e: \(\left(-4156+2021\right)-\left(119+2021-4156\right)\)

\(=-4156+2021-119-2021+4156\)

\(=\left(-4156+4156\right)+\left(2021-2021\right)-119\)

=0+0-119

=-119

g: \(315\cdot75-\left(15\cdot100-315\cdot25\right)\)

\(=315\cdot75-15\cdot100+315\cdot25\)

\(=315\left(75+25\right)-15\cdot100\)

\(=315\cdot100-15\cdot100=300\cdot100=30000\)

h: \(\left(-489\right)\cdot125-\left(125\cdot11-500\cdot25\right)\)

\(=-489\cdot125-125\cdot11+500\cdot25\)

\(=125\left(-489-11\right)+500\cdot25\)

\(=125\cdot\left(-500\right)+500\cdot25\)

\(=500\left(-125+25\right)\)

\(=500\cdot\left(-100\right)=-50000\)

Bài 2:

a: \(-415-3\left(2x-1\right)^2=-490\)

=>\(3\left(2x-1\right)^2+415=490\)

=>\(3\left(2x-1\right)^2=75\)

=>\(\left(2x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Câu 3: 

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)

Do đó: x=54; y=36

=>3x^2+1/6x-3x^2-x-2=3

=>-5/6x=5

=>x=-6

\(2,\\ a,=6+12-15=3\\ b,=\left(2\sqrt{2}-3\sqrt{2}\right)\sqrt{2}=-\sqrt{2}\cdot\sqrt{2}=-2\\ c,=\dfrac{2\left(\sqrt{3}-1\right)}{2}-\dfrac{\sqrt{3}+2}{-1}+\dfrac{6\left(3-\sqrt{3}\right)}{6}\\ =\sqrt{3}-1+\sqrt{3}+2+3-\sqrt{3}=4+\sqrt{3}\)