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TN
0
HN
2
KN
7 tháng 11 2019
a) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=x\left(x^2+x+1\right)\)\(-\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1\)
\(=x^3-1\)
Vậy \(\left(x-1\right)\left(x^2+x+1\right)\)\(=x^3-1\)(đpcm)
b) Ta có: \(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)\)
\(=x\left(x^3+x^2y+xy^2+y^3\right)\)\(-y\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4\)
Vậy\(\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)\)\(=x^4-y^4\)(đpcm)
B
8 tháng 9 2020
Bài làm :
\(\text{a) }\left(x-1\right)\left(x^2+x+1\right)\)
\(=x\left(x^2+x+1\right)-\left(x^2+x+1\right)\)
\(=x^3+x^2+x-x^2-x-1\)
\(=x^3-1\)
=> Điều phải chứng minh
\(\text{b)}\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)\)
\(=x\left(x^3+x^2y+xy^2+y^3\right)-y\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=x^4-y^4\)
=> Điều phải chứng minh
HP
15 tháng 6 2016
\(A=4x^2-2\left(y+2,5x^2\right)+x^2-4y\)
\(=4x^2-2y-5x^2+x^2-4y=-6y\)
\(B=\left(x+y\right).\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)-\left(x^5+y^5-8\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5-x^5-y^5+8\)
\(=8\)
Vậy BT B ko phụ thuộc vào biến
câu sau tương tự
\(5x\left(x+1\right)-3\left(x-5\right)+4\left(3x-6\right)=2x^2-7\)
\(\Rightarrow5x^2+5x-3x+15+12x-24=2x^2-7\)
\(\Rightarrow5x^2+14x-9=2x^2-7\Rightarrow5x^2+14x-9-2x^2+7=0\)
\(\Rightarrow3x^2+14x-2=0\)
\(\Rightarrow3\left(x^2+\frac{14}{3}x-\frac{2}{3}\right)=0\Rightarrow x^2+2.x.\frac{7}{3}+\frac{49}{9}-\frac{55}{9}=0\)
\(\Rightarrow\left(x+\frac{7}{3}\right)^2=\frac{55}{9}\Rightarrow x+\frac{7}{3}\in\left\{\sqrt{\frac{55}{9}};-\sqrt{\frac{55}{9}}\right\}\Rightarrow x\in\left\{\sqrt{\frac{55}{9}}-\frac{7}{3};-\sqrt{\frac{55}{9}}-\frac{7}{3}\right\}\)
M
30 tháng 10 2019
Câu 1 : Tìm x :
1. \(A=x^2+4x-2\)
\(A=x^2+2.x.2+2^2-2^2-2\)
\(A=\left(x^2+4x+2^2\right)-4-2\)
\(A=\left(x+2\right)^2-6\)
\(\left(x+2\right)^2-6\ge-6\)
MIn A= -6 khi \(\left(x+2\right)^2=0\)
=> \(x+2=0hayx=-2\)
Vậy x=2
những câu tiếp theo làm tg tự như thế nhé
30 tháng 10 2019
Câu 1:
a) Ta có: \(A=x^2+4x-2\)
\(=x^2+4x+4-6\)
\(=\left(x+2\right)^2-6\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: x=-2
b) Ta có: \(B=2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)
\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)
\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)
\(=2\left(x-1\right)^2+1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: x=1
c) Ta có: \(C=x^2+y^2-4x+2y+5\)
\(=x^2-4x+4+y^2+2y+1\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y+1\right)^2\ge0\forall y\)
Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy: x=2 và y=-1
Câu 2:
a) Ta có: \(A=-x^2+6x+5\)
\(=-\left(x^2-6x-5\right)\)
\(=-\left(x^2-6x+9-14\right)\)
\(=-\left[\left(x^2-6x+9\right)-14\right]\)
\(=-\left[\left(x-3\right)^2-14\right]\)
\(=-\left(x-3\right)^2+14\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)
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 \(A=-x^2+6x+5\) là 14 khi x=3
b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)
\(=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)
\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)
Ta có: \(\left(3y-1\right)^2\ge0\forall y\)
\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)
Từ (1) và (2) suy ra
\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)
\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)
Câu 3:
a) Ta có: \(x^2+y^2-2x+4y+5=0\)
\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy: x=1 và y=-2
b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy: x=3 và y=-2
1 tháng 2 2020
a/VT=x5+x^4.y+x^3.y^2+x^2.y^4+x.y^4-x^4.y-x^3.y^2-x^2.y^3-x.y^4-y^5
=x^5-y^5=VP
=>dpcm
S
27 tháng 7 2019
\(x-y=1\Rightarrow x^2-2xy+y^2=1\Rightarrow x^2+xy+y^2=19\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=1.19=19\)
\(2,a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)=2ab+2bc+2ca\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0ma:\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow a=b=c\)
S
27 tháng 7 2019
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)=4a^2b^2+4c^2a^2+4b^2c^2\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=\left(a^2+b^2+c^2\right)^2\left(dpcm\right)\)
22 tháng 7 2017
1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)
\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)
\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
4. \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
5. \(a^3x-ab+b-x\)
\(=a^3x-x-ab+b\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)
6. \(x^3-64=x^3-4^3\)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
7. \(0,125\left(a+1\right)^3-1\)
\(=\left[0,5\left(a+1\right)\right]^3-1^3\)
\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)
\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)
\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)
8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
11. \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)