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Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
a) \(\left(-\frac{5}{2}\right)^2:\left(-15\right)-\left(-0,45+\frac{3}{4}\right).\left(-1\frac{5}{9}\right)\)
= \(-\frac{25}{4}:\left(-15\right)-\left(\frac{9}{20}+\frac{15}{20}\right).\left(-\frac{14}{9}\right)\)
=\(-\frac{25}{4}.\frac{1}{-15}-\frac{6}{5}.\left(-\frac{14}{9}\right)\)
= \(\frac{-5}{12}-\frac{8}{5}\)
= \(\frac{\left(-25\right)-96}{60}\)
= \(\frac{\left(-25\right)+\left(-96\right)}{60}\)
=\(\frac{121}{60}\)
b) \(\left(\frac{-1}{3}\right)-\left(\frac{-3}{5}\right)^0+\left(1-\frac{1}{2}\right)^2:2\)
= \(\left(\frac{-1}{3}\right)-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}\)
=\(\left(\frac{-1}{3}\right)-\frac{3}{3}+\frac{1}{4}.\frac{1}{2}\)
= \(\frac{-4}{3}+\frac{1}{8}\)=\(\frac{-32+3}{24}\)
=\(\frac{-29}{24}\)
c) E=\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{2^{10}.3^8-2.6^9}{2^{10}.3^8+6^8.20}\)
=\(\frac{3}{5}\)
d)\(\frac{5^4.20^4}{25^5.4^5}\)
=\(\frac{\left(5.20\right)^4}{\left(25.4\right)^5}\)
=\(\frac{100^4}{100^5}\)
=\(\frac{1}{100}\)
Dài quá !!!
Giúp mk ,mk đag cần gấp