a) \(A=\dfrac{2^6\cdot9^2}{6^4\cdot8}\)

\(=\dfrac{2^6\cdot\left(3^2\right)^2}{3^4\cdot2^4\cdot2^3}\)

\(=\dfrac{2^6\cdot3^4}{3^4\cdot2^7}\)

\(=\dfrac{1}{2}\)

b) \(B=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot\left(3^2\right)^2}\)

\(=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}\)

\(=\dfrac{3}{2^2}\)

\(=\dfrac{3}{4}\)

\(3.2+\left(x+5^2\right)=10^2\)

\(6+\left(x+25\right)=100\)

\(x+25=100-6=94\)

\(x=94-25=69\)

Bạn xem lại đề

a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2ab+b^2\)

\(=a^2+2ab+b^2-4ab\)

\(=\left(a+b\right)^2-4ab=VP\left(dpcm\right)\)

b) Ta có:

\(VT=\left(x+y\right)^2+\left(x-y\right)^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2\)

\(=\left(x^2+y^2\right)+\left(x^2+y^2\right)\)

\(=2\left(x^2+y^2\right)=VP\left(dpcm\right)\)

a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)

\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)

\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)

\(=\left(10-4+6\right)\sqrt{2}\)

\(=12\sqrt{2}\)

b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)

\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)

\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)

\(=\left(8-15+15-3\right)\sqrt{5}\)

\(=5\sqrt{5}\)

c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)

\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)

\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)

\(\left(-x\right)^8:x^2\)

\(=x^8:x^2\)

\(=x^{8-2}\)

\(=x^6\)

_____

\(27^8:9^4\)

\(=\left(3^3\right)^8:\left(3^2\right)^4\)

\(=3^{24}:3^8\)

\(=3^{24-8}\)

\(=3^{16}\)

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(3-2x\right)\left(4x^2+6x+9\right)\)

\(M=\left(x^3+3^3\right)-\left[3^3-\left(2x\right)^3\right]\)

\(M=x^3+27-27+8x^3\)

\(M=9x^3\)

Thay x=20 vào M ta có:
\(M=9\cdot20^3=72000\)

Vậy: ...

\(N=\left(x-2y\right)\left(x^2+2xy+4y^2\right)+16y^3\)

\(N=x^3-\left(2y\right)^3+16y^3\)

\(N=x^3-8y^3+16y^3\)

\(N=x^3+8y^3\)

\(N=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

Thay \(x+2y=0\) vào N ta có:

\(N=0\cdot\left(x^2-2xy+4y^2\right)=0\)

Vậy: ...

a)\(x+\dfrac{2}{3}x\dfrac{5}{6}=7\)

\(x+\dfrac{10}{18}=7\)

\(x=7-\dfrac{10}{18}\)

\(x=\dfrac{58}{9}\)

b)\(9-xx\dfrac{1}{2}=\dfrac{3}{4}\)

\(xx\dfrac{1}{2}=9-\dfrac{3}{4}\)

\(xx\dfrac{1}{2}=\dfrac{33}{4}\)

\(x=\dfrac{33}{4}:\dfrac{1}{2}\)

\(x=\dfrac{33}{2}\)

c)\(18-x:\dfrac{2}{3}=\dfrac{4}{5}-\dfrac{1}{10}\)

\(18-x:\dfrac{2}{3}=\dfrac{7}{10}\)

\(x:\dfrac{2}{3}=18-\dfrac{7}{10}\)

\(x:\dfrac{2}{3}=\dfrac{173}{10}\)

\(x=\dfrac{173}{10}x\dfrac{2}{3}\)

\(x=\dfrac{173}{15}\)

\(\left(3x+1\right)^2-\left(3x-1\right)^2\)

\(=\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\)

\(=2\cdot6x\)

\(=12x\)

_________

\(\left(x+y\right)^2-\left(x-y\right)^2\)

\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)

\(=2x\cdot2y\)

\(=4xy\)

\(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=\left(x+y+x-y\right)\left[\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2x\cdot\left(x^2+2xy+y^2-x^2+y^2+x^2-2xy+y^2\right)\)

\(=2x\cdot\left(x^2+3y^2\right)\)

______

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x-y\right)+z^3+3xyz\)
\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x-y-z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3z\left(x+y\right)-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2-xy-xz-yz\right)\)