Câu 1: D
Câu 2: D
Câu 3: A
Câu 4: A
Câu 5: B
Câu 6: A
Câu 7:
Câu 8: A
TỰ LUẬN:
Bài 1:
\(a,\dfrac{1}{2}-\dfrac{3}{4}+2022^0\)
\(=\dfrac{1}{2}-\dfrac{3}{4}+1\)
\(=\dfrac{2}{4}-\dfrac{3}{4}+\dfrac{4}{4}\)
\(=\dfrac{3}{4}\)
\(b,\dfrac{3}{4}+\left(-2\dfrac{1}{2}\right).\left(\dfrac{-3}{5}\right)\)
\(=\dfrac{3}{4}-\dfrac{5}{2}.\dfrac{-3}{5}\)
\(=\dfrac{3}{4}-\dfrac{1.\left(-3\right)}{2.1}\)
\(=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(=\dfrac{3}{4}+\dfrac{6}{4}\)
\(=\dfrac{9}{4}\)
\(c,\left(\dfrac{1}{5}\right)^2+0,2-2.\left(\dfrac{-1}{2}\right)^3-0,5\)
\(=\dfrac{1}{25}+\dfrac{1}{5}-2.\dfrac{-1}{8}-\dfrac{1}{2}\)
\(=\dfrac{1}{25}+\dfrac{1}{5}-\dfrac{-1}{4}-\dfrac{1}{2}\)
\(=\left(\dfrac{1}{25}+\dfrac{1}{5}\right)-\left(\dfrac{-1}{4}-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{5}{25}\right)+\left(\dfrac{1}{4}-\dfrac{2}{4}\right)\)
\(=\dfrac{6}{25}+\dfrac{-1}{4}\)
\(=\dfrac{24}{100}+\dfrac{-25}{100}\)
\(=\dfrac{-1}{100}\)
\(d,\dfrac{2^6.3^3}{6^3.8^2}=\dfrac{2^6.3^3}{\left(2.3\right)^3.\left(2^3\right)^2}=\dfrac{2^6.3^3}{2^3.3^3.2^6}=\dfrac{1}{2^3}=\dfrac{1}{8}\)
Bạn chia nhỏ câu hỏi để người ta dễ làm hơn ạ.
Câu 1 : \(D.\dfrac{-4}{6}\)
Câu 2: \(D.\dfrac{-1}{3};\dfrac{2}{3}\)
Câu 3:
\(\left(\dfrac{-2}{5}\right)^2+2^{-2}\)
\(=\dfrac{4}{25}+\dfrac{1}{4}\)
\(=\dfrac{16}{100}+\dfrac{25}{100}\)
\(=\dfrac{41}{100}\)
`= 41/100 = 0,41`
\(\Rightarrow A.0,41\)
Câu 4 :
\(\dfrac{1}{4}-\dfrac{5}{3}x=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{1}{4}-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{1}{12}\)
\(\Rightarrow x=\dfrac{1}{12}:\dfrac{5}{3}\)
\(\Rightarrow x=\dfrac{1}{12}\cdot\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{1}{4}\cdot\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{20}\)
Vậy `x = 1/20`
`=>``A . 1/20`
Bài 2:
\(a,\left(2x-10\right).\left(5x+3\right)=0\)
\(2x-10=0\) hoặc \(5x+3=0\)
\(2x=0+10\) hoặc \(5x=0-3\)
\(2x=10\) hoặc \(5x=-3\)
\(x=10:2\) hoặc \(x=\left(-3\right):5\)
\(x=5\) hoặc \(x=\dfrac{-3}{5}\)
Vậy \(x\in\left\{5;\dfrac{-3}{5}\right\}\)
\(b,2^x+2^{x+4}=544\)
\(2^x+2^x.2^4=544\)
\(2^x.\left(1+2^4\right)=544\)
\(2^x.17=544\)
\(2^x=544:17\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
Vậy \(x=5\)
\(c.3^x+3^{x+2}=10^2-\left(3^2+10^0\right)\)
\(3^x+3^x.3^2=100-\left(9+1\right)\)
\(3^x.\left(1+3^2\right)=100-10\)
\(3^x.10=90\)
\(3^x=90:10\)
\(3^x=9\)
\(3^x=3^2\)
\(x=2\)
Vậy \(x=2\)
\(d,2^x+2^{x+3}=144\)
\(2^x+2^x.2^3=144\)
\(2^x.\left(1+2^3\right)=144\)
\(2^x.9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\)
\(x=4\)
Vậy \(x=4\)
Bài 4:
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
\(3A=\dfrac{3}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+...+\dfrac{3}{3^{99}}\)
\(3A=1+\dfrac{1}{3^{ }}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)
\(2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{99}}\)
\(2A=1-\dfrac{1}{3^{99}}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}\)
Vì \(\dfrac{1}{3^{99}.2}>0\Rightarrow A=\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
