a)267 + 125 - 278 = 392 - 278 = 114

b) 538 - 38 ´ 3 = 538 - 114 = 424

267  + 125 – 278 = 114

A

Chọn A

Câu 1: x = 50

Câu 2: ko bik

386 + 793 = 1179

234 + 673 = 907

573 + 456 = 1029

452 + 352 = 804

-.-

k mik

Sửa đề

\(A=cos^212+cos^223+cos^234+cos^245+cos^256+cos^267+\)

\(=\left(cos^212+cos^278\right)+\left(cos^223+cos^267\right)+\left(cos^234+cos^256\right)+cos^245\)

\(=\left(cos^212+sin^212\right)+\left(cos^223+sin^223\right)+\left(cos^234+sin^234\right)+cos^245\)

\(=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)

Ta có: \(\left(0,125\right)^3\cdot8^3+\left(\frac34-\frac56\right)^2-\frac{50^3}{125}+20^5\cdot\frac{5^{10}}{100^5}\)

\(=\left(0,125\cdot8\right)^3+\left(\frac{9}{12}-\frac{10}{12}\right)^2-\frac{\left(5^2\cdot2\right)^3}{5^3}+\frac{5^5\cdot2^{10}\cdot5^{10}}{\left(2^2\cdot5^2\right)^5}\)

\(=1+\left(-\frac{1}{12}\right)^2-\frac{5^6\cdot2^3}{5^3}+5^5=1+\frac{1}{144}-5^3\cdot2^3+5^5\)

\(=\frac{145}{144}+5^3\left(5^2-2^3\right)=\frac{145}{144}+125\cdot17=\frac{306145}{144}\)