5-0x3+9/3=?
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\(5-\left(0\times3+9\div3\right)\)
\(=5-\left(0+3\right)\)
\(=5-3\)
\(=2\)
Tất cả các bạn có kết quả bằng 8 đều sai.Vì các bạn viết biểu thức thiếu dấu ngoặc đơn.
5+0x3+9:3=8
ban a
tich minh nha ! minh tra loi nhanh nhat day !
1. (x + 5)2 - (x + 5)(x - 2) = 0
<=> (x + 5 - x + 2)(x + 5) = 0
<=> 7(x + 5) = 0
<=> x + 5 = 0
<=> x = -5
2. x3 + 7x2 + 6x = 0
<=> x3 + x2 + 6x2 + 6x = 0
<=> x2(x + 1) + 6x(x + 1) = 0
<=> (x2 + 6x)(x + 1) = 0
<=> x(x + 6)(x + 1) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+6=0\\x+1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-6\\x=-1\end{matrix}\right.\)
3. (x + 1)2 - (2x + 3)2 = 0
<=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0
<=> (3x + 4)(-2 - x) = 0
<=> \(\left[{}\begin{matrix}3x+4=0\\-2-x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-4}{3}\\x=-2\end{matrix}\right.\)
a)
\(=\left(x+2y\right)\left(x^2-xy+y^2\right)-3xy\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x^2-xy+y^2-3xy\right)\)
\(=\left(x+2y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+2\right)\left(x-2\right)^2\)
b)
\(3x\left(2x-1\right)\left(2x+1\right)=0\)
3x=0 =>x=0
hoặc 2x-1=0 => 2x=1=>x=1/2
hoặc 2x+1=0=>2x=-1=>x=-1/2
1.\(\left(x+2\right)\left(2x-3\right)=x^2-4\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
2.\(x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
3.\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4.\(x^3+x^2-12x=0\)
\(\Leftrightarrow x\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x\left(x+4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x+2\right)\left(2x-3\right)-\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-3-x+2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
b: =>(x+1)(x+2)=0
=>x=-1 hoặc x=-2
c: =>(2x+3)(x+1)=0
=>x=-1 hoặc x=-3/2
d: =>x(x+4)(x-3)=0
hay \(x\in\left\{0;-4;3\right\}\)
5-0x3 + \(\frac{9}{3}\)
= 5-3
=2
Nhầm 5 - 0 x 3 + 9 :3
5-0+3
8