Ôn tập chương 1: Căn bậc hai. Căn bậc ba

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

\(\Leftrightarrow x^2-4x+4+2x-3-2\sqrt{2x-3}+1=0\)

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

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

Câu 3: 

\(4x^2-6x+10\)

\(=4\left(x^2-\dfrac{3}{2}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{31}{16}\right)\)

\(=4\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{4}>=\dfrac{31}{4}\)

=>A>=căn 31/2

Dấu = xảy ra khi x=3/4

a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b: ĐểP<15/4 thì P-15/4<0

\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)

=>12 căn +32-15 căn x+30<0

=>-3 căn x<-62

=>căn x>62/3

=>x>3844/9

\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1+1}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}\le1+1=2\left(x\ge0\right)\)

Dấu " = " xảy ra khi : \(x=0\left(TM\right)\)

\(\Rightarrow A_{Max}=2\Leftrightarrow x=0\)

a: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{4}\)

\(=\dfrac{2\sqrt{10}+3\sqrt{2}-\sqrt{10}}{4}=\dfrac{\sqrt{10}+3\sqrt{2}}{4}\)

b: \(\dfrac{3+4\sqrt{3}}{\sqrt{6+2\sqrt{5}}}-\dfrac{2\sqrt{2}\left(\sqrt{3}+1\right)}{2}\)

\(=\dfrac{4\sqrt{3}+3}{\sqrt{5}+1}-\sqrt{2}\left(\sqrt{3}+1\right)\)

\(=\dfrac{\left(4\sqrt{3}+3\right)\left(\sqrt{5}-1\right)}{4}-\dfrac{4\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)

\(=\dfrac{4\sqrt{15}-4\sqrt{3}+3\sqrt{5}-3-4\sqrt{6}-4\sqrt{2}}{4}\)

Lời giải:
ĐKXĐ: $x\geq 1$

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

\(\sqrt{x^4-1}=\sqrt{(x-1)(x^3+x^2+x+1)}=ab\). PT đã cho trở thành:
\(a+b=1+ab\)

\(\Leftrightarrow ab+1-a-b=0\)

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

Nếu $a=1$: \(\Leftrightarrow \sqrt{x-1}=1\Rightarrow x=1\) (thỏa mãn)

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

\(\Rightarrow x^3+x^2+x=0\) (vô lý với mọi $x\geq 1$)

Vậy PT có nghiệm duy nhất $x=1$

Bài 1: 

a: ĐKXĐ: x>0; x<>1

b: \(M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)

c: \(M=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)

Dấu = xảy ra khi x=1/4

\(P=\dfrac{\sqrt{a}+1-\sqrt{a}+1+2\sqrt{a}}{2\left(a-1\right)}=\dfrac{2\sqrt{a}+2}{2\left(a-1\right)}=\dfrac{1}{\sqrt{a}-1}\)

P-|P|=0

=>|P|=P

=>P>=0

=>1/căn a-1>=0 với a>1(luôn đúng)

\(=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)

=4