\(\left|x-1.7\right|=2.3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1.7=2.3\\x-1.7=-2.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-0.6\end{matrix}\right.\)

\(A=\left(\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right):\dfrac{\sqrt{xy}}{x-y}\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-\left(\sqrt{x}-\sqrt{y}\right)^2}{x-y}\cdot\dfrac{x-y}{\sqrt{xy}}\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}-\sqrt{x}+\sqrt{y}\right)}{x-y}\cdot\dfrac{x-y}{\sqrt{xy}}\)

\(=\dfrac{2\sqrt{x}\cdot2\sqrt{y}}{x-y}\cdot\dfrac{x-y}{\sqrt{xy}}\)

\(=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{x-y}{\sqrt{xy}}\)

\(=4\)

a) ĐKXĐ: \(x^3+8\ne0\Leftrightarrow x^3\ne-8\Leftrightarrow x\ne-2\)

b) \(\dfrac{2x^2-4x+8}{x^3+8}=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)

c) Với x = 2 (t/m ĐKXĐ) ta có
\(\dfrac{2}{2+2}=\dfrac{1}{2}\)

d) \(\dfrac{2}{x+2}=2\)

\(\Leftrightarrow2\left(x+2\right)=2\)

\(\Leftrightarrow2x+4=2\)

\(\Leftrightarrow x=-1\) (t/m ĐKXĐ)

a) \(VT=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=VP\)

Vậy \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)=2^{32}-1\)

Bài 1:

a) \(6+7=13\) (số nguyên tố, vì chỉ có 2 ước là 1 và 13)

b) \(3\cdot10-2\cdot9=12\)(hợp số, vì có nhiều hơn 2 ước)

c) \(383+972=1355\) (hợp số, vì có nhiều hơn 2 ước)

d) \(11\cdot13\cdot17+19\cdot23\cdot29=14230\) (hợp số, vì có nhiều hơn 2 ước)

e) \(17\cdot5\cdot6-17\cdot29=17\) (số nguyên tố, vì có 2 ước là 1 và 17)

a) \(x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2+y\right)\left(x+2-y\right)\)

b) \(3x^2+6xy+3y^2-3z^2\)

\(=3\left(x^2+2xy+y^2-z^2\right)\)

\(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]\)

\(=3\left[\left(x+y\right)^2-z^2\right]\)

\(=3\left(x+y+z\right)\left(x+y-z\right)\)

c) \(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y+z-t\right)\left(x-y-z+t\right)\)

\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)

\(\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)

\(\sqrt{36}\sqrt{x-1}-\sqrt{9}\sqrt{x-1}-\sqrt{4}\sqrt{x-1}=16-\sqrt{x-1}\)

\(\left(6-3-2\right)\sqrt{x-1}=16-\sqrt{x-1}\)

\(\sqrt{x-1}=16-\sqrt{x-1}\)

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

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

\(\sqrt{x-1}=8\)

\(\Rightarrow x-1=64\)

\(x=65\)

Vậy \(x=65\)

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

\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}=\dfrac{7}{10}\)

\(x=\dfrac{7\cdot3}{10}=\dfrac{21}{10}=2.1\)