\(M=\left(1-2x+x^2\right)\left(1+2x\right)+\left(1+2x+x^2\right)\left(1-2x\right)-6\left(1-x^2\right)\\ =1-2x+x^2+2x-4x^2+2x^3+1+2x+x^2-2x-4x^2-2x^3-6+6x^2\\ =-4\)

Vậy biểu thức ko phụ thuộc vào biến x

\(A=\left(27+3\right)+\left(23+7\right)+\left(11+19\right)+15=30+30+30+15=105\\ ---------------\\ B,=19-10-\left(16+4\right)+\left(13+7\right)\\ =9-20+20\\ =9\)

\(a,=125.\left(93+8\right)\\=125.93+125.8=11625+1000=12625\\ b,=3.7.7.7=3.7^3=3.343=1029 \)

\(a,x=-\dfrac{7}{9}:\left(\dfrac{14}{27}\right)\\ x=-\dfrac{7}{9}.\dfrac{27}{14}\\ x=-\dfrac{3}{2}\\ ----------\\ b,x=\left(-\dfrac{5}{9}\right):\dfrac{2}{3}\\ x=-\dfrac{5}{9}.\dfrac{3}{2}\\ x=-\dfrac{5}{6}\\ ------------\\ c,\dfrac{2}{5}:x=\dfrac{1}{2}\\ x=\dfrac{2}{5}:\dfrac{1}{2}\\ x=\dfrac{2}{5}.\dfrac{2}{1}\\ x=\dfrac{4}{5}\\ -----------\\ d,-\dfrac{5}{12}x=\dfrac{4-3}{3}\\ -\dfrac{5}{12}x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:\left(-\dfrac{5}{12}\right)\\ x=-\dfrac{2}{5}\)

\(-T-X-G-A-T-X-X-A-\)

theo nguyên tắc : \(A=T;G=X\)

\(a,=1991\times\left(25\times4\right)=1991\times100=199100\\ b,=3025\times\left(125\times8\right)=3025\times1000=3025000\)

\(2,=\sqrt{64}=8\\ 6,=2\sqrt{3}+\left|2-\sqrt{3}\right|=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)

\(\left(x+\dfrac{3}{4}\right)=\dfrac{10}{9}:\dfrac{5}{7}\\ x+\dfrac{3}{4}=\dfrac{10}{9}.\dfrac{7}{5}\\ x+\dfrac{3}{4}=\dfrac{14}{9}\\ x=\dfrac{14}{9}-\dfrac{3}{4}\\ x=\dfrac{14.4-3.9}{36}\\ x=\dfrac{29}{36}\)

\(1,\\ C=6^2.\left(6^2\right)^4.\left(6^3\right)^2\\ =6^2.6^8.6^6=6^{16}\\ ----------\\ 2,a,=8=2^3\\ b,=3=3^1\\ c,=49=7^2\\ d,=100=10^2\\ e,=1331=11^3\\ f,=3=3^1\)

a, Điện thế tương đương là :

\(R_{Tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{20.30}{20+30}=12\left(\Omega\right)\)

b, Cường độ dòng điện là :

\(I=\dfrac{U}{R}=\dfrac{24}{12}=2\left(A\right)\)