ai cứu mình được ko ;-;
4)
a) Thay \(x=2022;y=-1\) vào biểu thức A, ta có:
\(A=\left(2022+1\right)-\left(2022-1\right)=2023-2021=2.\)
b) Thay \(a=-2;b=-3\) vào biểu thức B, ta có:
\(B=3.\left(-2\right)+\left(-3\right)^2=-6+9=3.\)
5)
\(a,-2x+15=7.\Leftrightarrow-2x=-8.\Leftrightarrow x=4.\\ b,\left(-25\right)-\left(2x-1\right)=-30.\Leftrightarrow2x-1=5.\Leftrightarrow x=3.\)
\(c,\left(2x+14\right).\left(-3x-15\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}2x+14=0.\\-3x-15=0.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7.\\x=-5.\end{matrix}\right.\)
\(d,\left(x+1\right)^2=4.\Leftrightarrow\left[{}\begin{matrix}x+1=2.\\x+1=-2.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-3.\end{matrix}\right.\)
\(e,\left(2x+1\right)^2=9.\Leftrightarrow\left[{}\begin{matrix}2x+1=3.\\2x+1=-3.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-2.\end{matrix}\right.\)
\(f,\left(x-2\right)^3=-27.\Leftrightarrow\left(x-2\right)^3=\left(-3\right)^3.\Leftrightarrow x-2=-3.\Leftrightarrow x=-1.\)
\(g,\left(3x-1\right)^3=-8.\Leftrightarrow\left(3x-1\right)^3=\left(-2\right)^3.\Leftrightarrow3x-1=-2.\Leftrightarrow x=-\dfrac{1}{3}.\)
\(h,\left(-5\right)^x=-125.\Leftrightarrow\left(-5\right)^x=\left(-5\right)^3.\Rightarrow x=3.\)
\(i,\left(-3\right)^{x-1}=9.\Leftrightarrow\left(-3\right)^{x-1}=\left(-3\right)^2.\Rightarrow x-1=2.\Leftrightarrow x=3.\)