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b) | x - 5 | = | -7 |
=> | x - 5 | = 7
\(\Rightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)
d) ( x : 3 - 4 ) . 5 = 15
=> x : 3 - 4 = 15 : 5 = 3
=> x : 3 = 3 + 4 = 7
=> x = 7 . 3 = 21
a/ x-7=-5
x=-5+7
x=2
câu này ai ko làm được ko phải người ^-^
i) \(2345-1000\div\left[19-2\left(21-18\right)^2\right]\)
\(=\)\(2345-1000\div\left[19-2.3^2\right]\)
\(=\)\(2345-1000\div\left[19-2.9\right]\)
\(=\)\(2345-1000\div\left[19-18\right]\)
\(=\)\(2345-1000\div1\)
\(=\)\(2345-1000\)
\(=\)\(1345\)
j) \(128-\left[68+8\left(37-35\right)^2\right]\div4\)
\(=\)\(128-\left[68+8.2^2\right]\div4\)
\(=\)\(128-\left[68+8.4\right]\div4\)
\(=\)\(128-\left[68+32\right]\div4\)
\(=\)\(128-100\div4\)
\(=\)\(128-25\)
\(=\)\(3\)
k) \(568-\left\{5\left[143-\left(4-1\right)^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-3^2\right]+10\right\}\div10\)
\(=\)\(568-\left\{5\left[143-9\right]+10\right\}\div10\)
\(=\)\(568-\left\{5.134+10\right\}\div10\)
\(=\)\(568-\left\{670+10\right\}\div10\)
\(=\)\(568-680\div10\)
\(=\)\(568-68\)
\(=\)\(500\)
a) \(107-\left\{38+\left[7.3^2-24\div6+\left(9-7\right)^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.3^2-24\div6+2^3\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[7.9-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+\left[63-4+8\right]\right\}\div15\)
\(=\)\(107-\left\{38+67\right\}\div15\)
\(=\)\(107-105\div15\)
\(=\)\(107-7\)
\(=\)\(7\)
b) \(307-\left[\left(180-160\right)\div2^2+9\right]\div2\)
\(=\)\(307-\left[20\div4+9\right]\div2\)
\(=\)\(307-\left[5+9\right]\div2\)
\(=\)\(307-14\div2\)
\(=\)\(307-7\)
\(=\)\(300\)
c) \(205-\left[1200-\left(4^2-2.3\right)^3\right]\div40\)
\(=\)\(205-\left[1200-\left(16-6\right)^3\right]\div40\)
\(=\)\(205-\left[1200-10^3\right]\div40\)
\(=\)\(205-\left[1200-1000\right]\div40\)
\(=\)\(205-200\div40\)
\(=\)\(205-5\)
\(=\)\(200\)
a=-3/4; b=4/7 =>ab=-3/7
\(a=\dfrac{5}{9};b=-\dfrac{18}{15}=-\dfrac{6}{5}\Leftrightarrow ab=-\dfrac{2}{3}\)
\(a=-\dfrac{7}{25};b=\dfrac{50}{21}\Leftrightarrow ab=-\dfrac{2}{3}\)
\(ab=1;b=-\dfrac{3}{7}\Leftrightarrow a=-\dfrac{7}{3}\)
\(a=\dfrac{4}{7};b=-\dfrac{3}{4}\Leftrightarrow ab=-\dfrac{3}{7}\)
\(a=\dfrac{-4}{19};ab=-\dfrac{4}{19}\Leftrightarrow b=1\)
\(a=-\dfrac{18}{15}=-\dfrac{6}{5};c=\dfrac{5}{9}\Leftrightarrow ab=-\dfrac{2}{3}\)
\(ab=0;b=\dfrac{6}{13}\Leftrightarrow a=0\)
là A okk
\(\left(7^{2018}+7^{2017}\right):7^{2017}\)
\(=7^{2018}:7^{2017}+7^{2017}:7^{2017}\)
\(=7+1\)
\(=8\)
Vậy đáp án là : B.8