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\(\left(x+y+z\right)^2-2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\)
= \(\left[\left(x+y+z\right)-\left(x+y\right)\right]^2\)
= \(z^2\)
Ta có:(x + y + z)2 - 2(x + y + z) (x + y) + (x + y)2
=[(x+y+z)-(x+y)]2=z2
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
Câu h đề không đẹp lắm, sửa thành-2x nha
f) x2-2x+5
=x2-2x+1+4
=(x-1)2+4
Vì: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)
Min = 4 khi x=1
g) 2x2-6x
= \(\sqrt{2x}^2-2.\sqrt{2x}.\dfrac{3\sqrt{2}}{2}+\left(\dfrac{3\sqrt{2}}{2}\right)^2-\left(\dfrac{3\sqrt{2}}{2}\right)^2\)
= \(\left(\sqrt{2x}-\dfrac{3\sqrt{2}}{2}\right)^2-\dfrac{9}{2}\)
Tương tự bài trên
h) x2+y2-2x+6y+10
=(x2-2x+1)+(y2+6y+9)
=(x-1)2+(y+3)2
Min=0 khi x=1; y=-3
nói thật bn xạo lz vc đề thế nào thì để đó chứ ko đẹp thì nó ko có Min à
\(x^2+xy+y^2=x^2y^2\)
\(\left(x+y\right)^2=xy\left(xy+1\right)\)
=> \(\left[{}\begin{matrix}x=y=0\\\left\{{}\begin{matrix}x=-y\\xy=-1\end{matrix}\right.\end{matrix}\right.\)=>(x,y)=(0,0);(1,-1);(-1,1);
a) Ta có:
\(x+y=3\)
\(\Rightarrow\left(x+y\right)^2=9\)
\(\Leftrightarrow x^2+2xy+y^2=9\)
\(\Leftrightarrow5+2.xy=9\)
\(\Leftrightarrow2xy=4\)
\(\Rightarrow xy=2\)
Ta có:
\(x^3+y^3=\left(x+y\right).\left(x^2-xy+y^2\right)\)
\(\Rightarrow x^3+y^3=3.\left(5-2\right)\)
\(\Rightarrow x^3+y^3=9\)
Bài 1:
a: \(\Leftrightarrow x^2-4x-x^2+8=0\)
=>-4x+8=0
hay x=2
b: \(\Leftrightarrow3x^2-3x+2x-2-3\left(x^2-x-2\right)=4\)
\(\Leftrightarrow3x^2-x-2-3x^2+3x+6=4\)
=>2x+4=4
hay x=0
a: \(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
=>x=1 hoặc x=3
b: \(x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=3 hoặc x=-4
c: \(3x^2+2x-5=0\)
\(\Leftrightarrow3x^2+5x-3x-5=0\)
=>(3x+5)(x-1)=0
=>x=1 hoặc x=-5/3
d: \(x^4-2x^2-3=0\)
\(\Leftrightarrow x^4-3x^2+x^2-3=0\)
\(\Leftrightarrow x^2-3=0\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(x+y\right)+\left(x-y\right)\right]^2=\left[x+y+x-y\right]^2=\left(2x\right)^2=4x^2\)
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
b) \(\left(x-y+z\right)^2+\left(z-y\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z\right)^2+\left(y-z\right)^2+2\left(x-y+z\right)\left(y-z\right)\)
\(=\left(x-y+z+y-z\right)^2=x^2\)
a) \(\left(x-y+x+y\right)^2=4x^2\)
b) \(\left(x-y+z+y-z\right)^2=x^2\)