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a)\(\sqrt{2x^2+x+6}+\sqrt{x^2+x+2}=x+\dfrac{4}{x}\)
\(pt\Leftrightarrow\sqrt{2x^2+x+6}-3+\sqrt{x^2+x+2}-2=x+\dfrac{4}{x}-5\)
Liên hợp quy đồng nốt
@Nguyễn Huy Thắng@Mysterious Person@bảo nam trần@Lightning Farron@Thiên Thảo@Sky SơnTùng
\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
\(\dfrac{3x+3}{\sqrt{x}}=4+\dfrac{x+1}{\sqrt{x^2-x+1}}\) ĐK: \(x\ge0\)
\(\Leftrightarrow\dfrac{3x+3}{\sqrt{x}}-6-\left(\dfrac{x+1}{\sqrt{x^2-x+1}}-2\right)=0\)
\(\Leftrightarrow\dfrac{3x+3-6\sqrt{x}}{\sqrt{x}}-\dfrac{x+1-2\sqrt{x^2-x+1}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(x-2\sqrt{x}+1\right)}{\sqrt{x}}-\dfrac{x^2-x+1-2\sqrt{x^2-x+1}+1-x^2+2x-1}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\dfrac{\left(\sqrt{x^2-x+1}-1\right)^2-\left(x-1\right)^2}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\dfrac{\left(\dfrac{x\left(x-1\right)}{\sqrt{x^2-x+1}+1}\right)^2-\left(x-1\right)^2}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-1\right)^2}{\sqrt{x}}-\left(x-1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2\left[\dfrac{3}{\sqrt{x}}-\left(\sqrt{x}+1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}\right]=0\)
Ta có \(\dfrac{3}{\sqrt{x}}-\left(\sqrt{x}+1\right)^2\dfrac{\left(\dfrac{x}{\sqrt{x^2-x+1}+1}\right)^2-1}{\sqrt{x^2-x+1}}\) vô nghiệm
Vậy x=1
b. \(\sqrt{x-4}+\sqrt{x^2-3x+4}=x\)
(ĐKXĐ: \(x\ge4\))
\(\Leftrightarrow\sqrt{x^2-3x+4}=x-\sqrt{x-4}\)
\(\Leftrightarrow x^2-3x+4=x^2+x-4-2\sqrt{x\left(x-4\right)}\)
\(\Leftrightarrow x^2-3x+4-x^2-x+4+2\sqrt{x^2-4x}=0\Leftrightarrow-4x+8+2\sqrt{x^2-4x}=0\Leftrightarrow-2\left(2x-4-\sqrt{x^2-4x}\right)=0\Leftrightarrow2x-4-\sqrt{x^2-4x}=0\Leftrightarrow\sqrt{x^2-4x}=2x-4\Leftrightarrow x^2-4x=4x^2+16-16x\Leftrightarrow x^2-4x^2-4x+16x-16=0\Leftrightarrow-3x^2+12x-16=0\Leftrightarrow3x^2-12x+16=0\)
Ta có: \(\Delta=b^2-4ac=\left(-12\right)^2-4.3.16=-48< 0\)
=> pt vô nghiệm.
Vậy pt đã cho vô nghiệm.
ĐK : x > 3/2
Đặt \(\sqrt{3x-2}=a\left(a>0\right)\) . Khi đó pt thành :
\(1+\dfrac{x}{a}=\dfrac{1+a}{x}\Leftrightarrow\dfrac{a+x}{a}=\dfrac{a+1}{x}\Leftrightarrow a^2+a=ax+x^2\Leftrightarrow x^2+a\left(x-1\right)-a^2=0\)
hay \(\sqrt{3x-2}\left(x-1\right)+x^2-3x+2=0\Leftrightarrow\left(\sqrt{3x-2}-1\right)\left(x-1\right)+x^2-2x+1=0\Leftrightarrow\dfrac{3x-3}{\sqrt{3x-2}+1}\left(x-1\right)+\left(x-1\right)^2=0\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x-2}+1}+\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x-2}+1}+1\right)=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)
Vì \(\dfrac{3}{\sqrt{3x-2}+1}+1>0\)
Vậy nghiệm của pt là x = 1