Lời giải:
a. 

$\frac{84}{108}+\frac{231}{189}=\frac{7}{9}+\frac{11}{9}=\frac{7+11}{9}=\frac{18}{9}=2$
b.

$\frac{234}{180}-\frac{272}{240}=\frac{13}{10}-\frac{17}{15}=\frac{39}{30}-\frac{34}{30}=\frac{39-34}{30}=\frac{5}{30}=\frac{1}{6}$

a.37,8

b.2283,625

e.43,125

g.5892,5

h.1432

Bài 5: Tính bằng cách thuận tiện nhất

a) 189 x 6 + 189 x 5 - 189

=189x(6+5-1)

=189x10

=1890

b) 25 x 86 x 4

=25x4x86

=100x86

=8600

HT

Ừ BẠN LÀM ĐÚNG RỒI

\(\left(35+x\right)-12=27\\ 35+x=27+12=39\\ x=39-35=4\\ ---\\ 2x-5=3^3:3^2=3\\ 2x=3+5=8\\ x=\dfrac{8}{2}=4\)

1) (35 + x) -12 =27

=>35 + x = 39 => x = 4

2) 2x -5 = 3³ : 3²

=> 2x -5 =3¹ => 2x = 8

=> x = 4

189-2.(100-x) =\(5^2\)

189-2.(100-x) =25

2.(100-x) =189 - 25

2.(100-x) =164

100-x=164:2

100-x=82

x=100-82

x=18

Trả lời :....................

x=18.....................

Hk tốt..........................

Thay cái [ ] thành || mới đúng chứ nhỉ :)

\(A=-\left|x+\frac{5}{937}\right|+\frac{144}{272}\)

\(\Leftrightarrow A=\frac{144}{272}-\left|x+\frac{5}{937}\right|\le\frac{144}{272}\forall x\)

Dấu '' = '' xảy ra \(\Leftrightarrow x+\frac{5}{937}=0\Leftrightarrow x=\frac{-5}{937}\)

Vậy \(maxA=\frac{144}{272}\Leftrightarrow x=\frac{-5}{937}\)

a) \(\left(x+1\right)\left(x+2\right)=272\)

\(\Rightarrow x^2+3x+2=272\)

\(\Rightarrow x^2+3x-270=0\)

\(\Rightarrow x^2+18x-15x-270=0\)

\(\Rightarrow x\left(x+18\right)-15\left(x+18\right)=0\)

\(\Rightarrow\left(x+18\right)\left(x-15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+18=0\\x-15=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-18\\x=15\end{matrix}\right.\)

d) \(\left(x+4\right)\left(x+5\right)=552\)

\(\Rightarrow x^2+9x+20=552\)

\(\Rightarrow x^2+9x-532=0\)

\(\Rightarrow x^2+28x-19x-532=0\)

\(\Rightarrow x\left(x+28\right)-19\left(x+28\right)=0\)

\(\Rightarrow\left(x+28\right)\left(x-19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+28=0\\x-19=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-28\\x=19\end{matrix}\right.\)

2 bạn sai rồi!!!