giải phương trình :\(\frac{x^2}{\sqrt{3x-2}}-\sqrt{3x-2}=1-x\)
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NN
9
4 tháng 10 2016
ĐKXĐ: z>0
pt<=> \(\frac{x^3+3x^2\sqrt[3]{3x-2}-12x+\sqrt{x}-\sqrt{x}-8}{x}=0\)
<=> \(x^3+3x^2\sqrt[3]{3x+2}-12x-8=0\)
<=> \(3x^2\sqrt[3]{3x-2}-6x^2+x^3-6x^2+12x-8=0\)
<=> \(3x^2\left(\sqrt[3]{3x-2}-2\right)+\left(x-2\right)^3=0\)
<=> \(3x^2\cdot\frac{3x-2-8}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^3=0\)
<=> \(\left(x-2\right)\left(\frac{9x^2}{\left(\sqrt[3]{3x-2}\right)^2+2\sqrt[3]{3x-2}+4}+\left(x-2\right)^2\right)=0\)
<=> \(x=2\)( vì cái trong ngoặc thứ 2 luôn dương vs mọi x>0)
vậy x=2
NH
0
DT
1
1 tháng 11 2016
\(\frac{x^2}{\sqrt{3x-2}}-\frac{\sqrt{\left(3x-2\right)\left(3x-2\right)}}{\sqrt{3x-2}}=1-x\Leftrightarrow\frac{x^2-3x+2}{\sqrt{3x-2}}-1+x=0\Leftrightarrow x^2-3x+2-\sqrt{3x-2}+x\sqrt{3x-2}=0\Leftrightarrow\left(x-2\right)\left(x-1\right)+\sqrt{3x-2}\left(x-1\right)=\left(x-1\right)\left(x-2+\sqrt{3x-2}\right)\Leftrightarrow\hept{\begin{cases}x-1=0\\x-2+\sqrt{3x-2}=0\end{cases}\Leftrightarrow}x=1\)
HP
0
HD
3
7 tháng 12 2017
ĐIều kiện x >2/3
\(\Leftrightarrow\frac{x^2+\left(\sqrt{3x-2}\right)^2}{x\sqrt{3x-2}}=2\)
\(\Leftrightarrow x^2+\left(\sqrt{3x-2}\right)^2=2x\sqrt{3x-2}\)
\(\Leftrightarrow x^2+\left(\sqrt{3x-2}\right)^2-2x\sqrt{3x-2}=0\)
\(\Leftrightarrow\left(x-\sqrt{3x-2}\right)^2=0\)
\(\Leftrightarrow x-\sqrt{3x-2}=0\Leftrightarrow x=\sqrt{3x-2}\)
vì ta bình phương 2 vế ta có:
x2 = 3x-2
,<=> x2-3x+2 = 0
ta có x1= 1 (thỏa mãn) ; x2 = 2 (thỏa mãn)
Vậy:......................................
IL
0
NN
1
TD
9 tháng 11 2016
\(ĐK:x\ne0v̀ax>\frac{2}{3}\)
đặt \(t=\frac{x}{\sqrt{3x-2}}\Rightarrow\frac{1}{t}=\frac{\sqrt{3x-2}}{x}\)
\(pt\Leftrightarrow t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1\Leftrightarrow t=1\)
\(\Leftrightarrow\frac{x}{\sqrt{3x-2}}=1\Leftrightarrow\sqrt{3x-2}=x\Leftrightarrow x^2=3x-2\left(vi.x>\frac{2}{3}\right)\)
\(\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(t.m\right)\\x=2\left(t.m\right)\end{array}\right.\)
9 tháng 11 2016
\(\frac{x}{\sqrt{3x-2}}+\frac{\sqrt{3x-2}}{x}=2\)
Đk:\(\sqrt{3x-2}\ge0\Rightarrow3x-2\ge0\Rightarrow x\ge\frac{2}{3}\)
\(\Leftrightarrow\frac{x^2}{x\sqrt{3x-2}}+\frac{3x-2}{x\sqrt{3x-2}}-\frac{2\left(x\sqrt{3x-2}\right)}{x\sqrt{3x-2}}=0\)
\(\Leftrightarrow\frac{x^2+3x-2-2\left(x\sqrt{3x-2}\right)}{x\sqrt{3x-2}}=0\)
\(\Leftrightarrow x^2+3x-2-2\left(x\sqrt{3x-2}\right)=0\)
\(\Leftrightarrow x^2+3x-2=2x\sqrt{3x-2}\)
\(\Leftrightarrow\left(x^2+3x-2\right)^2=\left(2x\right)^2\sqrt{\left(3x-2\right)^2}\)
\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=4x^2\left(3x-2\right)\)
\(\Leftrightarrow x^4+6x^3+5x^2-12x+4=12x^3-8x^2\)
\(\Leftrightarrow x^4-6x^3+13x^2-12x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-2\right)^2=0\\\left(x-1\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)(thỏa mãn)
Vậy pt có nghiệm là \(\left[\begin{array}{nghiempt}x=2\\x=1\end{array}\right.\)
ND
1
TT
16 tháng 9 2020
ĐKXĐ : \(x\ne-\frac{1}{3}\)
Ta có : \(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)
\(\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)
\(\Leftrightarrow\frac{x^2+x+2-4}{\sqrt{x^2+x+2}+2}=\frac{3x^2+3x+2-6x-2}{3x+1}\)
\(\Leftrightarrow\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}=\frac{3x^2-3x}{3x+1}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{3x\left(x-1\right)}{3x+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}\right]=0\)
\(\Leftrightarrow x=1\)( Thỏa mãn )
quy đồng lên bạn ơi , rồi đặt ẩn
\(\Leftrightarrow x^2-3x+2=\left(1-x\right)\sqrt{3x-2}\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=-\left(x-1\right)\sqrt{3x-2}\)
\(\Leftrightarrow x-2=-\sqrt{3x-2}\)
\(\Leftrightarrow x^2-4x+4=3x-2\Leftrightarrow x=6;x=1\left(\text{nhận cả 2}\right)\)
Vậy................