\(2^1+2^2+2^3+2^4=2+4+8+16=30\)

Ta có: \(\dfrac{AB}{AC}=\dfrac{7}{24}\Rightarrow AC=\dfrac{24}{7}AB\) (1)

Tam giác ABC vuông tại A có AH là đường cao nên

\(\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}\Leftrightarrow\dfrac{AB^2+AC^2}{AB^2.AC^2}=\dfrac{1}{33,6^2}\) (2)

Thay (1) vào (2) ta được :

\(\dfrac{AB^2+\dfrac{576}{49}AB^2}{\dfrac{AB^2.576}{49}.AB^2}=\dfrac{1}{33,6^2}\Leftrightarrow\dfrac{\dfrac{625}{49}.AB^2}{\dfrac{576}{49}AB^4}=\dfrac{1}{33,6^2}\Leftrightarrow\dfrac{625}{576.AB^2}=\dfrac{1}{33,6^2}\Leftrightarrow AB^2=1225\Leftrightarrow AB=35\)

=> AC = 24/7.35 = 120 (cm)

\(BC^2=AB^2+AC^2=35^2+120^2=15625\Leftrightarrow BC=125\left(cm\right)\)

a/ \(\sqrt{0,36.100}=\sqrt{36}=\sqrt{6^2}=6\)

b/ \(\sqrt[3]{-0,008}=\sqrt[3]{-\dfrac{8}{1000}}=\sqrt[3]{-\left(\dfrac{2}{10}\right)^3}=-\dfrac{2}{10}=-0,2\)

c/ \(\sqrt{12}+6\sqrt{3}+\sqrt{27}\) = \(2\sqrt{3}+6\sqrt{3}+3\sqrt{3}=11\sqrt{3}\)

\(3a^2-6ab-3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=\left(\sqrt{3}\left(a-b\right)\right)^2-\left(\sqrt{12}c\right)^2=\left(\sqrt{3}a-\sqrt{3}b-\sqrt{12}c\right)\left(\sqrt{3}a-\sqrt{3}b+\sqrt{12}c\right)\)

\(87^2+73^2-27^2-13^2=\left(87^2-13^2\right)+\left(73^2-27^2\right)=74.100+46.100=100\cdot\left(74+46\right)=12000\)

\(2x^2+\left(1-\sqrt{5}\right)x+\sqrt{5}-3=0\Leftrightarrow2x^2-2x+\left(3-\sqrt{5}\right)x-\left(3-\sqrt{5}\right)=0\Leftrightarrow2x\left(x-1\right)+\left(3-\sqrt{5}\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-\sqrt{5}\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+3-\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{\sqrt{5}-3}{2}\end{matrix}\right.\)

Vậy x = 1 ; x = \(\dfrac{\sqrt{5}-3}{2}\)

* Tìm MAX :

Áp dụng bđt bunhiacopxki:

ta có : \(\left(3\sqrt{x-2}+4\sqrt{10-x}\right)^2\le\left(3^2+4^2\right)\left(x-2+10-x\right)=1152\)

Dấu ''='' xảy ra khi : x = \(\dfrac{122}{25}\)

Vậy max của y là 1152 khi x = 122/25

* Tìm MIN:

Ta có bđt : \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge a+b\) (với a,b là các số không âm)

=> \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

Do đó:

\(3\sqrt{x-2}+4\sqrt{10-x}=3\left(\sqrt{x-2}+\sqrt{10-x}\right)+\sqrt{10-x}\ge3\left(\sqrt{x-2}+\sqrt{10-x}\right)\ge3\sqrt{x-2+10-x}=3\sqrt{8}\)

Dấu ''='' xảy ra khi x = 10

Vậy MIN y là 8 khi x = 10

KQ: 185183950600000