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

\(A\ge\frac{1}{2}\left(x-2+3-x\right)^2=\frac{1}{2}\)

\(A_{min}=\frac{1}{2}\) khi \(x-2=3-x\Leftrightarrow x=\frac{5}{2}\)

Hoặc: \(A=x^2-4x+4+x^2-6x+9=2x^2-10x+13=2\left(x-\frac{5}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\)

b/

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{b+c+a+c+a+b}=\frac{a+b+c}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

a, Ta thấy : \(\sqrt{x-1}\ge0\)

=> \(-\sqrt{x-1}\le0\)

=> \(\sqrt{3}-\sqrt{x-1}\le\sqrt{3}\)

Vậy Max_A = \(\sqrt{3}\) khi x - 1 = 0 <=> x = 1 .

b, Ta có : \(B=6\sqrt{x}-x-1\)

=> \(B=-\left(x-2.3\sqrt{x}+9\right)+8\)

=> \(B=-\left(\sqrt{x}-3\right)^2+8\)

- Ta thấy : \(\left(\sqrt{x}-3\right)^2\ge0\)

=> \(-\left(\sqrt{x}-3\right)^2\le0\)

=> \(8-\left(\sqrt{x}-3\right)^2\le8\)

Vậy Max_B = 8 khi \(\sqrt{x}-3=0\) <=> x = 9 .

c, Ta có : \(C=\frac{1}{x-\sqrt{x}+1}\)

=> \(C=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\)

- Ta thấy : \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\)

=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

=> \(\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{4}{3}\)

Vậy Max_C = \(\frac{4}{3}\) khi \(\sqrt{x}-\frac{1}{2}=0\) <=> x = 1/4 .

Câu 2: 

a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)

\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)

b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)

\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)

\(M=\dfrac{a}{\sqrt{a^2-b^2}}+\dfrac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\cdot\dfrac{a-\sqrt{a^2-b^2}}{b}\)

\(=\dfrac{a}{\sqrt{a^2-b^2}}+\dfrac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)

\(=\dfrac{a+b}{\sqrt{a^2-b^2}}=\dfrac{\sqrt{a+b}}{\sqrt{a-b}}\)

Để M<1 thì (a+b)/(a-b)<1

=>(a+b-a+b)/(a-b)<0

=>2b/(a-b)<0

=>0<a<b

a: ĐKXĐ: a>=0; a<>1; a<>4

b: \(A=\dfrac{\sqrt{a}+1+\sqrt{a}-1}{a-1}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{2\sqrt{a}}{a-1}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{2\sqrt{a}\left(\sqrt{a}-2\right)}{3\left(\sqrt{a}+1\right)}\)

c: Để A>0 thì căn a-2>0

=>a>4

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(C^2=(\sqrt{x}+\sqrt{2-x})^2\leq (x+2-x)(1+1)=4\)

\(\Rightarrow C\leq 2\)

Vậy \(C_{\max}=2\Leftrightarrow \frac{\sqrt{x}}{1}=\frac{\sqrt{2-x}}{1}\Leftrightarrow x=1\)

\(\sqrt{4x^2-4x+1}=x-16\)

⇔\(\sqrt{\left(2x-1\right)^2}=x-16\)

⇔\(\left|2x-1\right|\) = \(x-16\)

⇔\(\left[{}\begin{matrix}2x-1=x-16\\2x-1=16-x\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}2x-x=-16+1\\2x+x=16+1\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-15\\x=\dfrac{17}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-15;\dfrac{17}{3}\right\}\)

Ta có: \(4x^2-4x+1=\left(2x-1\right)^2\ge0\forall x\)

ĐKXĐ: Với mọi giá trị thực của x.

\(\sqrt{4x^2-4x+1}=x-16\) (1)

\(\Leftrightarrow\) \(\sqrt{\left(2x-1\right)^2}=x-16\)

\(\Leftrightarrow\) \(\left|2x-1\right|=x-16\) (2)

- Nếu \(x\ge\dfrac{1}{2}\), hay \(2x-1\ge0\) thì ta có:

(2) \(\Leftrightarrow\) \(2x-1=x-16\)

\(\Leftrightarrow\) \(x=-15\) (loại vì \(x\ge\dfrac{1}{2}\) )

- Nếu \(x< \dfrac{1}{2}\), hay \(2x-1< 0\) thì ta có:

(2) \(\Leftrightarrow\) \(1-2x=x-16\)

\(\Leftrightarrow\) \(3x=17\)

\(\Leftrightarrow\) \(x=\dfrac{17}{3}\) (loại vì \(x< \dfrac{1}{2}\) )

Vậy phương trình (1) vô nghiệm.

<=>x(x+1)=y²

^{Do x,y nguyên nên x(x+1) là số chính phương}

^{Mà x, x+1 là 2 số nguyên liên tiếp nên x(x+1)=0}

<=>x=0 hoặc x=-1. Khi đó y=0