a: \(\left(2x-3\right)^3=27\)

=>\(\left(2x-3\right)^3=3^3\)

=>2x-3=3

=>2x=6

=>x=3

b: \(\left(90:30\right)^2+2^{x}=2^4\cdot7-87\)

=>\(2^{x}+3^2=16\cdot7-87\)

=>\(2^{x}=112-87-9=112-96=16=2^4\)

=>x=4

c: \(234-30^0\cdot\left(x^2-5\right)=230\)

=>\(234-\left(x^2-5\right)=230\)

=>\(x^2-5=234-230=4\)

=>\(x^2=9\)

=>x=3 hoặc x=-3

d: \(144-44\cdot\left(3^{x+1}-3^5\right)=144\)

=>\(44\cdot\left(3^{x+1}-3^5\right)=0\)

=>\(3^{x+1}-3^5=0\)

=>\(3^{x+1}=3^5\)

=>x+1=5

=>x=4

e: \(\left(x+1\right)\left(x+2\right)=444222\)

=>\(x^2+3x+2-444222=0\)

=>\(x^2+3x-444220=0\)

=>(x+668)(x-665)=0

=>x=-668 hoặc x=665

f: \(\left(3-x\right)^{15x}:\left(3-x\right)^{30}=1\)

=>\(\left(3-x\right)^{15x-30}=1\)

=>15x-30=0

=>15x=30

=>x=2

e: Ta có: \(\left(x+1\right)\left(x+2\right)=444222\)

\(\Leftrightarrow x^2+3x-444220=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-444220\right)=1776889\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-1333}{2}=-668\\x_2=\dfrac{-3+1333}{2}=665\end{matrix}\right.\)

lớp 6 be like ;-;

\(g,4=2^2;6=2.3\\ \Rightarrow BCNN\left(4,6\right)=2^2.3=12\\ \Rightarrow x\in BC\left(4,6\right)=B\left(12\right)=\left\{0;12;24;36;48;60;...\right\}\\ \text{Mà }0< x< 50\\ \Rightarrow x\in\left\{12;24;36;48\right\}\\ h,12=2^2.3;18=2.3^2\\ \Rightarrow BCNN\left(12,18\right)=2^2.3^2=36\\ \Rightarrow x\in BC\left(12,18\right)=B\left(36\right)=\left\{0;36;72;108;144;180;216;252;...\right\}\\ \text{Mà }x< 250\\ \Rightarrow x\in\left\{0;36;72;108;144;180;216\right\}\)

g,\(x⋮4,x⋮6\Rightarrow x\in BC\left(4,6\right)=\left\{\pm0;\pm12;\pm24;\pm36;\pm48;\pm60;...\right\}\)

Mà \(0< x< 50\Rightarrow x\in\left\{12;36;48\right\}\)

h,\(x⋮12,x⋮18\Rightarrow x\in BC\left(12,18\right)=\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216;\pm252;...\right\}\)

Mà \(x< 50\Rightarrow x\in\left\{0;\pm36;\pm72;\pm108;\pm144;\pm180;\pm216\right\}\)

120 + 2. (8x-17) = 0

122 . (8x-17) = 0

(8x-17) = 0

8x = 17

x = 17/8

70=2.5.7

140=2².5.7

280=2³.5.7

ƯC(70,140,280)=2.5.7=70

bài này còn một cách nữa nhưng mình ko nhớ

10 nhé bn

\(6\dfrac{1}{2}-2\dfrac{1}{3}:\dfrac{7}{15}\)

\(=\dfrac{13}{2}-\dfrac{7}{3}:\dfrac{7}{15}\)

\(=\dfrac{13}{2}-\dfrac{7}{3}\cdot\dfrac{15}{7}\)

\(=\dfrac{13}{2}-\dfrac{15}{3}\)

\(=\dfrac{13}{2}-5\)

\(=\dfrac{3}{2}\)

CÁC BẠN GIÚP MÌNH VỚI HIHIHIHI

ko hiểu

uk 276 nha mik tính là vậy còn ko biết đúng ko nữa cho mik 1 k nha hihi / HT/