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

\(\Rightarrow\sqrt{2}M=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

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

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

= 2

\(\Rightarrow M=\sqrt{2}\)

\(M=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Rightarrow M^2=\left(4+\sqrt{10+2\sqrt{5}}\right)+2\sqrt{\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}\right)\left(\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)}+\left(4-\sqrt{10+2\sqrt{5}}\right)\)

\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

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

\(=8+2\sqrt{5}-2\)

\(=\left(\sqrt{5}+1\right)^2\)

\(\Rightarrow\sqrt{M^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(\Rightarrow\left|M\right|=\sqrt{5}+1\) mà M > 0

\(\Rightarrow M=\sqrt{5}+1\)

Kẻ đường cao CK (K thuộc AB)

\(\Rightarrow\Delta HAD=\Delta KBC\left(ch-gn\right)\)

=> AH = BK

\(\Delta\)HAD vuông tại H

(+) \(\Rightarrow\sin A=\dfrac{HD}{AD}\)

\(\Rightarrow HD\approx7,25\left(cm\right)\)

(+) \(\Rightarrow\cos A=\dfrac{AH}{AD}\)

\(\Rightarrow AH\approx3,38\left(cm\right)\)

\(\widehat{CDH}=\widehat{DHK}=\widehat{HKC}=90^0\)

=> HDCK là hcn

=> DC = HK = AB - AH - BK = AB - 2AH ~ 13,238 (cm)

~ ~ ~ ~ ~

HB = HK + KB = DC + AH ~ 16,619 (cm)

\(\Delta\)HBD vuông tại H

\(\Rightarrow\tan\widehat{HDB}=\dfrac{HB}{HD}\approx2,292\)

\(\Rightarrow\widehat{HDB}\approx66^025'\)

\(\Delta\)HAD vuông tại H

=> \(\widehat{HAD}+\widehat{HDA}=90^0\)

\(\Rightarrow\widehat{HDA}=25^0\)

\(\Rightarrow\widehat{ADB}=\widehat{HDB}+\widehat{HDA}\approx91^025'\)

\(\Delta\)HDB vuông tại H

\(\Rightarrow BD^2=HB^2+HD^2\left(ptg\right)\)

=> BD ~ 18,13 (cm)

\(\Delta\)HAB vuông tại H

\(\Rightarrow AB^2=AH^2+BH^2\left(ptg\right)\)

\(\Rightarrow AH=6\sqrt{3}\left(cm\right)\)

\(\Delta\)ABC vuông tại A có AH là đc

\(\Rightarrow AB^2=BH\times BC\left(htl\right)\)

=> BC = 24 (cm)

=> HC = BC - BH = 18 (cm)

gọi số cần tìm là a ta có 

a là số có 2 chữ số

a chia 2;3;5 dư lần lượt là 1;2;4

=>a+1 chia hết cho 2;3;5

=>a+1 thuộc BC(2;3;5)

BCNN(2;3;5)=2.3.5=30

=>a +1 thuộc B(30)

=>a+1 thuộc {0;30;60;90;120;...} mà a có hai chữ số 

=>a thuộc { 29;59;89}

\(\Delta\)HAB vuông tại H có HD là đc

\(\Rightarrow AH^2=AD\times AB\left(htl\right)\left(1\right)\)

\(\Delta\)HAC vuông tại H có HE là đc

\(\Rightarrow AH^2=AE\times AC\left(htl\right)\left(2\right)\)

(1) và (2) => đpcm

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

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

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

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

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

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

(x² + 5x + 6)(x² + 9x + 20) = 24

<=> (x + 2)(x + 3)(x + 4)(x + 5) - 24 = 0

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0 (1)

Đặt x² + 7x + 11 = t, ta có:

(1) <=> (t - 1)(t + 1) - 24 = 0

<=> t² - 1 - 24 = 0

<=> (t - 5)(t + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}t-5=0\\t+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+7x+11-5=0\\x^2+7x+11+5=0\end{matrix}\right.\)

<=> (x + 1)(x + 6) = 0 (vì \(x^2+7x+16\ge\dfrac{15}{4}>0\))

<=> x = - 1 hoặc x = - 6

~ ~ ~ ~ ~

x⁴ - 24x = 32

<=> x⁴ - 24x - 32 = 0

<=> (x² - 2x - 4)(x² + 2x + 8) = 0

<=> \(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\) (vì \(x^2+2x+8\ge7>0\))

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

\(\sqrt{x-\dfrac{1}{x}}-\sqrt{1-\dfrac{1}{x}}=\dfrac{x-1}{x}\)

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

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{x-1}{x}=0\)

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

Pt \(\dfrac{1}{\sqrt{x-\dfrac{1}{x}}+\sqrt{1-\dfrac{1}{x}}}-\dfrac{1}{x}=0\) vô n₀

=> x - 1 = 0

<=> x = 1 (nhận)