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\(A=1500-\left\{5^3.2^3-11.\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=1500-\left\{125.8-11.\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=1500-\left\{1000-11\left[49-40+8.0\right]\right\}\)
\(A=1500-\left\{1000-11.9\right\}\)
\(A=1500-\left\{1000-99\right\}\)
\(A=1500-901=599\)
\(1000-\left\{\left(-5\right)^3.\left(-2\right)^3-11.[7^2-5.2^3+8.\left(11^2-121\right)]\right\}\)
=\(1000-\left\{[\left(-5\right).\left(-2\right)]^3-11.[7^2-5.2^3+2^3.\left(11^2-11^2\right)]\right\}\)
= \(1000-\left\{1000-11.[7^2-2^3.\left(5+0\right)]\right\}\)
= \(1000-[1000-11.\left(7^2-2^3.5\right)\)
= \(1000-[1000-11.\left(49-40\right)]\)
= \(1000-\left(1000-11.9\right)\)
= \(1000-\left(1000-99\right)=1000-1000+99\)
= 0 + 99 = 99
1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)
\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)
\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)
\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)
2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)
\(=1500-\left\{1000-11\cdot9\right\}\)
\(=1500-1000+99=599\)
\(2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
Đặt : \(A=2017-\left\{5^2.2^2-11\left[7^2-5.2^3+8\left(11^2-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+8\left(121-121\right)\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-5.8+0\right]\right\}\)
\(A=2017-\left\{25.4-11\left[49-40\right]\right\}\)
\(A=2017-\left\{25.4-11.9\right\}\)
\(A=2017-\left\{25.4-99\right\}\)
\(A=2017-\left\{100-99\right\}\)
\(A=2017-1=2016\)
Vậy A = 2016
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)
\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)
\(B=\frac{7}{2.7}\)
\(B=\frac{1}{2}\)
\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)
\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)
\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)
\(C=\frac{-2}{9}\)
\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)
\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)
\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)
\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)
\(D=1\)
Ta tính các số mũ thành số hết
\(A=-1500-\left\{125.8-11.\left[49-40+8\left(121-121\right)\right]\right\}.2\)
\(A=-1500-\left\{1000-11.\left(9+0\right)\right\}.2\)
\(A=-1500-\left(1000-99\right).2\)
\(A=-1500-901.2\)
\(A=-1500-1802=-3302\)
sai rồi nhé Hậu Trần Công đề bài là -2 chứ ko phải là 2