2647 - 13590 : 45

= 2647 - 302

= 2345

22776 : 78 x 25

= 292 x 25

= 7300

= 6831

-\(\frac{955416}{89}\)

42731−(7502−356×4)

=42731−(7502−1424)

=42731−6078

=36653

Đáp án B

\(P\left(x\right)=\dfrac{4x^4+16x^3+56x^2+80x+356}{x^2+2x+5}\\ P\left(x\right)=\dfrac{4x^2\left(x^2+2x+5\right)+8x\left(x^2+2x+5\right)+20\left(x^2+2x+5\right)+256}{x^2+2x+5}\\ P\left(x\right)=4\left(x^2+2x+5\right)+\dfrac{256}{x^2+2x+5}\\ \ge2\sqrt{\dfrac{4\left(x^2+2x+5\right)\cdot256}{x^2+2x+5}}=2\sqrt{1024}=64\left(BĐTcosi\right)\)

Dấu \("="\Leftrightarrow4\left(x^2+2x+5\right)=\dfrac{256}{x^2+2x+5}\)

\(\Leftrightarrow x^2+2x+5=8\Leftrightarrow x^2+2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

P(x)=\(\dfrac{\text{(4x^2+8x^3+20x^2)+(8x^3+16x^2+40x)+(20x^2+40x+100)+256}}{x^2+2x+5}\)

      =(4x^2+8x+20x) +\(\dfrac{256}{x^2+2x+5}\)

áp dụng BĐT Cosi a+b≥\(2\sqrt{ab}\)

=>P(x)≥64

Dấu = xảy ra khi x=-1 hoặc x=3

Ta có m bằng 12. Thay số vào, ta có:

356 - 432 : 12

= 356 - 36

= 320

Vậy 356 - 432 : m với m bằng 12 là 320 (Đ)

41912

giúp mik vs

\(a)245^2+490\cdot54+54^2-199^2\\=(245^2+2\cdot245\cdot54+54^2)-199^2\\=(245+54)^2-199^2\\=299^2-199^2\\=(299-199)(299+199)\\=100\cdot498\\=49800\\---\\b)356^2-356\cdot246+123^2-133^2\\=(356^2-2\cdot356\cdot123+123^2)-133^2\\=(356-123)^2-133^2\\=233^2-133^2\\=(233-133)(233+133)\\=100\cdot366\\=36600\)

\(---\)

\(c)468^2-412^2-110\cdot412-55^2\\=468^2-(412^2+110\cdot412+55^2)\\=468^2-(412^2+2\cdot412\cdot55+55^2)\\=468^2-(412+55)^2\\=468^2-467^2\\=(468-467)(468+467)\\=1\cdot935\\=935\\---\)

\(d)615^2+250\cdot615+125^2-540^2\\=(615^2+2\cdot615\cdot125+125^2)-540^2\\=(615+125)^2-540^2\\=740^2-540^2\\=(740-540)(740+540)\\=200\cdot1280\\=256000\)

#\(Toru\)

a.

\(\frac{4^2\times4^3}{2^{10}}=\frac{4^2\times4^3}{2^{2\times5}}=\frac{4^2\times4^3}{\left(2^2\right)^5}=\frac{4^5}{4^5}=1\)

b.

\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\times3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5\times3^5}{\left(0,2\right)^6}=\frac{243}{0,2}=1215\)

c.

\(\frac{2^7\times9^3}{6^5\times8^2}=\frac{2^7\times\left(3^2\right)^3}{\left(2\times3\right)^5\times\left(2^3\right)^2}=\frac{2^7\times3^6}{2^5\times3^5\times2^6}=\frac{3}{2^4}=\frac{3}{16}\)

d.

\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{\left(2\times3\right)^3+3\times\left(3\times2\right)^2+3^3}{-13}=\frac{2^3\times3^3+3\times3^2\times2^2+3^3}{-13}=\frac{8\times3^3+3^3\times4+3^3}{-13}\)\(=\frac{3^3\times\left(8+4+1\right)}{-13}=\frac{27\times13}{-13}=-27\)

a/  \(\frac{4^2.4^3}{2^{10}}=\frac{4^5}{2^{10}}=\frac{\left(2^5\right)^2}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)

b/\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,6\right)^5}{\left(0,2\right)^5.\left(0,2\right)}=\left(\frac{0,6}{0,2}\right)^5.\frac{1}{\frac{1}{5}}=3^5.5=243.5=1215\)

c/ \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3.3^5}{2^7.2^4.3^5}=\frac{3}{16}\)

d/ \(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(2.3\right)^3+3.\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{27.13}{-13}=-27\)