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 \(x=\dfrac{52}{5}\).

\(Toru\)

-15

1275

nho lk nhé

a) 8.(-5).(-4).2 = 8.20.2 = 8.40 = 320

b) \(1\frac{3}{7}+\left(-\frac{1}{3}+2\frac{4}{7}\right)\)

\(=1\frac{3}{7}-\frac{1}{3}+2\frac{4}{7}\)

\(=\left(1\frac{3}{7}+2\frac{4}{7}\right)-\frac{1}{3}=\left(1+2\right)+\left(\frac{3}{7}+\frac{4}{7}\right)-\frac{1}{3}=4-\frac{1}{3}=\frac{11}{3}\)

c) \(\frac{8}{5}\cdot\frac{-2}{3}+\frac{-5\cdot5}{3\cdot5}\)

\(=\frac{8}{5}\cdot\frac{-2}{3}+\frac{-25}{15}=\frac{-16}{15}+\frac{-25}{15}=\frac{-41}{15}\)

d) \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)

\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{13}{56}\)

=[\(\left(\frac{2}{3}.\frac{5}{7}+\frac{2}{3}.\frac{2}{7}\right).\left(\frac{-1}{3}\right)\)]

=[\(\left(\frac{2}{3}\right).\left(\frac{5}{7}+\frac{2}{7}\right).\left(\frac{-1}{3}\right)\)]

=[\(\left(\frac{2}{3}.1\right).\frac{-1}{3}\)]

=\(\frac{-1}{3}\)

\(a,\frac{3}{4}+\frac{2}{7}+\frac{1}{4}+\frac{5}{7}+\frac{3}{5}\)

\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{3}{5}\)

\(=1+1+\frac{3}{5}\)

\(=2+\frac{3}{5}\)

\(=\frac{13}{5}\)

\(b,\frac{7}{3}.\frac{5}{7}.\frac{21}{5}=\frac{7.5.3.7}{3.7.5}=7\)

học tốt

a,\(\frac{3}{4}+\frac{2}{7}+\frac{1}{4}+\frac{5}{7}+\frac{3}{5}\)

\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{3}{5}\)

\(=1+1+\frac{3}{5}\)

\(=\frac{13}{5}\)

b,\(\frac{7}{3}\times\frac{5}{7}\times\frac{21}{5}=\frac{7\times5\times21}{3\times7\times5}=\frac{735}{105}=7\)

\(\dfrac{-2}{3}\cdot\dfrac{5}{7}+\dfrac{5}{7}:\dfrac{3}{4}-1=\dfrac{-2}{3}\cdot\dfrac{5}{7}+\dfrac{5}{7}\cdot\dfrac{4}{3}-1=\left(\dfrac{-2}{3}+\dfrac{4}{3}\right)\cdot\dfrac{5}{7}-1=\dfrac{2}{3}\cdot\dfrac{5}{7}-1=\dfrac{10}{21}-1=-\dfrac{11}{21}\)

−2 / 3⋅5 / 7+5 / 7:3 / 4−1=−2 / 3⋅5 / 7+5 / 7⋅4 / 3−1=(−2 / 3+4 / 3)⋅5 /. 7−1=2 / 3⋅5 / 7−1=10 / 21−1=−11 / 21

nha