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\(\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14}\)
\(=\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{18}{7}\)
\(=\frac{1.5}{3.7}-\frac{7.18}{27.7}\)
\(=\frac{5}{21}-\frac{2}{3}\)
\(=\frac{5}{21}-\frac{14}{21}=\frac{-9}{21}=\frac{-3}{7}\)
Chúc bn học tốt
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: \(=\dfrac{5}{21}-\dfrac{7}{14}\cdot\dfrac{36}{27}=\dfrac{5}{21}-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{5}{21}-\dfrac{2}{3}=\dfrac{5-14}{21}=\dfrac{-9}{21}=\dfrac{-3}{7}\)
\(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)=\frac{1}{2}\cdot\left(\frac{6}{27}-\frac{5}{27}+\frac{3}{7}\right)=\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{2}\cdot\frac{1}{27}=\frac{3}{14}+\frac{1}{54}=\frac{44}{189}\)
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)=\frac{1}{2}+\frac{3}{4}-\frac{3}{4}+\frac{4}{5}=\frac{1}{2}+\frac{4}{5}=\frac{13}{10}\)
\(\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{36}{14}=\frac{1}{3}\cdot\frac{5}{7}-\frac{7}{27}\cdot\frac{18}{7}=\frac{1}{3}\cdot\frac{5}{7}-\frac{6}{7}=\frac{5}{21}-\frac{14}{21}=\frac{-3}{7}\)
a) \(227+50+23=\left(227+23\right)+50=250+50=300\)
b) \(135+360+65+40=\left(135+65\right)+\left(360+40\right)=200+400=600\)
c) \(1+2+3+4+5+...+97+98+99+100\)
\(=\left(100+1\right)+\left(99+2\right)+...+\left(50+51\right)\)
\(=101+101+101+...+101\)
\(=101\cdot50\)
\(\Leftrightarrow5050\)
d) \(115\cdot13-13\cdot15=13\cdot\left(115-15\right)=13\cdot100=1300\)
e) \(50-49+48-47+...+4-3+2-1\)
\(=\left(50-49\right)+\left(48-47\right)+...+\left(2-1\right)\)
\(=1+1+1+1+..+1\)
\(=1\cdot25\)
\(=25\)
f) \(30\cdot40\cdot50\cdot60=10\cdot3+10\cdot4+10\cdot5+10\cdot6\)
\(=10\cdot10\cdot10\cdot10\cdot3\cdot4\cdot5\cdot6\)
\(=10000\cdot360\)
\(=3600000\)
g) \(27\cdot36+27\cdot64=27\cdot\left(36+64\right)=27\cdot100=2700\)
h) \(5\cdot2^2-18:3=5\cdot4-18:3=20-6=14\)
i) \(13\cdot17-256:16+14:7-2021^0\)
\(=13\cdot17-4^4:4^2+2-1\)
\(=13\cdot17-16+2-1\)
\(=13\cdot17-17\)
\(=17\cdot\left(13-1\right)\)
\(=204\)
j) \(7^2-36:3=49-12=37\)
\(160:\left\{17+\left[3^3\cdot5-\left(14+2^7:2^4\right)\right]\right\}\)
\(=160:\left\{17+\left[27\cdot5-\left(14+2^3\right)\right]\right\}\)
\(=160:\left\{17+\left[135-\left(14+8\right)\right]\right\}\)
\(=160:\left[17+\left(135-22\right)\right]\)
\(=160:\left(17+113\right)\)
\(=160:130\)
\(=\dfrac{16}{13}\)
#Urushi☕
160 : { 17 + [ 3 3 ⋅ 5 − ( 14 + 2 7 : 2 4 ) ] } = 160 : { 17 + [ 27 ⋅ 5 − ( 14 + 2 3 ) ] } = 160 : { 17 + [ 135 − ( 14 + 8 ) ] } = 160 : [ 17 + ( 135 − 22 ) ] = 160 : ( 17 + 113 ) = 160 : 130 = 16 13
(2/3.5/7 + 2/3: 7/2). -1/3
=(10/21 + 2/3x2/7 ) . -1/3
=(10/21 + 4/21) . -1/3
=-10/63 - 4/63
=-14/63
(2/3.5/7+2/3:7/2)-1/3
=(2/3.5/7+2/3x2/7)-1/3
=[2/3.(5/7+2/7)]-1/3
=[2/3x1]-1/3
=2/3-1/3
=1/3
HT
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 \(x=\dfrac{52}{5}\).
\(Toru\)
bằng -3/14
Nhớ k cho mình nha :3
cho mjk hoi ca cach lam ban a