Giải pt:
\(x-3\sqrt{x-3}-3=0\)
giải pt \(x^3-3x^2+2\sqrt{x+3}^3-9x=0\)
giải PT \(x^3+\sqrt{3}x^2+x-\sqrt{3}=0\)
\(\Leftrightarrow3\sqrt{3}x^3+9x^2+3\sqrt{3}x-9=0\)
\(\Leftrightarrow\left(\sqrt{3}x\right)^3+3.\left(\sqrt{3}x\right)^2.1+3.\left(\sqrt{3}x\right).1^2+1^3-10=0\)
\(\Leftrightarrow\left(\sqrt{3}x+1\right)^3=10\)
\(\Leftrightarrow\sqrt{3}x+1=\sqrt[3]{10}\)
\(\Leftrightarrow x=\frac{\sqrt[3]{10}-1}{\sqrt{3}}\)
giải pt
`a,(x+\sqrt{3})+4(x^2-3)=0`
`a,(x+\sqrt{3})+4(x^2-3)=0`
`<=>(x+\sqrt{3})+4(x-\sqrt{3})(x+\sqrt{3})=0`
`<=>(x+\sqrt{3})[4(x-\sqrt{3}+1]=0`
`<=>(x+\sqrt{3})(4x-4\sqrt{3}+1)=0`
`<=>` \(\left[ \begin{array}{l}x+\sqrt{3}=0\\4x-4\sqrt{3}+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\4x=4\sqrt{3}-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\x=\sqrt{3}-\dfrac{1}{4}\end{array} \right.\)
Vậy phương trình có tập nghiệm `S={-\sqrt{3},\sqrt{3}-1/4}`
\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)
a) Ta có: \(\left(x+\sqrt{3}\right)+4\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=0\\4x+1-4\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\4x=4\sqrt{3}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{3};\dfrac{4\sqrt{3}-1}{4}\right\}\)
giải pt:
\(\sqrt[3]{3-x}+\sqrt[3]{x-1}=0\)
\(\sqrt[3]{3-x}+\sqrt[3]{x-1}=0\)
\(\Leftrightarrow\sqrt[3]{3-x}=-\sqrt[3]{x-1}\)
\(\Leftrightarrow3-x=1-x\)
\(\Leftrightarrow0x=2\left(voli\right)\)
Vậy \(ptvn\)
\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
Giải PT trên
Giải pt
\(5\sqrt[3]{x+1}+1\sqrt{x+2}+5\sqrt[3]{x+3}=0\)
b5: giải pt ;
a, \(\sqrt{49\left(1-2x+x^2\right)}-35=0\)
b, \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
c, \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)
\(\Leftrightarrow7\left|x-1\right|=35\)
\(\Leftrightarrow\left|x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)
c) ĐKXĐ: \(x\ge0\)
Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
\(\Leftrightarrow x-1=x+\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}-6=-1\)
\(\Leftrightarrow\sqrt{x}=5\)
hay x=25(nhận)
1. Giải pt:
\(\sqrt{x^2-4x+1}-\sqrt{x+1}=0\)
2. Giải pt:
\(\sqrt{x^2-4x+3}+\sqrt{x-1}=0\)
a) Đk: \(\hept{\begin{cases}x^2-4x+1\ge0\\x+1\ge0\end{cases}}\)
\(\sqrt{x^2-4x+1}=\sqrt{x+1}\)
\(\Leftrightarrow x^2-4x+1=x+1\)
\(\Leftrightarrow x^2-4x-x=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)thỏa mãn điều kiện
Vậy x=0 hoặc x=5
2)\(\sqrt{\left(x-1\right)\left(x-3\right)}+\sqrt{x-1}=0\)(1)
Đk: x>=3 hoặc x=1
pt (1)<=> \(\sqrt{x-1}\left(\sqrt{x-3}+1\right)=0\)
<=> \(\sqrt{x-1}=0\)(vì\(\sqrt{x-3}+1>0\)mọi x )
<=> x-1=0
<=> x=1 ( thỏa mãn điều kiện)
Giải pt: \(\sqrt[3]{x^2+1}-\sqrt{x^3-2}+x=0\)