a. 6 và 5/3 + 7 và 13/8 +3 và 1/3- 4 và 5/8

=23/3+69/8+4/3-37/8

=391/24+4/3+37/8=141/8-37/8=103/8

b.=(27/5+69/11)-(17/5-52/11)

=642/55+73/55=13

lm dùm tui bài này cái

\(\frac{5}{3}+\frac{5}{6}-x=\frac{2}{9}\)

\(\frac{30}{18}+\frac{15}{18}-x=\frac{4}{18}\)

\(x=\frac{30}{18}+\frac{15}{18}-\frac{4}{18}\)

\(x=\frac{41}{18}\)

\(\frac{5}{3}+\frac{5}{6}-x=\frac{2}{9}\)

\(\frac{30}{18}+\frac{15}{18}-x=\frac{2}{9}\)

\(\frac{45}{18}-x=\frac{2}{9}\)

             \(x=\frac{45}{18}-\frac{2}{9}\)

             \(x=\frac{45}{18}-\frac{4}{18}\)

              \(x=\frac{41}{18}\)

Vậy \(x=\frac{41}{18}\)

tiểu học mà bảo toán lớp 8

bạn đã suy nghĩ chưa

cvfbhm,

Xin lỗi em ko biết làm , em vẫn chưa lên lớp 9

\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)

\(=\left(5x^4+4x^4\right)+\left(-3x^3+3x^3\right)+\left(x^2+x^2\right)-x+5\)

\(=9x^4+2x^2-x+5\)

\(Q\left(x\right)=x-5x_{}^3-x^2-x^4+4x^3-x^2+3x-1\)

\(=\left(-5x^4-x^4\right)+4x^3+\left(-x^2-x^2\right)+\left(x+3x\right)-1\)

\(=-6x^4+4x^3-2x^2+4x-1\)

P(x)+Q(x)

\(=9x^4+2x^2-x+5-6x^4+4x^3-2x^2+4x-1\)

\(=3x^4+4x^3+3x+4\)

Đặt P(x)+Q(x)=0

=>\(3x^4+4x^3+3x+4=0\)

=>\(x^3\left(3x+4\right)+\left(3x+4\right)=0\)

=>\(\left(3x+4\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}3x+4=0\\ x^3+1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=-\frac43\\ x=-1\end{array}\right.\)

a: 3x-2=2x-3

=>x=-1

b: 2x+3=5x+9

=>-3x=6

=>x=-2

c: 5-2x=7

=>2x=-2

=>x=-2

d: 10x+3-5x=4x+12

=>5x+3=4x+12

=>x=9

e: 11x+42-2x=100-9x-22

=>9x+42=78-9x

=>18x=36

=>x=2

f: 2x-(3-5x)=4(x+3)

=>2x-3+5x=4x+12

=>7x-3=4x+12

=>3x=15

=>x=5

\(a=4^5.9^4-2.\dfrac{6^9}{2^{10}}.3^8+6^8.20\)

Đề là như vầy đúng ko bn?

$\textbf{A)}$

$\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}$

$=\dfrac{(2^2)^5(3^2)^4-2(2\cdot3)^9}{2^{10}\cdot3^8+2^2\cdot5(2\cdot3)^8}$

$=\dfrac{2^{10}3^8-2^{10}3^9}{2^{10}3^8+2^{10}\cdot5\cdot3^8}$

$=\dfrac{2^{10}3^8(1-3)}{2^{10}3^8(1+5)}$

$=\dfrac{-2}{6}$

$=-\dfrac{1}{3}.$