Ta có:

(x-78).(x+54)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x-78=0\\x+54=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=78\\x=-54\end{matrix}\right.\)

Vậy x \(\in\left\{78;-54\right\}\)

Nhớ like cho mình nhé

ta có:

(x-78).(x+54)=0

Vậy x

Nhớ like cho mình nhé

\(30+\left\{96-3.\left[54-2\left(3^2+6\right)\right]\right\}.x=78\)

\(30+\left\{96-3.\left[56-2.15\right]\right\}.x=78\)

\(30+\left\{96-3.\left[56-30\right]\right\}.x=78\)

\(30+\left\{96-3.26\right\}.x=78\)

\(30+\left\{96-78\right\}.x=78\)

\(30+18.x=78\)

\(\Rightarrow18.x=48\)

\(\Rightarrow x=\frac{8}{3}\)

học tốt

\(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)

\(B=\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)

\(B=\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)

\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)

\(B=\dfrac{2^3\cdot5^3}{1\cdot1}\)

\(B=\left(2\cdot5\right)^3\)

\(B=10^3\)

\(B=1000\)

bạn ơi ( 2² )⁵ là sao hả bạn

a) 40 - x = 54

x = 40 - 54

x = -14

b) 36 - x = -11

x = 36 - (-11)

x = 47

c) 2.x - 3 = -14

2x = -14 + 3

2x = -11

x = -11/2

Ta có:\(\left(x-4\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow x-4=0\) hoặc \(\left(x-1\right)^2=0\)

\(\Leftrightarrow x=4\) hoặc \(x-1=0\)

\(\Leftrightarrow x=4\) hoặc x=1

\(\left(2x-6\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow2x-6=0\) hoặc \(x^2+1=0\)(vô lí)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Xét VT ta có:1+2+3+4+...+x=\(\frac{x\left(x+1\right)}{2}\)

Mà VP=78

\(\Rightarrow\frac{x\left(x+1\right)}{2}=78\)

\(\Rightarrow x\left(x+1\right)=156\)

\(x\left(x+1\right)=12\cdot13\)

\(\Rightarrow x=12\)

vt la gi va vp la gi

\(\left(1-2x\right).5^4=3.5^5\)
\(1-2x=3.5^5:5^4\)
\(1-2x=3.5=15\)
\(2x=\left(-14\right)\)
\(x=\left(-7\right)\)

\(\left(1-2x\right).5^4=3.5^5\)

\(\Rightarrow1-2x=3.\left(5^5:5^4\right)\)

\(\Rightarrow1-2x=3.5\)

\(\Rightarrow1-2x=15\)

\(\Rightarrow2x=-14\)

\(\Rightarrow x=-7\)

Vậy ...

\(13.\left(23+22\right)-3.\left(17+28\right)\)

\(=13.23+22-3.17+28\)

\(=13.\left(23+17\right)-3\left(22+28\right)\)

\(=13.40-3.50\)

\(=520-150=370\)

\(-48+48.\left(-78\right)+48.\left(-21\right)\)

\(=48.\left[-78+\left(-21\right)-1\right]\)

\(=48.\left(-100\right)=-4800\)

\(\frac{7}{x}=\frac{y}{27}=\frac{-42}{54}\)

+) => \(\frac{y\cdot2}{27\cdot2}=\frac{-42}{54}\)

=> \(y\cdot2=-42\)

\(\Rightarrow y=-21\)

+) \(\frac{7}{x}=\frac{-21}{27}\)

=> \(\frac{7\cdot\left(-3\right)}{x\cdot\left(-3\right)}=\frac{-21}{27}\)

=> \(x\cdot\left(-3\right)=27\)

=> \(x=-9\)

* Lớp 6 chưa học đến tỉ lệ thức nên đây là cách đơn giản nhất r *

\(\frac{y}{27}\)=\(\frac{-42}{54}\)=>\(\frac{2y}{54}\)=\(\frac{-42}{54}\)=>2y= -42                                                                                                                                                                                                                   y= \(\frac{-42}{2}\)= -21                                                                                                                           \(\frac{7}{x}\)=\(\frac{-21}{27}\)=> -21x=7*27=189                                                                                                                                                                                                     x=\(\frac{189}{-21}\)= -9                                                                                                                                                              Vậy x= -9, y= -21