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<=> (x2 +2)2 =( \(2\sqrt{x^3+1}\)) 2
<=> x4 +4x2 +4 = 4(x3+1 )
<=> x4 +4x2 +4- 4x3 -4=0
<=> x4 +4x2 - 4x3 =0
<=> x2( x2 - 4x + 4 ) = 0
<=> \(\orbr{\begin{cases}x^2=0\\x^2-4x+4=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=0\\\left(x-2\right)^2=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy nghiệm của pt là x=0 hoặc x=2
Đk: tự tìm
\(pt\Leftrightarrow\sqrt{\left(x-4\right)\left(x+4\right)}+\sqrt{x-4}=0\)
\(\Leftrightarrow\sqrt{x-4}\left(\sqrt{x+4}+1\right)=0\)
Dễ thấy: \(\sqrt{x+4}\ge0\forall x\)
\(\Rightarrow\sqrt{x+4}+1\ge1>0\forall x\) (vô nghiệm)
\(\Rightarrow\sqrt{x-4}=0\Rightarrow x-4=0\Rightarrow x=4\)
ĐKXĐ: \(x\ge2\)
\(6x-3\sqrt{3x-6}=12\Leftrightarrow3\left(2x-\sqrt{3x-6}\right)=12\Leftrightarrow2x-\sqrt{3x-6}=4\)
<=>\(2x-4=\sqrt{3x-6}\Leftrightarrow\left(2x-4\right)^2=\left(\sqrt{3x-6}\right)^2\Leftrightarrow4x^2-16x+16=3x-6\)
<=>\(4x^2-19x+22=0\Leftrightarrow4x^2-8x-11x+22=0\Leftrightarrow4x\left(x-2\right)-11\left(x-2\right)=0\)
<=>\(\left(4x-11\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}4x-11=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=2\end{cases}}\)
Đk: tự xác định
\(pt\Leftrightarrow\sqrt{x+3}-\left(\frac{1}{3}x+1\right)+\sqrt{6-x}-\left(-\frac{1}{3}x+2\right)-\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\frac{x+3-\left(\frac{1}{3}x+1\right)^2}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{6-x-\left(-\frac{1}{3}x+2\right)^2}{\sqrt{6-x}-\frac{1}{3}x+2}-\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\frac{-\frac{1}{9}\left(x+3\right)\left(x-6\right)}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}\left(x+3\right)\left(x-6\right)}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{\left(x+3\right)\left(x-6\right)}{\sqrt{-\left(x+3\right)\left(x-6\right)}}=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-6\right)\left(\frac{-\frac{1}{9}}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{1}{\sqrt{-\left(x+3\right)\left(x-6\right)}}\right)=0\)
Dễ thấy:\(\frac{-\frac{1}{9}}{\sqrt{x+3}+\frac{1}{3}x+1}+\frac{-\frac{1}{9}}{\sqrt{6-x}-\frac{1}{3}x+2}-\frac{1}{\sqrt{-\left(x+3\right)\left(x-6\right)}}< 0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-6=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-3\\x=6\end{cases}}\)
a: Ta có: x+17<10
nên x<-7
b: Ta có: 9-2x<0
\(\Leftrightarrow2x>9\)
hay \(x>\dfrac{9}{2}\)
c: Ta có: \(-3x-11\ge0\)
\(\Leftrightarrow-3x\ge11\)
hay \(x\le-\dfrac{11}{3}\)
\(\sqrt{x-2}-\sqrt{4-x}=0\)
\(\Leftrightarrow\sqrt{x-2}=\sqrt{4-x}\)
\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=\left(\sqrt{4-x}\right)^2\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=4+2\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
c: Ta có: \(\sqrt{2x}=\sqrt{5}\)
\(\Leftrightarrow2x=5\)
hay \(x=\dfrac{5}{2}\)
d: Ta có: \(\sqrt{3x-1}=4\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
hay \(x=\dfrac{17}{3}\)
Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}=6\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(\sqrt{16x}=8\Leftrightarrow16x=64\Leftrightarrow x=4\)
Ta có: \(\sqrt{16x}=8\)
\(\Leftrightarrow16x=64\)
hay x=4