ghê dữ zậy cho tớ xin in4 với :>

=> khomg cạu cúc đi 

\(a,9.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{7}\right)^0-1\\ =9.\dfrac{1}{9}+1-1\\ =\dfrac{9}{9}+1-1\\ =1+1-1\\ =1\\ b,7.\left(5^2-5\right)+\left(-1\right)^2\\ =7.\left(25-5\right)+1\\ =7.20+1\\ =140+1\\ =141\)

\(a,\left(\dfrac{5}{7}\right)^2.\left(\dfrac{5}{7}\right)^5.\dfrac{5}{7}=\left(\dfrac{5}{7}\right)^{2+5+1}=\left(\dfrac{5}{7}\right)^8\\ b,\left(\dfrac{3}{7}\right)^2.\dfrac{9}{49}=\left(\dfrac{3}{7}\right)^2.\left(\dfrac{3}{7}\right)^2=\left(\dfrac{3}{7}\right)^{2+2}=\left(\dfrac{3}{7}\right)^4\)

\(a,\left(\dfrac{3}{7}\right)^2=\dfrac{3^2}{7^2}=\dfrac{9}{49}\\ \left(-\dfrac{1}{2}\right)^4=\dfrac{\left(-1\right)^4}{2^4}=\dfrac{1}{16}\\ \left(-\dfrac{3}{4}\right)^3=\dfrac{\left(-3\right)^3}{4^3}=\dfrac{-27}{16}\\ \left(-\dfrac{27}{100}\right)^0=1\)

\(b,\left(\dfrac{1}{2}\right)^3.16-\dfrac{1}{4}\\ =\dfrac{1}{8}.16-\dfrac{1}{4}\\ =\dfrac{16}{8}-\dfrac{1}{4}\\ =2-\dfrac{1}{4}\\ =\dfrac{2.4-1}{4}=\dfrac{7}{4}\)

\(3,\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\\ =\sqrt{\dfrac{9}{4}.6}+\sqrt{4.\dfrac{2}{3}}-\sqrt{16.\dfrac{3}{2}}\\ =\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\\ =\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-2\sqrt{6}\\ =\dfrac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)

\(4,5\sqrt{x}+6\sqrt{\dfrac{x}{4}}-x\sqrt{\dfrac{4}{x}}\\ =5\sqrt{x}+\sqrt{\dfrac{36x}{4}}-\sqrt{\dfrac{4x^2}{x}}\\ =5\sqrt{x}+\sqrt{9x}-\sqrt{4x}\\ =5\sqrt{x}+3\sqrt{x}-2\sqrt{x}\\ =6\sqrt{x}\)

\(\left(3x+4\right)^2+\left(5-3x\right)^2+2.\left(3x+4\right)\left(5-3x\right)\\ =\left[\left(3x+4\right)+\left(5-3x\right)\right]^2\\ =\left(3x+4+5-3x\right)^2\\ =9^2=81\)

 Hình Như Vỏ cà chua không tiêu hóa được

\(1,5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{15}\\ =\sqrt{\dfrac{25}{5}}+\sqrt{\dfrac{20}{4}}+\sqrt{5}.\sqrt{3}\\ =\sqrt{5}+\sqrt{5}+\sqrt{5}.\sqrt{3}\\ =2\sqrt{5}+\sqrt{3}.\sqrt{5}\\ =\left(2+\sqrt{3}\right).\sqrt{5}\\\)

\(2,\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\\ =\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{25}{2}}\\ =\dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{2}+\dfrac{5\sqrt{2}}{2}\\ =\dfrac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\dfrac{9\sqrt{2}}{2}\)