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Bài 1:
a) -16 +(x-3)2
<=> (x-3)2-16
<=> (x-3)2 -42
<=> (x-3-4)(x-3+4)
<=> (x-7)(x+1)
b) 64+16y+y2
<=> y2 + 2.8.y + 82
<=> (y+8)2
c) \(\dfrac{1}{8}-8x^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^3-\left(2x\right)^3\)
\(\Leftrightarrow\left(\dfrac{1}{2}-2x\right)\left(\dfrac{1}{4}+x+4x^2\right)\)
d)\(x^2-x+\dfrac{1}{4}\)
\(\Leftrightarrow x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2\)
e) x4 + 4x2 + 4
<=> (x2)2 + 2.2.x2 +22
<=> (x2 + 2)2
g)\(8x^3+60x^2y+150xy^2+125y^3\)
\(\Leftrightarrow\left(2x+5y\right)^3\)
a) Ta có:
(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2
=x^2 - y^2 - 2yz - z^2.
b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2
=x^2 +2xz- y^2 +z^2.
c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)
= -16 + x^2 - 6x + 9
= x^2 - 6x - 7.
\(a,\left(x+y+z\right)\left(x-y-z\right)\)
\(=x\left(x-y-z\right)+y\left(x-y-z\right)+z\left(x-y-z\right)\)
\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)
\(=x^2-y^2-2yz-z^2\)
\(b,\left(x-y+z\right)\left(x+y+z\right)\)
\(=x\left(x+y+z\right)-y\left(x+y+z\right)+z\left(x+y+z\right)\)
\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)
\(=x^2+2xz-y^2+z^2\)
\(c,-16+\left(x-3\right)^2\)
\(=-16+x^2-6x+9\)
\(=x^2-6x-7\)
1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
\(a.\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2xz\\ b.\left(x-y+z\right)^2=x^2+y^2+z^2-2xy-2yz+2xz\\ c.\left(x-y-z\right)^2=x^2+y^2+z^2-2xy+2yz-2xz\)
a, \(\left(x+y+z\right)^2=x^2+y^2+c^2+2xy+2yz+2zx\)
b, \(\left(x-y-z\right)^2=x^2+y^2+z^2-2xy+2yz-2zx\)
c, \(\left(x-y+z\right)^2=x^2+y^2+z^2-2xy-2yz+2xz\)
a, \(\left(x-y+z\right)\left(x-y-z\right)\)
\(=\left(x-y\right)^2-z^2\)(hằng đẳng thức số 3)
b, Sửa đề:\(\left(\dfrac{1}{2}x+y-z\right)\left(\dfrac{1}{2}x+y+z\right)\)
\(=\left(\dfrac{1}{2}x+y\right)^2-z^2\)(hằng đẳng thức số 3)
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