GPT: x^3+(x-1)^3=(2x-1)^3
GPT sau: \(\sqrt[3]{x+4}=\sqrt{x-1}+2x-3\)
GPT: 2x(x-1)-3(x^2-4x)+x(x+2)=-3
GPT sau:
a) 5/( x^2 +x -6 ) - 2/( x^2 + 4x + 3 ) = -3/( 2x-1 )
GPT :
\(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
GPT :
\(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
x=0 ko là nghiệm
chia cả hai vê cho x<>0, ta được:
\(x-\dfrac{1}{x}+\sqrt[3]{x-\dfrac{1}{x}}=2\)
Đặt \(\sqrt[3]{x-\dfrac{1}{x}}=a\)
=>a^3+a=2
=>a=1
=>x-1/x=1
=>\(x=\dfrac{1\pm\sqrt{5}}{2}\)
GPT :
\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)
\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)
\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)
\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)
\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)
\(\Leftrightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)
Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)
Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.
x^3-4x^2+5x-1-căn 2x-3=0
=>\(x^3-4x^2+5x-2-\left(\sqrt{2x-3}-1\right)=0\)
=>\(\left(x-1\right)\left(x-2\right)^2-\dfrac{2x-3-1}{\sqrt{2x-3}+1}=0\)
=>\(\left(x-2\right)\left[\left(x-1\right)\left(x-2\right)-\dfrac{2}{\sqrt{2x-3}+1}\right]=0\)
=>x-2=0
=>x=2
GPT sau: \(x^2+\sqrt[3]{x^4-x^2}=2x+1\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(x+\sqrt[3]{x-\dfrac{1}{x}}=2+\dfrac{1}{x}\)
\(\Leftrightarrow x-\dfrac{1}{x}+\sqrt[3]{x-\dfrac{1}{x}}-2=0\)
Đặt \(\sqrt[3]{x-\dfrac{1}{x}}=t\)
\(\Rightarrow t^3+t-2=0\Leftrightarrow\left(t-1\right)\left(t^2+t+2\right)=0\)
\(\Leftrightarrow t=1\Rightarrow x-\dfrac{1}{x}=1\)
\(\Leftrightarrow x^2-x-1=0\Leftrightarrow...\)
GPT: 2x+3/x+1 - 6/x=2
2x+3/x+1 - 6/x=2
<=>x(2x+3)/x(x+1) - 6(x+1)/x(x+1)=2x(x+1)/x(x+1)
<=>x(2x+3) - 6(x+1)=2x(x+1)
<=>2x^2+3x - 6x -6=2x^2 + 2x
<=>(2x^2 - 2x^2) + (3x - 6x - 2x)=6
<=>-5x =6
<=> x =-6/5
GPT: (2x + 1)(x + 1)2(2x + 3) = 18