tìm x:
( x^2 - 16 )( x + 3 ) = 0
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\(x-5.\left(-6\right)=-4.\left(-9\right)\)
\(x-5\left(-6\right)-4\left(-9\right)\)
\(x+30==-4\left(-9\right)\)
\(x+30=36\)
\(x=36-30\)
\(x=6\)
\(7.\left(-8\right)+x=4.\left(-12\right)\)
\(7\left(-8\right)+x=4\left(-12\right)\)
\(-56+x=4\left(-12\right)\)
\(x-56=4\left(-12\right)\)
\(x-56=-48\)
\(x=-48=56\)
\(x=8\)
\(\left(-23\right).\left(-5\right)-x=49.\left(-2\right)\)
\(-23\left(-5\right)-x=49\left(-2\right)\)
\(115-x=49.\left(-2\right)\)
\(-x+115=49.\left(-2\right)\)
\(-x+115=-98\)
\(-x=-98-115\)
\(-x=-213\)
\(x=213\)
1) -35 . 1753 + (-35) . 247 + (-35) . (-1996)
= [ (-35) - (-35) ] . ( 1753 + 247 ) + (-1996)
= 0 . 2000 + (-1996)
= 0 + (-1996)
= -1996
2) 4 . 125 . (-25) . (-8)
= [ ( -25) . 4 ] . [ 125 . (-8)]
= -100 . -1000
= 100 000
1.
\(-35.1753+\left(-35\right).247+\left(-35\right).\left(-1996\right)=-35.\left(1753+247-1996\right)\)
\(=-35.4=-140\)
2. \(4.125.\left(-25\right).\left(-8\right)=\left(-25.4\right).\left(-8.125\right)=\left(-100\right).\left(-1000\right)=100\text{ }000\)
a. ta có \(2019.\left(-2\right)< 0\)
b. \(\left(-2018\right).\left(-2019\right)>0\)
c. \(\left(-1\right).\left(-2\right)\left(-3\right)..\left(-2020\right)>0\)
( x-3) 3 = 24 - ( -3)
( x-3) 3 = 27
( x-3) 3 = 33
x-3 = 3
x = 3+3
x=6
Bài 1: Tính hợp lý (nếu có thể)
a) 5.(-8).(-2).(-3)\(=\left(-2.5\right).\left(\left(-3\right).\left(-8\right)\right)=-10.24=-240\)
c) 147.333+233.(-147)\(=147\left(333-233\right)=147.100=14700\)
b) (-125).8.(-2).5.19\(=\left(-125.8\right).\left(-2.5\right).19=-1000.\left(-10\right).19=190\text{ }000\)
d) (-115).27+33.(-115)\(=-115.\left(27+33\right)=-115.60=-6900\)
Bài 2: Tìm số nguyên x, biết:
a) 2x+19=15\(\Leftrightarrow2x=15-19=-4\Leftrightarrow x=-2\)
c) 24-(x-3)^3=-3\(\Leftrightarrow\left(x-3\right)^3=27=3^3\Leftrightarrow x-3=3\Leftrightarrow x=6\)
12987 -x- [(-720) + 1247 - 247] = 12987
\(\Leftrightarrow\text{12987 -x- [280] = 12987}\)
\(\Leftrightarrow-x-280=0\Leftrightarrow x=-280\)
Ta có:
\(\left(x^2-16\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}0.\left(x+3\right)=0\\\left(x^2-16\right).0=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-16=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0-3\\x^2=0+16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-3\right)\\x^2=16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-3\right)\\x^2=\left(\pm4\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-3\right)\\x=\left(\pm4\right)\end{cases}}\)
Vậy \(x\in\left\{\pm4;-3\right\}\)
@Nghệ Mạt
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