Câu 1 dòng hai \(\left(-1+\dfrac{2}{3}\right):\dfrac{1}{2}\)

\(a,\left(-0,75\right)-\left(-1+\dfrac{2}{3}\right):0,5-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(\dfrac{-3+2}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right):\dfrac{1}{2}-\dfrac{1}{4}\\ =-\dfrac{3}{4}-\left(-\dfrac{1}{3}\right)\times2-\dfrac{1}{4}\\ =-\dfrac{3}{4}+\dfrac{2}{3}-\dfrac{1}{4}\\ =\left(-\dfrac{3}{4}-\dfrac{1}{4}\right)+\dfrac{2}{3}\\ =-\dfrac{4}{4}+\dfrac{2}{3}\\ =-1+\dfrac{2}{3}\\ =\dfrac{-3+2}{3}=-\dfrac{1}{3}\)

\(b,\left[\left(-\dfrac{3}{2}\right)+\dfrac{2}{3}\right]^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(\dfrac{-3\times3+2\times2}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\left(-\dfrac{5}{6}\right)^2\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{25}{36}\times\dfrac{24}{25}-\dfrac{1}{5}\\ =\dfrac{2}{3}-\dfrac{1}{5}\\ =\dfrac{2\times5-3}{15}=\dfrac{7}{15}\)

\(đk:x\ge0;x\ne1\)

\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ \Rightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)\\ \Rightarrow x-2\sqrt{x}+\sqrt{x}-2=x-\sqrt{x}\\ \Rightarrow-\sqrt{x}-2+\sqrt{x}=0\\ \Rightarrow-2=0\left(voli\right)\)

Vậy phương trình vô nghiệm

Ta chỉ dùng từ thỏa mãn khi có đặt điều kiện xác định nhé em!

=> Tại đề cho là tìm x thuộc N đó ạ 

\(a,27.3^x=243\\ \Rightarrow3^x=243:27\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2\left(thoaman\right)\)

\(b,49.7^x=2041\\ \Rightarrow7^x=2041:49\\ \Rightarrow7^x=7^4:7^2\\ \Rightarrow7^x=7^{4-2}\\ \Rightarrow7^x=7^2\\ \Rightarrow x=2\left(thoaman\right)\)

\(c,64.4^x=4^5\\ \Rightarrow4^x=4^5:4^3\\ \Rightarrow4^x=4^2\\ \Rightarrow x=2\left(thoaman\right)\)

\(d,3^x=243\\ \Rightarrow3^x=3^5\\ \Rightarrow x=5\left(thoaman\right)\)

\(e,2^{x+5}=128\\ \Rightarrow2^x.2^5=2^7\\ \Rightarrow2^x=2^7:2^5\\ \Rightarrow2^x=2^2\\ \Rightarrow x=2\left(thoaman\right)\)

\(f,3.3^x=81\\ \Rightarrow3^x=81:3\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\left(thoaman\right)\)

\(g,25+5^x.5^x=650\\ \Rightarrow5^{2x}=650-25\\ \Rightarrow5^{2x}=625\\ \Rightarrow5^{2x}=5^4\\ \Rightarrow2x=4\\ \Rightarrow x=4:2=2\left(thoaman\right)\)

câu h không rõ đề ? 

@seven 

đề bài là \(-\dfrac{\sqrt{2}}{\sqrt{4+\sqrt{15}}}\) anh ơi 

câu c cả ba số đều thỏa mãn nha 

\(a,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+\dfrac{3.\left(\sqrt{6}-\sqrt{5}\right)}{6-5}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+3\left(\sqrt{6}-\sqrt{5}\right)\\ =\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\\ =4\sqrt{6}-2\sqrt{5}\)

\(b,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\sqrt{4+\sqrt{15}}}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\dfrac{2}{\sqrt{8+2.\sqrt{3}.\sqrt{5}}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\left|\sqrt{3}-\sqrt{2}\right|}-\dfrac{2}{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\left|\sqrt{5}+\sqrt{3}\right|}\)

\(=\sqrt{5}+\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{3-2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{3}\\ =0\)

\(\dfrac{2}{5}giờ=\dfrac{2}{5}\times60=24\left(phút\right)\)