đây mà là toán 8 hả bn

1.

145+55=200

13+50+37=100

92+56-48=100

2.\(6x+x^2=-9\Leftrightarrow x\left(6+x\right)=-9\Leftrightarrow\hept{\begin{cases}x=-9\\6+x=-9\end{cases}\Leftrightarrow\hept{\begin{cases}x=-9\\x=-15\end{cases}}}\)

k mk nhé

1. 

\(145+55=200\)

\(13+50+37=63+37=100\)

\(92+56-48=148-48=100\)

2.

\(6x+x^2=-9\)

\(x^2+6x+9=0\)

\(\left(x+3\right)^2=0\)

\(\Leftrightarrow x+3=0\)

\(x=-3\)

1) \(\dfrac{1}{4}x-\dfrac{1}{3}=\dfrac{-5}{9}\)

\(\Rightarrow\dfrac{1}{4}x=-\dfrac{2}{9}\Rightarrow x=-\dfrac{8}{9}\)

2) \(2^{x-3}-3.2^x=-92\)

\(\Rightarrow2^x\left(2^{-3}-3\right)=-92\)

\(\Rightarrow2^x.\dfrac{-23}{9}=-92\)

\(\Rightarrow2^x=32\Rightarrow x=5\)

 (1/95+1/5).18/95:3=24/1805

a ,   x = 7 4 b ,   36 15 c ,   x = - 2   h o ặ c   x   =   3

c) x = - 2 hoặc x = 3

f) \(-\dfrac{1}{2}+\dfrac{11}{4}-\left(\dfrac{11}{4}-\dfrac{1}{2}\right)=-\dfrac{1}{2}+\dfrac{11}{4}-\dfrac{11}{4}+\dfrac{1}{2}=0\)

g) \(\left(\dfrac{92}{9}+\dfrac{13}{5}\right)-\dfrac{47}{9}=\left(\dfrac{92}{9}-\dfrac{47}{9}\right)+\dfrac{13}{5}=5+\dfrac{13}{5}=\dfrac{25}{5}+\dfrac{13}{5}=\dfrac{38}{5}\)

h) \(\left(\dfrac{44}{7}-\dfrac{32}{9}\right)-\left(\dfrac{37}{7}+\dfrac{4}{9}\right)=\dfrac{44}{7}-\dfrac{32}{9}-\dfrac{37}{7}-\dfrac{4}{9}=\left(\dfrac{44}{7}-\dfrac{37}{7}\right)-\left(\dfrac{32}{9}+\dfrac{4}{9}\right)=1-4=-3\)

Bài 5 :

S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399

S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)

S = 1 + 0 + 0 + ... + 0 + (- 401)

S = 1 - 401

S = - 400

Bài 5

A= 1+3-5-7+9+11-13-15+...-397-399

A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)

A= -8 -8 -...-8

A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)

A= -400

Bài 6: 

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: S={0;-3}

c) Ta có: 52-(x+21)=19-x

\(\Leftrightarrow52-x-21-19+x=0\)

\(\Leftrightarrow12=0\)(vô lý)

Vậy: \(x\in\varnothing\)

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...