Tính giá trị biểu thức sau:
B= \(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
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46.95 + 69.120 / 84.312 - 611
= 212.315 + 212.310.5 / 212.312 - 211.311
= 212. 310.( 35 + 5 ) / 211.311( 2.3 - 1 )
= 2 . 248 / 3 . 5 = 496/15
\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{23}.3^{23}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{23}.3^{23}}=\frac{6}{2^{11}.3^{13}}=\frac{2.3}{2^{11}.3^{12}}=\frac{1}{2^{10}.3^{11}}=\frac{1}{6^{10}.3}\)
\(A=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{8}{11}-\dfrac{5}{7}.\dfrac{2}{11}\)
\(A=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{8}{11}-\dfrac{2}{11}\right)\)
\(A=\dfrac{5}{7}.\dfrac{5+8-2}{11}\)
\(A=\dfrac{5}{7}.\dfrac{11}{11}\)
\(A=\dfrac{5}{7}.1=\dfrac{5}{7}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{63}\)
\(B=\dfrac{95}{72}\)
\(C=\dfrac{4^6.9^5+6^9.120}{8^4-3^{12}-6^{11}}\)
\(C=\dfrac{\left(2^2\right)^3.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(C=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(C=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(C=\dfrac{2.6}{5.3}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{4}{5}\)
Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}\)
\(=\frac{2\cdot6}{3\cdot5}=\frac{4}{5}\)
Tham khảo :
Câu hỏi của ๖ۣۜ๖ۣۜHσàηɠ ๖ۣۜTử ๖ۣۜGĭóッ - Toán lớp 7 - Học trực tuyến OLM