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Đặt A=\(\frac{16.17-5}{16.16+11}\)
\(\Rightarrow A=\frac{16.\left(16+1\right)-5}{16.16+11}\)
\(\Rightarrow A=\frac{16.16+16-5}{16.16+11}\)
\(\Rightarrow A=\frac{16.16+11}{16.16+11}\)
=> A=1
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3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1
3/14 × 5/17 + 11/14 × 5/17 + 12/17 × 5/16 + 12/17 × 11/16
= 5/17 × ( 3/14 + 11/14) + 12/17 × ( 5/16 + 11/16)
= 5/17 × 1 + 12/17 × 1
= 5/17 + 12/17
= 1
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\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^3\cdot2^{22}-2^{36}}\)
\(=\dfrac{3^2\cdot2^{36}}{2^{25}\left(11-2^{11}\right)}=\dfrac{3^2\cdot2^{11}}{11-2048}=\dfrac{-3\cdot2^{11}}{679}\)
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Cây a, bạn nhân cả 2 vế với 3
Lấy vế nhân với 3 trừ đi ban đầu tất cả chia 2
b) Tính như bình thường
Câu c hình như sai đề
\(\frac{272-5}{256+11}\)= \(\frac{267}{267}\) = \(1\)
Vậy tổng của bài bằng 1