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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(30+\left\{96-3.\left[54-2\left(3^2+6\right)\right]\right\}.x=78\)
\(30+\left\{96-3.\left[56-2.15\right]\right\}.x=78\)
\(30+\left\{96-3.\left[56-30\right]\right\}.x=78\)
\(30+\left\{96-3.26\right\}.x=78\)
\(30+\left\{96-78\right\}.x=78\)
\(30+18.x=78\)
\(\Rightarrow18.x=48\)
\(\Rightarrow x=\frac{8}{3}\)
học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(4^5\cdot10\cdot5^6+25^5\cdot2^8\right):\left(2^8\cdot5^4+5^7\cdot2^5\right)\)
\(B=\dfrac{4^5\cdot10\cdot5^6+25^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{\left(2^2\right)^5\cdot2\cdot5\cdot5^6+\left(5^2\right)^5\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^{11}\cdot5^7+5^{10}\cdot2^8}{2^8\cdot5^4+5^7\cdot2^5}\)
\(B=\dfrac{2^8\cdot5^7\cdot\left(2^3\cdot1+5^3\cdot1\right)}{2^5\cdot5^4\cdot\left(2^3\cdot1+5^3\cdot1\right)}\)
\(B=\dfrac{2^8\cdot5^7}{2^5\cdot5^4}\)
\(B=\dfrac{2^3\cdot5^3}{1\cdot1}\)
\(B=\left(2\cdot5\right)^3\)
\(B=10^3\)
\(B=1000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\Rightarrow\frac{x\left(x+1\right)}{2}=78\)
\(\Rightarrow x\left(x+1\right)=156\)
\(x\left(x+1\right)=12\cdot13\)
\(\Rightarrow x=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-2x\right).5^4=3.5^5\)
\(1-2x=3.5^5:5^4\)
\(1-2x=3.5=15\)
\(2x=\left(-14\right)\)
\(x=\left(-7\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(13.\left(23+22\right)-3.\left(17+28\right)\)
\(=13.23+22-3.17+28\)
\(=13.\left(23+17\right)-3\left(22+28\right)\)
\(=13.40-3.50\)
\(=520-150=370\)
\(-48+48.\left(-78\right)+48.\left(-21\right)\)
\(=48.\left[-78+\left(-21\right)-1\right]\)
\(=48.\left(-100\right)=-4800\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{7}{x}=\frac{y}{27}=\frac{-42}{54}\)
+) => \(\frac{y\cdot2}{27\cdot2}=\frac{-42}{54}\)
=> \(y\cdot2=-42\)
\(\Rightarrow y=-21\)
+) \(\frac{7}{x}=\frac{-21}{27}\)
=> \(\frac{7\cdot\left(-3\right)}{x\cdot\left(-3\right)}=\frac{-21}{27}\)
=> \(x\cdot\left(-3\right)=27\)
=> \(x=-9\)
* 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
Bài làm
\(( x - 78).(x + 54)=0 \)
\(\Rightarrow\left[{}\begin{matrix}x - 78=0\\x+ 54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=78\\x=-54\end{matrix}\right.\)
\(\Rightarrow x\in\left\{78\text{ ; }-54\right\}\)