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a: \(=\dfrac{1}{2}\cdot\dfrac{2\cdot21+3\cdot27-5\cdot7}{189}\)
\(=\dfrac{1}{2}\cdot\dfrac{88}{189}=\dfrac{44}{189}\)
b: \(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}=\dfrac{1}{2}+\dfrac{4}{5}=\dfrac{13}{10}\)
c: \(=\dfrac{5}{21}-\dfrac{7}{14}\cdot\dfrac{36}{27}=\dfrac{5}{21}-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{5-14}{21}=\dfrac{-9}{21}=\dfrac{-3}{7}\)
\(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)=\frac{1}{2}\cdot\left(\frac{6}{27}-\frac{5}{27}+\frac{3}{7}\right)=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{1}{27}=\frac{3}{14}+\frac{1}{54}=\frac{44}{189}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}=\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
\(\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{36}{14}=\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{18}{7}=\frac{1}{3}\cdot\frac{5}{7}-\frac{6}{7}=\frac{5}{21}-\frac{14}{21}=\frac{-3}{7}\)
Bài 2:
a: 2/5=14/35
3/7=15/35
b: -3/4=-9/12
-7/-12=7/12
c: 5/9=60/108
-11/-12=11/12=99/108
d: -4/7=-36/63
8/9=56/63
-10/21=-30/63
Bài 2:
a: 2/5=14/35
3/7=15/35
b: -3/4=-9/12
-7/-12=7/12
c: 5/9=60/108
-11/-12=11/12=99/108
d: -4/7=-36/63
8/9=56/63
-10/21=-30/63
Bài 1.
\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)
\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)
\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)
\(=1200:\left[450:\left(450-300\right)\right]\)
\(=1200:\left(450:150\right)\)
\(=1200:3\)
\(=400\)
\(---\)
\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)
\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)
\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)
\(=675-20\left[90-\left(164-100\right)\right]\)
\(=675-20\left(90-64\right)\)
\(=675-20\cdot26\)
\(=675-520\)
\(=155\)
\(---\)
\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)
\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)
\(=\left(1\cdot2022-1986\right)\cdot5-169\)
\(=\left(2022-1986\right)\cdot5-169\)
\(=36\cdot5-169\)
\(=180-169\)
\(=11\)
Bài 2.
\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)
Vậy \(x=2\).
\(---\)
\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)
Vậy \(x=\dfrac{52}{5}\).
\(Toru\)
thôi chịu nhiều quá ai mà làm đc tự đi mà làm hỏi thì hỏi thì hỏi ít thôi người ta còn trả lời đc .
1275
nho lk nhé