![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
k: \(\left(4x-16\right)\left(-72+9x\right)=0\)
=>\(4\cdot\left(x-4\right)\cdot9\left(x-8\right)=0\)
=>\(36\left(x-4\right)\left(x-8\right)=0\)
=>\(\left(x-4\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=8\end{matrix}\right.\)
m: \(\left(20+5x\right)\left(4x-8\right)=0\)
=>\(5\cdot\left(x+4\right)\cdot4\left(x-2\right)=0\)
=>\(\left(x+4\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
n: \(\left(-4x+48\right)\left(2x-24\right)=0\)
=>\(-4\left(x-12\right)\cdot2\left(x-12\right)=0\)
=>\(\left(x-12\right)^2=0\)
=>x-12=0
=>x=12
o: \(\left(4x+16\right)\left(-2x+20\right)\left(-40+x\right)=0\)
=>\(4\cdot\left(x+4\right)\cdot\left(-2\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left(x+4\right)\left(x-10\right)\left(x-40\right)=0\)
=>\(\left[{}\begin{matrix}x+4=0\\x-10=0\\x-40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=10\\x=40\end{matrix}\right.\)
p: \(\left(-5x+40\right)\left(-x+2023\right)\left(2x-2\right)=0\)
=>\(-5\left(x-8\right)\cdot\left(-1\right)\cdot\left(x-2023\right)\cdot2\left(x-1\right)=0\)
=>\(\left(x-8\right)\left(x-2023\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x-8=0\\x-1=0\\x-2023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=1\\x=2023\end{matrix}\right.\)
q: \(2024x\left(4x-8\right)\left(5+5x\right)=0\)
=>\(x\cdot4\left(x-2\right)\cdot5\left(x+1\right)=0\)
=>\(x\left(x-2\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-1\end{matrix}\right.\)
r: \(-4x\left(3x+9\right)\left(2x-16\right)=0\)
=>\(-4x\cdot3\left(x+3\right)\cdot2\left(x-8\right)=0\)
=>\(x\left(x+3\right)\left(x-8\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=8\end{matrix}\right.\)
s: \(\left(-100+5x\right)\left(2x-10\right)\left(6x+6\right)=0\)
=>\(5\cdot\left(x-20\right)\cdot2\left(x-5\right)\cdot6\left(x+1\right)=0\)
=>\(\left(x-20\right)\left(x-5\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-20=0\\x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=5\\x=-1\end{matrix}\right.\)
t: \(\left(-2x+4\right)\left(2x+16\right)\cdot\left(7-x\right)=0\)
=>\(-2\left(x-2\right)\cdot2\left(x+8\right)\cdot\left(-1\right)\cdot\left(x-7\right)=0\)
=>\(\left(x-2\right)\left(x+8\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-2=0\\x-7=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-8\\x=7\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x : 3 dư 2
x : 5 dư 1
→ x + 4 chia hết cho 3 và 5
→ x + 4 € BC ( 3, 5 )
Ta có: 3 . 5 = 15
→ BC ( 3, 5 ) = B ( 15 ) = {0;15;30;45;...}
Dựa vào các điều kiện trên, ta kết luận: Vậy x € { 15;30 }
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
1: \(\Leftrightarrow x\cdot\dfrac{7}{2}=\dfrac{9}{2}+3=\dfrac{15}{2}\)
hay x=15/7
2: \(\Leftrightarrow x=\dfrac{5}{2}\cdot\dfrac{8}{5}=4\)
3: \(\Leftrightarrow x=\dfrac{-11\cdot10}{5}=-11\cdot2=-22\)
4: =>2x=90
hay x=45
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 21:
Gọi độ dài cạnh hình vuông là x(cm)
(Điều kiện: x>0)
Chiều dài hình chữ nhật mới là x+5(cm)
Chiều rộng hình chữ nhật mới là x-5(cm)
Vì chiều dài gấp 2 lần chiều rộng nên x+5=2(x-5)
=>2x-10=x+5
=>2x-x=5+10
=>x=15(nhận)
Chu vi hình vuông là \(15\cdot4=60\left(cm\right)\)
Diện tích hình vuông là \(15^2=225\left(cm^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + ... + x ) + ( 1 + 2 + ... + 100 ) = 5750
SSH là : ( 100 - 1) : 1 + 1 = 100 ( số )
Tổng là : ( 100 + 1 ) . 100 : 2 = 5050
=> 100x + 5050 = 5750
=> 100x = 700
=> x = 7
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(156:\left(7x+135\right):x=\frac{156}{\frac{7x+135}{x}}=156.\frac{x}{7x+135}\)
\(\Rightarrow156.\frac{x}{7x+135}=13\)
\(\Rightarrow\frac{x}{7x+135}=\frac{13}{156}\)
\(\Rightarrow\frac{x}{7x+135}=\frac{1}{12}\)
\(\Rightarrow12x=7x+135\)
\(\Rightarrow5x=135\)
\(\Rightarrow x=27\)
\(156:\left(7x+135\right):x=13\)
\(\left(7x+135\right):x=12\)
\(7x+135=12x\)
\(7x-12x=-135\)
\(-5x=-135\)
\(x=\frac{-135}{-5}\)
\(x=27\)
hok tốt!!
![](https://rs.olm.vn/images/avt/0.png?1311)
e: \(\left(-4156+2021\right)-\left(119+2021-4156\right)\)
\(=-4156+2021-119-2021+4156\)
\(=\left(-4156+4156\right)+\left(2021-2021\right)-119\)
=0+0-119
=-119
g: \(315\cdot75-\left(15\cdot100-315\cdot25\right)\)
\(=315\cdot75-15\cdot100+315\cdot25\)
\(=315\left(75+25\right)-15\cdot100\)
\(=315\cdot100-15\cdot100=300\cdot100=30000\)
h: \(\left(-489\right)\cdot125-\left(125\cdot11-500\cdot25\right)\)
\(=-489\cdot125-125\cdot11+500\cdot25\)
\(=125\left(-489-11\right)+500\cdot25\)
\(=125\cdot\left(-500\right)+500\cdot25\)
\(=500\left(-125+25\right)\)
\(=500\cdot\left(-100\right)=-50000\)
Bài 2:
a: \(-415-3\left(2x-1\right)^2=-490\)
=>\(3\left(2x-1\right)^2+415=490\)
=>\(3\left(2x-1\right)^2=75\)
=>\(\left(2x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a: \(\left(x+10\right)\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x+10=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=5\end{matrix}\right.\)
b: \(\left(2x+10\right)\left(4+x\right)=0\)
=>\(\left[{}\begin{matrix}2x+10=0\\4+x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-4\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)
c: \(\left(4x+20\right)\left(12x-24\right)=0\)
=>\(\left[{}\begin{matrix}4x+20=0\\12x-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-20\\12x=24\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d: \(\left(x-2024\right)\left(4x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-2024=0\\4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2024\\4x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-1\\x=2024\end{matrix}\right.\)
e: \(\left(2x-6\right)\left(7+x\right)=0\)
=>\(\left[{}\begin{matrix}2x-6=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\x=-7\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)
g: (4x+8)(6-x)=0
=>\(\left[{}\begin{matrix}4x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-2\\x=6\end{matrix}\right.\)
h: (2x+2)(4x-8)=0
=>2(x+1)*4*(x-2)=0
=>(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
i: (2x-2024)(8x-16)=0
=>\(2\left(x-1012\right)\cdot8\cdot\left(x-2\right)=0\)
=>\(\left(x-1012\right)\left(x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-1012=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1012\\x=2\end{matrix}\right.\)