\(A=16,2\times3,7+5,7\times16,2-7,8\times4,8-4,6\times7,8\\ =16,2\times\left(3,7+5,7\right)-7,8\times\left(4,8+4,6\right)\\ =16,2\times9,4-7,8\times9,4\\ =9,4\times\left(16,2-7,8\right)\\ =9,4\times8,4\\ =78,96\)

\(B=11,2+12,3+13,4-12,6-11,5-10,4\\ =11+0,2+12+0,3+13+0,4-12-0,6-11-0,5-10-0,4\\ =\left(11+12+13-12-11-10\right)+\left(0,2+0,3+0,4-0,5-0,6-0,4\right)\\ =3-0,6\\ =2,4\)

\(\left\{{}\begin{matrix}2\left(m+1\right)x-7\left(n-2\right)y=6\\\left(m+1\right)x+\left(n-2\right)y=12\end{matrix}\right.\left(m\ne-1;n\ne2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(m+1\right)x-7\left(n-2\right)y=6\\2\left(m+1\right)x+2\left(n-2\right)y=24\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}9\left(n-2\right)y=-18\\\left(m+1\right)x+\left(n-2\right)y=12\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-18}{9\left(n-2\right)}=\dfrac{-2}{n-2}\\\left(m+1\right)x+\left(n-2\right)\cdot\dfrac{-2}{n-2}=24\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{n-2}\\\left(m+1\right)x-2=24\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-2}{n-2}\\\left(m+1\right)x=26\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-2}{n-2}\\x=\dfrac{26}{m+1}\end{matrix}\right.\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\ \Rightarrow\dfrac{x+y}{xy}=\dfrac{1}{6}\\ \Rightarrow xy=6\left(x+y\right)\\ \Rightarrow xy-6x-6y=0\\ \Rightarrow x\left(y-6\right)-6\left(y-6\right)-36=0\\ \Rightarrow\left(y-6\right)\left(x-6\right)=36\)

Ta có bảng: 

Mà x,y nguyên dương nên (bạn tự chọn lại nhé) 

`1/x + 1/y = 1/6`

`<=> (x+y)/(xy) = 1/6`

`<=> xy = 6x + 6y`

`<=> xy - 6x - 6y = 0`

`<=> x(y-6) - 6(y-6) = 36`

`<=> (x-6)(y-6) = 36`

Do `x-6, y-6 in ZZ` nên `(x-6) in Ư(36)`.

Đến đây bạn tự chia trường hợp và làm nhé.

a) Ta có: +) \(\widehat{zOt}+\widehat{xOz}=\widehat{xOt}\) (hai góc kề nhau)

Mà \(\widehat{xOz}=50^o;\widehat{xOt}=110^o\) (gt) nên:

\(\widehat{zOt}+50^o=110^o\)

\(\widehat{zOt}=110^o-50^o=60^o\)

+) \(\widehat{yOt}+\widehat{xOy}=\widehat{xOt}\) (hai góc kề nhau)

Mà \(\widehat{xOy}=30^o;\widehat{xOt}=110^o\) (gt) nên;

\(\widehat{zOt}+30^o=110^o\)

\(\widehat{zOt}=110^o-30^o=80^o\)

Vậy....

b) Ta có: 

+) Ba tia \(Oy;Oz;Ot\) cùng nằm trên một nửa mặt phẳng có bờ \(Ox\)

+) \(\widehat{xOy}=30^o< \widehat{xOz}=50^o< \widehat{xOt}=110^o\)

Do đó: Tia \(Oz\) nằm giữa hai tia \(Oy\) và \(Ot\)

Vậy...

Mọi người giải giúp mình với

\(6x^4+7x^3-37x^2-8x+12\\ =6x^4-3x^3+10x^3-5x^2-32x^2+16x-24x+12\\ =3x^3\left(2x-1\right)+5x^2\left(2x-1\right)-16x\left(2x-1\right)-12\left(2x-1\right)\\ =\left(2x-1\right)\left(3x^3+5x^2-16x-12\right)\\ =\left(2x-1\right)\left(3x^3-6x^2+11x^2-22x+6x-12\right)\\ =\left(2x-1\right)\left[3x^2\left(x-2\right)+11x\left(x-2\right)+6\left(x-2\right)\right]\\ =\left(2x-1\right)\left(x-2\right)\left(3x^2+11x+6\right)\\ =\left(2x-1\right)\left(x-2\right)\left(3x^2+9x+2x+6\right)\\ =\left(2x-1\right)\left(x-2\right)\left[3x\left(x+3\right)+2\left(x+3\right)\right]\\ =\left(2x-1\right)\left(x-2\right)\left(x+3\right)\left(3x+2\right)\)

Chón ý B 5/8. Mình trả lời nhanh nhất nè bạn. Tick cho mình đi!

Bạn xem lại các đáp án nhé, mình tính không ra bạn ạ!

 a : b = 9 (dư 8)

a = b x 9 + 8

a - b = 88

b x 9 + 8 - b = 88

b x 8 = 80

b = 80 : 8

b = 10

⇒ a = b x 9 + 8 = 10 x 9 + 8 = 98

Vậy số bị chia là: 98

       số chia là: 10

Gọi số bị chia và số chia lần lượt là a và b 

Ta có: 

\(a=b\times9+8\) 

\(a-b-8=b\times8\)

Mà hiệu số bị chia và số chia là 88 nên:

\(88-8=b\times8\)

\(b\times8=80\)

\(b=10\)

hay số chia là \(10\); số bị chia là \(88+10=98\) 

Đáp số:...

\(a.\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{8}{9}\right)^2\\ =\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{\left(2\sqrt{2}\right)^2}{3^2}\right)^2\\ =\left(\dfrac{3}{2}\right)^4\cdot\left(\dfrac{2\sqrt{2}}{3}\right)^4\\ =\left(\dfrac{3}{2}\cdot\dfrac{2\sqrt{2}}{3}\right)^4\\ =\left(\sqrt{2}\right)^4\\ =4\\ b.\left(\dfrac{-3}{5}\right)^6\cdot\left(\dfrac{-5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot\left(-\dfrac{3}{5}\right)^5\cdot\left(\dfrac{-5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot\left(-\dfrac{3}{5}\cdot-\dfrac{5}{3}\right)^5\\ =\left(-\dfrac{3}{5}\right)\cdot1\\ =-\dfrac{3}{5}\\ c.\left(\dfrac{4}{7}\right)^3\cdot\left(\dfrac{4}{7}\right)^5\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)^8\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)\cdot\left(\dfrac{4}{7}\right)^7\cdot\left(\dfrac{7}{4}\right)^7\\ =\left(\dfrac{4}{7}\right)\cdot\left(\dfrac{4}{7}\cdot\dfrac{7}{4}\right)^7\\ =\dfrac{4}{7}\)

a: \(\left(\dfrac{3}{4}\right)^4\cdot\left(\dfrac{8}{9}\right)^2=\dfrac{3^4}{4^4}\cdot\dfrac{8^2}{9^2}\)

\(=\dfrac{3^4}{3^4}\cdot\dfrac{2^6}{2^8}=\dfrac{1}{2^2}=\dfrac{1}{4}\)

b: \(\left(-\dfrac{3}{5}\right)^6\cdot\left(-\dfrac{5}{3}\right)^5\)

\(=\left(-\dfrac{3}{5}\right)^5\cdot\left(-\dfrac{5}{3}\right)^5\cdot\dfrac{-3}{5}=\left(-\dfrac{3}{5}\cdot\dfrac{-5}{3}\right)^5\cdot\dfrac{-3}{5}\)

\(=1^5\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\)

c: \(\left(\dfrac{4}{7}\right)^3\cdot\left(\dfrac{4}{7}\right)^5\cdot\left(\dfrac{7}{4}\right)^7=\left(\dfrac{4}{7}\right)^8\cdot\left(\dfrac{7}{4}\right)^7\)

\(=\left(\dfrac{4}{7}\cdot\dfrac{7}{4}\right)^7\cdot\dfrac{4}{7}=1^7\cdot\dfrac{4}{7}=\dfrac{4}{7}\)

d: \(\dfrac{8^{14}}{4^4\cdot64^5}=\dfrac{2^{42}}{2^8\cdot2^{30}}=2^4=16\)

e: \(\dfrac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\dfrac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\dfrac{3^{41}}{3^{43}}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

\(x^4+x^3+2x^2+x+1\\= x^4+x^3+x^2+x^2+x+1\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+1\right)\left(x^2+x+1\right)\)