+\sqrt{x^2-4x+1}}>0\)với mọi x thỏa mãn điều kiện nên 4x - 1 = 0 hay x = 1/4
Vậy phương trình có tập nghiệm S = {4; 1/4}
e làm câu dễ nhất ^^
\(\sqrt{x+1}+\sqrt{4-x}+\sqrt{\left(x+1\right)\left(4-x\right)}=5\left(đk:-1\le x\le4\right)\)
\(< =>\left(\sqrt{x+1}-1\right)+\left(\sqrt{4-x}-2\right)+\left(\sqrt{\left(x+1\right)\left(4-x\right)}-2\right)=0\)
\(< =>\frac{x}{\sqrt{x+1}+1}-\frac{x}{\sqrt{4-x}+2}+\frac{x\left(3-x\right)}{\sqrt{\left(x+1\right)\left(4-x\right)+2}}=0\)
\(< =>x=0\)
a) Đk: \(-\dfrac{1}{3}\le x\le2\)
\(\sqrt{3x+1}+\sqrt{2-x}=1\Leftrightarrow\sqrt{-3x^2+5x+2}=-x-1\)
Ta có: \(VT\ge0\) ; \(VP< 0\forall-\dfrac{1}{3}\le x\le2\)
Kl: ptvn
b) \(x^2+5x+9=\left(x+5\right)\left(\left|x\right|+9\right)\) (*)
Th1: x >/ 0
(*) \(\Leftrightarrow x^2+5x+9=\left(x+5\right)\left(x+9\right)\)
\(\Leftrightarrow x^2+5x+9=x^2+14x+45\)
\(\Leftrightarrow9x=36\Leftrightarrow x=4\left(N\right)\)
Th2: x \< 0
(*) \(\Leftrightarrow x^2+5x+9=\left(x+5\right)\left(9-x\right)\)
\(\Leftrightarrow2x^2+x-36=0\Leftrightarrow\left[{}\begin{matrix}x=4\left(L\right)\\x=-\dfrac{9}{2}\left(N\right)\end{matrix}\right.\)
Kl: x=4 , x= - 9/2
c) Đk: \(x\ge-\dfrac{1}{3}\)
\(\sqrt{3x+1}=3x+1\Leftrightarrow\sqrt{3x+1}\left(\sqrt{3x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=0\\\sqrt{3x+1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\left(N\right)\\x=0\left(N\right)\end{matrix}\right.\)
Kl: x= -1/3 , x=0