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câu 1 :
\(A=\frac{-7}{12}:\frac{49}{11}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:8-3:\frac{3}{4}\cdot-2^3\)
\(A=\frac{-11}{84}\cdot\frac{5}{121}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{8}{7}:\frac{8}{1}-\frac{3}{1}:\frac{3}{4}\cdot\left(-2^3\right)\)
\(A=\frac{-5}{924}-\frac{7}{6}\) \(B=\frac{1}{8}-\frac{1}{7}-\left(-32\right)\)
\(A=\frac{-361}{308}\) \(B=\frac{-1}{56}-\left(-32\right)\)
\(B=\frac{1791}{56}\)
Câu 2 :
a)\(\frac{22}{7}:x=\frac{11}{7}\) b)\(\left(1-3x\right)\cdot\frac{4}{3}=-2^3\)
\(x=\frac{22}{7}:\frac{11}{7}\) \(\left(1-3x\right)\cdot\frac{4}{3}=-8\)
\(x=2\) \(\left(1-3x\right)=-8:\frac{4}{3}\)
\(\left(1-3x\right)=-6\)
\(3x=-6-1=7\)
\(3x=7:3=\frac{7}{3}\)
c ) bằng \(\frac{27}{5}\)nhé
2. a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(3^3\right)^{25}=27^{25}\)
Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)
c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)
Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Bài làm :
a) Ta có :
\(\left(x-\frac{2}{3}\right)^2=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=\sqrt{\frac{5}{6}}\end{cases}}\)
\(\Leftrightarrow\frac{x=-\sqrt{\frac{5}{6}}+\frac{2}{3}}{x=\sqrt{\frac{5}{6}}+\frac{2}{3}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4-\sqrt{30}}{6}\\x=\frac{4+\sqrt{30}}{6}\end{cases}}\)
b) Ta có :
\(\left(\frac{3}{4}-x\right)^2\ge0\text{ với mọi x}\)
\(\Rightarrow\left(\frac{3}{4}-x\right)^2\ne-8\)
=> Không tồn tại x
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a.
\(x-\frac{2}{3}=\pm\sqrt{\frac{5}{6}}\)
\(\orbr{\begin{cases}x-\frac{2}{3}=\sqrt{\frac{5}{6}}\\x-\frac{2}{3}=-\sqrt{\frac{5}{6}}\end{cases}}\)
\(\orbr{\begin{cases}x=\sqrt{\frac{5}{6}}+\frac{2}{3}\\x=\frac{2}{3}-\sqrt{\frac{5}{6}}\end{cases}}\)
b.
Ta có : \(\left(\frac{3}{4}-x\right)^2\ge0\forall x\)
\(\Rightarrow\) không có x
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
a)38:3^4+2^2.2^3=3^8-4+2^2+3=3^4+2^5=81+32=113
b)3.4^2-2.3^2=3(4^2-2.3)=3(16-6)=3.10=30
c)4^6.3^4.9^5/6^12=(2^2)^6.3^4.(3^2)^5/(2.3)^12=2^12.3^4.3^10/2^12.3^12=2^12.3^14/2^12.3^12=3^14/3^12=3^2=9
e)45^3.20^4.18^2/180^5=(5.3^2)^3.(2^2.5)^4.(2.3^2)^2/(2^2.3^2.5)^5=5^3.3^6.2^8.5^4.2^2.3^4/2^10.3^10.5^5=5^9.3^10.2^10/2^10.3^10.5^5
=5^4/1=5^4=625
d)21^2.14.125/35^3.6=(2.7)^2.2.7.5^3/(5.7)^3.2.3=2^2.7^2.2.7.5^3/5^3.7^3.2.3=2^3.7^3.5^3/5^3.7^3.2.3=2^3/2.3=8/6=4/3
g)2^13+2^5/2^10+2^2=2^5(2^8+1)/2^2(2^8+1)=2^5/2^2=32/4=8
Chúc bạn học tốt !!!!
=36+\(\sqrt{17}\)
=40.12310563
k nhé