các bn giải giúp mk với các bn ơiiiiiii

hỏi 1 tháng chưa ai trả lời ._.

Tập hợp K có số phần tử là

   (300 -2 ) : 2 + 1 = 150 ( phần tử )

Đúngggggg

so phan tu cua tap hop K la : (300-2):2+1=150(phan tu)

Số phần tử là:

(300-2):2+1=150(pt)

Vậy tập hợp K có 150 phần tử

\(M=\sqrt{x-2}+\sqrt{4-x}\Rightarrow M^2=x-2+4-x+2\sqrt{\left(x-2\right)\left(4-x\right)}=2+2\sqrt{\left(x-2\right)\left(4-x\right)}\)

Áp dụng bđt Cauchy, ta có ; \(2\sqrt{\left(x-2\right)\left(4-x\right)}\le x-2+4-x=2\)

\(\Rightarrow M^2\le2+2=4\Rightarrow M\le2\)

Vậy Max M = 2 \(\Leftrightarrow\hept{\begin{cases}2\le x\le4\\x-2=4-x\end{cases}\Leftrightarrow}x=3\)

GTLN của M=2 tại x=3

đặt A=

\(\sqrt{x^2-6x+13}=\sqrt{x^2-2.x.3+3^2-3^2+13}=\sqrt{\left(x-3\right)^2+4}>=2\)

Min A=2<=> x-3=0<=> x=3

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\(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)

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

→ \(3x-2+3+x+2\sqrt{\left(2x-2\right)\left(3+x\right)}=5x+4\)

➝ \(4x+3+2\sqrt{6x+2x^2-6-2x}=5x+4\)

→ \(2\sqrt{2x^2+4x-6}=5x+4-4x-3\)

→ \(2\sqrt{2x^2+4x-6}=x+1\)

→ \(\left(2\sqrt{2x^2+4x-6}\right)^2=\left(x+1\right)^2\)

→ \(4\left(2x^2+4x-6\right)=x^2+2x+1\)

→ \(8x^2+16x-24=x^2+2x+1\)

→ \(8x^2+16x-24-x^2-2x-1=0\)

→ \(7x^2+14x-25=0\)

→ \(x_1=\frac{-7+4\sqrt{14}}{7}\)

\(x_2=\frac{-7-4\sqrt{14}}{7}\)

\( \sqrt {3x - 2} + \sqrt {3 + x} = \sqrt {5x + 4} \left( {x \ge \dfrac{2}{{3}}} \right)\\ \Leftrightarrow 3x - 2 + 2\sqrt {\left( {3x - 2} \right)\left( {3 + x} \right)} + 3 + x = 5x + 4\\ \Leftrightarrow 2\sqrt {7x + 3{x^2} - 6} = x + 3\\ \Leftrightarrow 4\left( {7x + 3{x^2} - 6} \right) = {x^2} + 6x + 9\\ \Leftrightarrow 28x + 12{x^2} - 24 = {x^2} + 6x + 9\\ \Leftrightarrow {x^2} + 2x - 3 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 1\left( {tm} \right)\\ x = - 3\left( {ktm} \right) \end{array} \right. \)

ĐKXĐ : \(\left\{{}\begin{matrix}3x-2\ge0\\3+x\ge0\\5x+4\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge\frac{2}{3}\\x\ge-3\\x\ge-\frac{4}{5}\end{matrix}\right.\)

=> \(x\ge\frac{2}{3}\) (1)

Ta có : \(\sqrt{3x-2}+\sqrt{3+x}=\sqrt{5x+4}\)

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

<=> \(\left(3x-2\right)+2\sqrt{\left(3x-2\right)\left(3+x\right)}+\left(3+x\right)=5x+4\)

<=> \(3x-2+2\sqrt{\left(3x-2\right)\left(3+x\right)}+3+x=5x+4\)

<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=5x+4+2-3-x-3x\)

<=> \(2\sqrt{\left(3x-2\right)\left(3+x\right)}=x+3\)

<=> \(\sqrt{\left(3x-2\right)\left(3+x\right)}=\frac{x+3}{2}\)

ĐKXĐ : \(\frac{x+3}{2}\ge0\)

=> \(x+3\ge0\)

=> \(x\ge-3\) (2)

Từ (1) và (2)

=> \(x\ge\frac{2}{3}\)

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

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

<=> \(9x-6+3x^2-2x=\frac{x^2+6x+9}{4}\)

<=> \(\frac{4\left(9x-6+3x^2-2x\right)}{4}=\frac{x^2+6x+9}{4}\)

<=> \(4\left(9x-6+3x^2-2x\right)=x^2+6x+9\)

<=> \(36x-24+12x^2-8x=x^2+6x+9\)

<=> \(36x-24+12x^2-8x-x^2-6x-9=0\)

<=> \(22x-33+11x^2=0\)

<=> \(11x^2+33x-11x-33=0\)

<=> \(11x\left(x-1\right)+33\left(x-1\right)=0\)

<=> \(\left(11x+33\right)\left(x-1\right)=0\)

<=> \(\left\{{}\begin{matrix}11x+33=0\\x-1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-3\left(L\right)\\x=1\left(TM\right)\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là x = 1 .