a) đkxđ: \(\left\{{}\begin{matrix}2x+1\ge0\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\ne0\end{matrix}\right.\)
b) đkxđ: \(2x^2+1\ge0\) (luôn thỏa mãn \(\forall x\in R\) )
c) đkxđ: \(\left\{{}\begin{matrix}x-1>0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>-3\end{matrix}\right.\) \(\Leftrightarrow x>1\)
d) đkxđ: \(\left\{{}\begin{matrix}x^2-4\ne0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ge-1\end{matrix}\right.\)

a) đk \(\left\{{}\begin{matrix}2x+1\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\ne0\end{matrix}\right.\)

b) đk \(x+3>0\Leftrightarrow x>-3\)

c) \(\left\{{}\begin{matrix}x-1>0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x\ge0\end{matrix}\right.\Leftrightarrow x>1\)

d) đk \(\left\{{}\begin{matrix}x^2-4\ne0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne\pm2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)

Giải ra dùm mình lun nha. Cảm ơn nhìu

a: ĐKXĐ: x-1>0 và x+2<>0

=>x>1

b: DKXĐ: (x-2)căn x-1<>0

=>x-1>0 và x-2<>0

=>x>1 và x<>2

c: ĐKXĐ: 2x-1>=0 và 3-x>0

=>x>=1/2 và x<3

d: ĐKXĐ: x-1>0 và x-2<>0

=>x>1 và x<>2

e: ĐKXĐ: x³+1>=0

=>x>=-1

a)
Đkxđ: \(\left\{{}\begin{matrix}-3x+2\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-2}{3}\\x\ne-1\end{matrix}\right.\)
b)
Đkxđ: \(\left\{{}\begin{matrix}x-2\ge0\\-x-4\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le4\end{matrix}\right.\)\(\Leftrightarrow2\le x\le4\).
c)
Đkxđ: \(\left\{{}\begin{matrix}3x^2+6x+11>0\\2x+1\ge0\end{matrix}\right.\)\(\Leftrightarrow2x+1\ge0\)\(\Leftrightarrow x\ge-\dfrac{1}{2}\).
d)
Đkxđ: \(\left\{{}\begin{matrix}x+4\ge0\\x^2-9\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\ne3\\x\ne-3\end{matrix}\right.\).

a)TXĐ:\(x-1>0\Rightarrow x>1\)

đkxđ:\(\left\{{}\begin{matrix}x\sqrt{x}-1\ne0\\\sqrt{x}\ge0\\2x+\sqrt{x}-1\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\ge0\\x\ne\dfrac{1}{4}\end{matrix}\right.\)

\(\Rightarrow x\ge1\)

sory, x>1 bạn nhé

a) \(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

Đặt \(\sqrt{x^2-3x+3}=a;\sqrt{x^2-3x+6}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\b^2-a^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=3\\\left(b+a\right)\left(b-a\right)=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+a=3\\b-a=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=2\\a=1\end{matrix}\right.\) (nhận)

\(\Rightarrow\sqrt{x^2-3x+3}=1\)

\(\Leftrightarrow x^2-3x+3=1\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\) (nhận)

b) \(\sqrt{3-x+x^2}-\sqrt{2+x-x^2}=1\)

Đặt \(\sqrt{3-x+x^2}=a;\sqrt{2+x-x^2}=b\left(a;b>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\a^2+b^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\\left(b^2+2b+1\right)+b^2-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b+1\\2\left(b-1\right)\left(b+2\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) (vì \(b+2>0\)) (nhận)

\(\Rightarrow\sqrt{2+x-x^2}=1\)

\(\Leftrightarrow2+x-x^2=1\)

\(\Leftrightarrow x^2-x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)

d) \(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}=2x+\dfrac{1}{2x}+4\)

\(\Leftrightarrow2\left(x+\dfrac{1}{4x}\right)+4=5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)\)

\(\Leftrightarrow2\left[\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-1\right]-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+4=0\)

\(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-5\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)+2=0\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Rightarrow2a^2-5a+2=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(\text{nhận}\right)\\a=\dfrac{1}{2}\left(\text{loại}\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{2+\sqrt{2}}{2}\\\sqrt{x}=\dfrac{2-\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+2\sqrt{2}}{2}\\x=\dfrac{3-2\sqrt{2}}{2}\end{matrix}\right.\) (nhận)

a, ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\\x=2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

b, ĐKXĐ: \(x\ge-1\)

\(pt\Leftrightarrow\sqrt{x+1}\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\x+1=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ: \(x>2\)

\(pt\Leftrightarrow\frac{x}{\sqrt{x-2}}=\frac{3-x}{\sqrt{x-2}}\)

\(\Leftrightarrow x=3-x\)

\(\Leftrightarrow x=\frac{3}{2}\left(l\right)\)

\(\Rightarrow\) Phương trình vô số nghiệm

d, ĐKXĐ: \(x>-1\)

\(pt\Leftrightarrow\frac{x^2-4}{\sqrt{x+1}}=\frac{x+3+x+1}{\sqrt{x+1}}\)

\(\Leftrightarrow x^2-4=2x+4\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

a) TXĐ: \(D=R\).
b) \(TXD=D=R\backslash\left\{4\right\}\)
c) Đkxđ: \(\left\{{}\begin{matrix}4x+1\ge0\\-2x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{4}\\x\le\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{-1}{4}\le x\le\dfrac{1}{2}\).
TXĐ: D = \(\left[\dfrac{-1}{4};\dfrac{1}{2}\right]\)

a) Đkxđ: \(\left\{{}\begin{matrix}x+9\ge0\\x^2+8x-20\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-9\\\left\{{}\begin{matrix}x\ne2\\x\ne-10\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-9\\x\ne2\end{matrix}\right.\)
Txđ: D = [ - 9; 2) \(\cup\) \(\left(2;+\infty\right)\)
b) Đkxđ: \(\left\{{}\begin{matrix}2x+1\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{2}\\x\ne3\end{matrix}\right.\)
Txđ: \(D=R\backslash\left\{\dfrac{-1}{2};3\right\}\)
c) \(x^2+2x-5\ne0\Leftrightarrow\left\{{}\begin{matrix}x\ne-1+\sqrt{6}\\x\ne-1-\sqrt{6}\end{matrix}\right.\)
Txđ: \(D=R\backslash\left\{-1+\sqrt{6};-1-\sqrt{6}\right\}\)