`7xx8xx12/6xx21xx16`

`=7xx8xx2xxx21xx16`

`=56xx2xx21xx16`

`=112xx21xx16`

`=2352xx16`

`=37632`

\(\dfrac{7\times8\times12}{6\times21\times16}=\dfrac{7\times8\times3\times2\times2}{2\times3\times3\times7\times8\times2}=\dfrac{1}{3}\)

Cảm ơn bạn nha không là mình chết mất

các bạn ơi giúp mình với

\(\dfrac{5}{6}-\dfrac{5}{8}+\dfrac{7}{12}=\dfrac{10}{48}+\dfrac{7}{12}=\dfrac{38}{48}\)

5/6-5/8+7/12=19/24

12/24+6/18=5/6

11/72

13/10

k mik

Mọi người ơi, giúp em với ạ

a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)

=\(9^4.13:9^3=13.9=117\)

b) 100-(75-25)=100-50=50

\(a,\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(< =>\left(8\cdot6561+729\cdot45\right):\left(81\cdot10-81\right)\)

\(< =>\left(52488+32805\right):\left(810-81\right)\)

\(< =>85293:729\)

\(< =>117\)

\(b,100-\left[75-\left(7-2\right)^2\right]\)

\(< =>100-\left[75-\left(5\right)^2\right]\)

\(< =>100-\left[75-25\right]\)

\(< =>100-50\)

\(< =>50\)

\(c,\left(20\cdot2^4+12\cdot2^4\right)-\left(48\cdot2^2\right):8^2\)

\(< =>\left[2^4\cdot\left(20+12\right)\right]-\cdot\left(48\cdot4\right):8^2\)

\(< =>\left[2^4\cdot32\right]-192:8^2\)

\(< =>\left[16\cdot32\right]-192:64\)

\(< =>512-3\)

\(< =>509\)

Ta có: \(A=1-2+3-4+5-6+7-8+9\)

\(=(1+9)-(2+8)+(3+7)-(4+6)+5\)

\(=10-10+10-10+5\)

\(=5\)

Vậy  \(A=5\) 

B = 12 - 14 + 16 - 18 + ... + 2008 - 2010

B = -2 + (-2)+ (-2)+ (-2) + ...+ (-2)

B = -2 . 100

B = -200

Bạn muốn kết bạn chứ

a \(\Leftrightarrow3x=12\Leftrightarrow x=4\)

b \(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c \(ĐKXĐ:x\ne2;x\ne-2\)

\(\Rightarrow\left(x+2\right)^2-6\left(x-2\right)=x^2\Leftrightarrow x^2+4x+4-6x+12=x^2\Leftrightarrow-2x+16=0\Leftrightarrow-2x=-16\Leftrightarrow x=8\left(TM\right)\)

a) Ta có: 3x-12=0

\(\Leftrightarrow3x=12\)

hay x=4

Vậy: S={4}

b) Ta có: (x-2)(2x+3)=0

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

Vậy: \(S=\left\{2;\dfrac{-3}{2}\right\}\)

mk ko viết lại đề

\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)

\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)

\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

Vậy A= \(\frac{1}{2}\)

\(=\frac{5}{6}x\left(\frac{7}{12}+\frac{5}{12}\right)+\frac{1}{6}=\frac{5}{6}x1+\frac{1}{6}=1\)

a) \(\frac{5}{6}x\frac{7}{12}+\frac{5}{12}x\frac{5}{6}+\frac{1}{6}=\frac{5}{6}x\left(\frac{7}{12}+\frac{5}{12}\right)+\frac{1}{6}=\frac{5}{6}x1+\frac{1}{6}=\frac{5}{6}+\frac{1}{6}=1\)

b) \(\left(\frac{8}{15}x\frac{3}{4}x\frac{5}{4}\right):\frac{7}{8}=\frac{4x2x3x5}{5x3x4x4}x\frac{8}{7}=\frac{2}{4}x\frac{8}{7}=\frac{2x8}{4x7}=\frac{2x4x2}{4x7}=\frac{4}{7}\)

Chúc bạn học tốt.

😁😁😁

=1 thôi nha