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a: \(\left(x+1\right)^2\)
b: \(\left(x^2+x+1\right)\left(x-1\right)\)
c: \(\left(x^2+2x+4\right)^2\)
d: \(\left(x-2\right)\left(x+2\right)\)
e: \(x^2+2x+1\)
a)\(\dfrac{x+5}{3x-2}=\dfrac{x\left(x+5\right)}{x\left(3x-2\right)}\) b)\(\dfrac{2x-1}{4}=\dfrac{\left(2x-1\right)\left(2x+1\right)}{8x+4}\) c)\(\dfrac{2x\left(x-2\right)}{x^2-4x+4}=\dfrac{2x}{x-2}\) d) \(\dfrac{5x^2+10x}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x}{x-2}\)
ý mình là vì sao được kết quả đó , giải thích ra giúp mình nha
a) \(2x\left(2x+5\right)-4x\left(x-3\right)=7\)
\(4x^2+10x-4x^2+12x=7\)
\(22x=7\Rightarrow x=0,31\)
b) \(\left(x+2\right)\left(x-2\right)-\left(x+1\right)^2=2\)
\(\left(x^2-4\right)-\left(x^2+2x+1\right)=2\)
\(x^2-4-x^2-2x-1=2\)
\(-2x=7\Rightarrow x=-3,5\)
c) \(\left(x+2\right)\left(x-1\right)-\left(x+3\right)\left(x-2\right)=0\)
\(x^2-x+2x-2-x^2+2x+3x-6=0\)
\(6x=8\Rightarrow x=1,3\)
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
B1:
\(=x^2+2x-5x-10+3\left(x^2-2^2\right)-\left(9x^2-2.3x.\frac{1}{2}+\frac{1}{4}\right)+5x^2\)
\(=-10-12-\frac{1}{4}=-22\frac{1}{4}\)
Bài 1.
( x - 5 )( x + 2 ) + 3( x - 2 )( x + 2 ) - ( 3x - 1/2 )2 + 5x2
= x2 - 3x - 10 + 3( x2 - 4 ) - ( 9x2 - 3x + 1/4 ) + 5x2
= 6x2 -- 3x - 10 + 3x2 - 12 - 9x2 + 3x - 1/4
= -89/4 không phụ thuộc vào biến
=> đpcm
Bài 2 < mình viết luôn nhé >
a) ( x + 2y2 )2 = x2 + 4xy2 + 4y4
b) ( a - 5/2b )2 = a2 - 5ab + 25/4b2
c) ( m + 1/2 )2 = m2 + m + 1/4
d) x2 - 16y4 = ( x + 4y2 )( x - 4y2 )
e) 25a2 - 1/4b2 = ( 5a + 1/2b )( 5a - 1/2b )
Bài 3:
a) ta có: \(A=x^2+4x+9\)
\(=x^2+4x+4+5=\left(x+2\right)^2+5\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+5\ge5\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của đa thức \(A=x^2+4x+9\) là 5 khi x=-2
b) Ta có: \(B=2x^2-20x+53\)
\(=2\left(x^2-10x+\frac{53}{2}\right)\)
\(=2\left(x^2-10x+25+\frac{3}{2}\right)\)
\(=2\left[\left(x-5\right)^2+\frac{3}{2}\right]\)
\(=2\left(x-5\right)^2+2\cdot\frac{3}{2}\)
\(=2\left(x-5\right)^2+3\)
Ta có: \(\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-5\right)^2+3\ge3\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-5\right)^2=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)
Vậy: GTNN của đa thức \(B=2x^2-20x+53\) là 3 khi x=5
c) Ta có : \(M=1+6x-x^2\)
\(=-x^2+6x+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left[\left(x-3\right)^2-10\right]\)
\(=-\left(x-3\right)^2+10\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(M=1+6x-x^2\) là 10 khi x=3
Bài 2:
a) \(\left(x+y\right)^2+\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2+\left(x-y\right).\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y+x-y\right)\)
\(=\left(x+y\right).2x\)
c) \(x^2-2xy+y^2-z^2+2zt-t^2\)
\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left[x-y-\left(z-t\right)\right].\left(x-y+z-t\right)\)
\(=\left(x-y-z+t\right).\left(x-y+z-t\right)\)
Chúc bạn học tốt!
Câu a : \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
Câu b : \(\left(x^2+x+1\right)\left(x-1\right)=x^3-1\)
Câu c : \(\left(x^2+2x+4\right)\left(x-2\right)=x^3-8\)
Câu d : \(\left(x-2\right)\left(x^2+2x+4\right)=x^3-8\)
Câu e : \(x^2+2x+1=\left(x+1\right)^2\)
Câu f : \(4x^2+8x+4=\left(2x+2\right)^2\)
Chúc bạn học tốt
cậu giải thích rõ hơn được không?