Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a/ ĐKXĐ : \(-2x+3\ge0\)
\(\Leftrightarrow x\le\dfrac{3}{2}\)
b/ ĐKXĐ : \(3x+4\ge0\)
\(\Leftrightarrow x\ge-\dfrac{4}{3}\)
c/ Căn thức \(\sqrt{1+x^2}\) luôn được xác định với mọi x
d/ ĐKXĐ : \(-\dfrac{3}{3x+5}\ge0\)
\(\Leftrightarrow3x+5< 0\)
\(\Leftrightarrow x< -\dfrac{5}{3}\)
e/ ĐKXĐ : \(\dfrac{2}{x}\ge0\Leftrightarrow x>0\)
P.s : không chắc lắm á!
1.
Áp dụng BĐT dạng $|a|+|b|\geq |a+b|$ ta có:
$A=|x+2|+|x+3|=|x+2|+|-x-3|\geq |x+2-x-3|=1$
Vậy GTNN của $A$ là $1$. Giá trị này đạt tại $(x+2)(-x-3)\geq 0$
$\Leftrightarrow (x+2)(x+3)\leq 0$
$\Leftrightarrow -3\leq x\leq -2$
2. ĐKXĐ: $x\geq 1$
\(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
\(=|\sqrt{x-1}+1|+|1-\sqrt{x-1}|\geq |\sqrt{x-1}+1+1-\sqrt{x-1}|=2\)
Vậy gtnn của $B$ là $2$. Giá trị này đạt tại $(\sqrt{x-1}+1)(1-\sqrt{x-1})\geq 0$
$\Leftrightarrow 1-\sqrt{x-1}\geq 0$
$\Leftrightarrow 0\leq x\leq 2$
a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)
a, \(x+1\ge0\Leftrightarrow x\ge-1\)
b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)
d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)
e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
a,\(A=2\sqrt{x^2+x+\dfrac{1}{2}}=2\sqrt{x^2+x+\dfrac{1}{4}+\dfrac{1}{4}}=2\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)
\(=\sqrt{4\left(x+\dfrac{1}{2}\right)^2+1}\ge1\) dấu"=" xảy ra<=>x=-1/2
\(B=\sqrt{2\left(x^2-2x+\dfrac{5}{2}\right)}=\sqrt{2\left[x^2-2x+1+\dfrac{3}{2}\right]}\)
\(=\sqrt{2\left(x-1\right)^2+3}\ge\sqrt{3}\) dấu"=" xảy ra<=>x=1
\(C=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\ge\dfrac{-2}{-\sqrt{2}}=\sqrt{2}\) dấu"=" xảy ra<=>x=1
\(D=x-2\sqrt{x+2}\ge-2\) dấu"=" xảy ra<=>x=-2
a) Ta có: \(\sqrt{3x-2}=x+1\) ( ĐK: \(x\ge\frac{2}{3}\))
\(\Leftrightarrow\left(\sqrt{3x-2}\right)^2=\left(x+1\right)^2\)
\(\Leftrightarrow3x-2=x^2+2x+1\)
\(\Leftrightarrow x^2-x+3=0\)
\(\Leftrightarrow\left(x^2-x+\frac{1}{4}\right)+\frac{11}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\forall x\)
mà \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\)
\(\Rightarrow\)\(S=\varnothing\)
b) Ta có: \(\sqrt{x^2-2x+1}=x-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=x-1\)
\(\Leftrightarrow\left|x-1\right|=x-1\)
+ Với \(x< 1\)\(\Rightarrow\)\(\left|x-1\right|=1-x\)
Ta có: \(1-x=x-1\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\left(L\right)\)
+ Với \(x\ge1\)\(\Rightarrow\)\(\left|x-1\right|=x-1\)
Ta có: \(x-1=x-1\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow\)\(x\inℝ\)
Vậy \(S=ℝ\)
c) Ta có: \(\sqrt{2x+1}=x-2\) ( ĐK: \(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(\sqrt{2x+1}\right)^2=\left(x-2\right)^2\)
\(\Leftrightarrow2x+1=x^2-4x+4\)
\(\Leftrightarrow x^2-6x+3=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)-6=0\)
\(\Leftrightarrow\left(x-3\right)^2=6\)
\(\Leftrightarrow x-3=\pm\sqrt{6}\)
+ \(x-3=\sqrt{6}\)\(\Leftrightarrow\)\(x=3+\sqrt{6}\approx5,45\)\(\left(TM\right)\)
+ \(x-3=-\sqrt{6}\)\(\Leftrightarrow\)\(x=3-\sqrt{6}\approx0,55\)\(\left(TM\right)\)
Vậy \(S=\left\{5,45;0,55\right\}\)
d) Ta có: \(\sqrt{x^2-3}=x^2-3\) ( ĐK: \(x\ge\pm\sqrt{3}\))
\(\Leftrightarrow\sqrt{x^2-3}-\left(\sqrt{x^2-3}\right)^2=0\)
\(\Leftrightarrow\sqrt{x^2-3}.\left(1-\sqrt{x^2-3}\right)=0\)
+ \(\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(x^2-3=0\)\(\Leftrightarrow\)\(x^2=3\)\(\Leftrightarrow\)\(x=\pm\sqrt{3}\)\(\left(TM\right)\)
+ \(1-\sqrt{x^2-3}=0\)\(\Leftrightarrow\)\(\sqrt{x^2-3}=1\)
\(\Leftrightarrow\)\(x^2-3=1\)
\(\Leftrightarrow\)\(x^2=4\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\left(TM\right)\\x=-2\left(L\right)\end{cases}}\)
Vậy \(S=\left\{-\sqrt{3};\sqrt{3};2\right\}\)
e) Ta có: \(\sqrt{x^2-6x+9}=6-x\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=6-x\)
\(\Leftrightarrow\left|x-3\right|=6-x\)
+ Với \(x< 3\)\(\Leftrightarrow\)\(\left|x-3\right|=3-x\)
Ta có: \(3-x=6-x\)
\(\Leftrightarrow0x=3\)( vô nghiệm )
+ Với \(x\ge3\)\(\Leftrightarrow\)\(\left|x-3\right|=x-3\)
Ta có: \(x-3=6-x\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)\(\left(TM\right)\)
Vậy \(S=\left\{\frac{9}{2}\right\}\)
g) Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\) ( ĐK: \(x\inℝ\))
\(\Leftrightarrow\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
+ Với \(x< 2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1< 0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)
Ta có: \(1-\sqrt{x-1}=\sqrt{x-1}-1\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)\(\left(L\right)\)
+ Với \(x\ge2\)\(\Leftrightarrow\)\(\sqrt{x-1}-1\ge0\)\(\Leftrightarrow\)\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}-1\)
Ta có: \(\sqrt{x-1}-1=\sqrt{x-1}-1\)
\(\Leftrightarrow0x=0\)( vô số nghiệm )
Vậy \(S=ℝ\)