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\(DK:x\ge-\frac{1}{3}\)
\(\Leftrightarrow\frac{2x-1}{\sqrt{3x+1}+\sqrt{x+2}}\left(\sqrt{3x^2+7x+2}+4\right)-2\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{\sqrt{3x^2+7x+2}+4}{\sqrt{3x+1}+\sqrt{x+2}}-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(1\right)\\\frac{\sqrt{3x^2+7x+2}+4}{\sqrt{3x+1}+\sqrt{x+2}}=2\left(2\right)\end{cases}}\)
Xet PT(2)
Dat \(\hept{\begin{cases}\sqrt{3x+1}=a\\\sqrt{x+2}=b\end{cases}\left(a,b\ge0\right)}\)
PT(2)\(\Leftrightarrow\frac{ab+4}{a+b}=2\)
\(\Leftrightarrow2a+2b-ab-4=0\)
\(\Leftrightarrow\left(a+2\right)\left(2-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(3\right)\\b=2\left(4\right)\end{cases}}\)
Xet PT(3)
Ta co:\(a\ge0\)
Nen PT vo nghiem
Xet PT (4)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=2\)
Vay PT co 2 nghiem la \(x_1=\frac{1}{2};x_2=2\)
a)Đk:\(x\ge\frac{1}{2}\)
\(pt\Leftrightarrow4x^2-12x+4+4\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(2x-1\right)^2-4\left(2x-1\right)-1+4\sqrt{2x-1}=0\)
Đặt \(t=\sqrt{2x-1}>0\Rightarrow\hept{\begin{cases}t^2=2x-1\\t^4=\left(2x-1\right)^2\end{cases}}\)
\(t^4-4t^2+4t-1=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}t-1=0\\t^2+2t-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}t=1\\t=\sqrt{2}-1\end{cases}\left(t>0\right)}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2-\sqrt{2}\end{cases}}\) là nghiệm thỏa pt
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
Đặt \(\sqrt{5x+7}=a\)
\(\Rightarrow\left(\frac{3\left(a^2-7\right)}{5}+1\right)^2+\left(\frac{4\left(a^2-7\right)}{5}+3\right)^2=a+5\)
\(\Leftrightarrow a^4-8a^2-a+12=0\)
\(\Leftrightarrow\left(a^2-a-4\right)\left(a^2+a-3\right)=0\)
dk \(x\ge0;2x+1\ge0< =>x\ge0\)
2(x+1)\(\sqrt{x}+\sqrt{3\left(x+1\right)^2\left(2x+1\right)}=\left(x+1\right)\left(5x^2-8x+8\right)< =>\)
\(2\sqrt{x}+\sqrt{3\left(2x+1\right)}=5x^2-8x+8\)(x+1>0 với x\(\ge0\)) <=>
2\(\sqrt{x}-2+\sqrt{6x+3}-3=5x^2-8x+3\) <=>\(\frac{2\left(x-1\right)}{\sqrt{x}+1}+\frac{6\left(x-1\right)}{\sqrt{6x+3}+3}=\left(x-1\right)\left(5x-3\right)< =>\)x-1=0 <=>x= 1 hoặc
\(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+3}=5x-3\)
x>1 thì \(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+3}< \frac{2}{1+1}+\frac{6}{3+3}=2\) hay 5x- 3<2 <=> x<1( vô lý)
x<1 thì \(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{6x+3}+}>2\) hay 5x-3>2 <=> x>1 (vô lý)
x=1 thỏa mãn
vậy pt có nghiệm duy nhất x=1