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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
mk k bít mới hỏi chứ 
Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
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) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
a)\(\left|2x+3\right|=x+2\)
\(\Leftrightarrow\left(\left|2x+3\right|\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+12x+9=x^2+4x+4\)
\(\Leftrightarrow3x^2+8x+5=0\)
\(\Leftrightarrow3x^2+3x+5x+5=0\)
\(\Leftrightarrow3x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b)\(x^2-9x+8=0\)
\(\Leftrightarrow x^2-8x-x+8=0\)
\(\Leftrightarrow x\left(x-8\right)-\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
c)\(x^2-2\left(x-2\right)=4\)
\(\Leftrightarrow\left(x^2-4\right)-2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b/ \(x^2-9x+8=0\)
Ta có: a = 1 ; b = -9 ; c = 8
\(\Delta=b^2-4ac=\left(-9\right)^2-4.1.8=49\)
\(\Rightarrow\sqrt{\Delta}=7\)
Pt có 2 nghiệm:
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{9+7}{2.1}=8\)
\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{9-7}{2.1}=1\)
Vậy.......................................
a , \(x^2+2xy+y^2+1=\left(x+y\right)^2+1>0\) , \(\forall x,y\)
b , \(x^2-x+1=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)
c , \(x-1-x^2=-\left(x^2-x+1\right)\)
vì \(x^2-x+1>0\) (c.m b)
nên -(\(x^2-x+1\) ) < 0 , \(\forall x\)
Câu a :
\(x^2+2xy+y^2+1=\left(x+y\right)^2+1\ge1\) nên bất kì giá trị nào của x thì biểu thức trên luôn lớn hơn 0
Câu b :
\(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
nên bất kì giá trị của x thì biểu thức trên luôn lớn hơn 0
Câu c :
\(x-1-x^2=-\left(x^2-x+1\right)=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\le-\dfrac{3}{4}\)
nên bất kì giá trị nào của x thì biểu thức luôn nhỏ hơn 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 \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
Bài 1:
a, \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy x = 0 hoặc x = -1
b, \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
Do \(x^2+1>0\)
\(\Rightarrow x=0\)
Vậy x = 0
c, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x=5\)
Vậy x = 5
Bài 2: ấn câu hỏi tương tự ý
bài 1:
a)
\(x+1=\left(x+1\right)^2\\ \Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)x=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
b)\(x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Rightarrow x=0\)
c)\(x^2-10x=-25\\ \Leftrightarrow x^2-2.5.x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Rightarrow x=5\)
bài 2:
a)
\(x^6-y^6=\left(x^3+y^3\right)\left(x^3-y^3\right)\\ =\left(x+y\right)\left(x^2+y^2+xy\right)\left(x-y\right)\left(x^2+y^2-xy\right)\)
b)
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y\right)-3xzy\\ =\left(x+y+z\right)^3-\left(3xz+3zy\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2zy-3xz-3zy-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)