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BT
2
1 tháng 7 2019
a, x^2 - 10x + 2
= x^2 - 10x + 25 - 23
= (x - 5)^2 - 23
(x - 5)^2 > 0 => (x - 5)^2 - 23 > -23
dấu "=" xảy ra khi :
(x - 5)^2 = 0
=> x - 5 = 0
=> x = 5
vậy_
b, không biết
2 tháng 7 2019
TL:
B)\(4x^2-12x-3=4x^2-12x+9-12\)
\(=\left(2x-3\right)^2-12\)
Mà \(\left(2x-3\right)^2\ge0\forall x\in R\Rightarrow\left(2x-3\right)^2-12\ge-12\forall x\in R\)
Dấu "=" xảy ra <=>\(\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Vậy..........
hc tốt
C
29 tháng 6 2017
1) \(4x^2+4x+1=\left(2x+1\right)^2\)
2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
3)\(-x^2+10x-25=-\left(x-5\right)^2\)
4)\(1+12x+36x^2=\left(1+6x\right)^2\)
5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)
6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
ND
11 tháng 6 2018
1) \(4x^2-12x+y^2-4y+13\)
\(=\left(4x^2-12x+9\right)+\left(y^2-4y+4\right)\)
\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(y^2-2.2y+4\right)\)
\(=\left(2x-3\right)^2+\left(y-2\right)^2\)
2) \(x^2+y^2+2y-6x+10\)
\(=\left(x^2+2y+1\right)+\left(y^2-6x+9\right)\)
\(=\left(x+1\right)^2+\left(y-3\right)^2\)
3) \(4x^2+9y^2-4x+6y+2\)
\(=\left(4x^2-4x+1\right)+\left(9y^2+6y+1\right)\)
\(=\left(2x-1\right)^2+\left(3y+1\right)^2\)
4) \(y^2+2y+5-12x+9x^2\)
\(\left(y^2+2y+1\right)+\left(9x^2-12x+4\right)\)
\(=\left(y+1\right)^2+\left(3x-2\right)^2\)
5) \(x^2+26+6y+9y^2-10x\)
\(=\left(x^2-10x+25\right)+\left(9y^2+6y+1\right)\)
\(=\left(x-5\right)^2+\left(3y+1\right)^2\)
25 tháng 5 2022
4: \(3x^3-5x^2+5x-2\)
\(=3x^3-2x^2-3x^2+2x+3x-2\)
\(=x^2\left(3x-2\right)-x\left(3x-2\right)+\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x^2-x+1\right)\)
5: \(5x^3-12x^2+14x-4\)
\(=5x^3-2x^2-10x^2+4x+10x-4\)
\(=\left(5x-2\right)\left(x^2-2x+2\right)\)
4 tháng 7 2019
a) 5.(x^2-3x+1)+x.(1-5x)=x-2
\(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
\(\Leftrightarrow-14x-x=-2-5\)
\(\Leftrightarrow-15x=-7\)
\(\Leftrightarrow x=\frac{7}{15}\)
b\(,3x.\left(\frac{4}{3}+1\right)-4x\left(x-2\right)=10\)
\(\Leftrightarrow4x+3x-4x^2+8x-10=0\)
\(\Leftrightarrow-4x^2+15x-10=0\)
Đề sai???
\(c,12x^2-4x\left(3x-5\right)=10x-17\)
\(\Leftrightarrow12x^2-12x^2+20x-10x=-17\)
\(\Leftrightarrow10x=-17\)
\(\Leftrightarrow x=-\frac{17}{10}\)
\(d,4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{3}{2}\)
AN
7 tháng 10 2018
a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)
b)\(a^4+a^2-2=a^4-a^2+2a^2-2=a^2\left(a^2-1\right)+2\left(a^2-1\right)=\left(a^2+2\right)\left(a^2-1\right)\)
c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)
\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)
e)\(6x^3-17x^2+14x-3\)
Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)
\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)
\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)
Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)
Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)
AN
7 tháng 10 2018
h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)
18 tháng 10 2019
2. \(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(Q=4x^2+5x-12x-15+2019\)
\(Q=4x^2-7x+2004\)
\(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\)
\(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)
\(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)
\(\Rightarrow Q\ge\frac{32255}{16}\)
\(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)
3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\) (do a+b=1)
\(T=4a^2-4ab+4a^2-6a^2-6b^2\)
\(T=-2a^2-4ab-2b^2\)
\(T=-2\left(a^2+2ab+b^2\right)\)
\(T=-2\left(a+b\right)^2\)
\(T=-2.1^2=-2.1=-2\) (do a+b=1)
KN
10 tháng 9 2019
1)
a) \(2x^2-12x+18+2xy-6y\)
\(=2x^2-6x-6x+18+2xy-6y\)
\(=\left(2xy+2x^2-6x\right)-\left(6y+6x-18\right)\)
\(=x\left(2y+2x-6\right)-3\left(2y+2x-6\right)\)
\(=\left(x-3\right)\left(2y+2x-6\right)\)
\(=2\left(x-3\right)\left(y+x-3\right)\)
b) \(x^2+4x-4y^2+8y\)
\(=x^2+4x-4y^2+8y+2xy-2xy\)
\(=\left(-4y^2+2xy+8y\right)+\left(-2xy+x^2+4x\right)\)
\(=2y\left(-2y+x+4\right)+x\left(-2y+x+4\right)\)
\(=\left(2y+x\right)\left(-2y+x+4\right)\)
2) \(5x^3-3x^2+10x-6=0\)
\(\Leftrightarrow x^2\left(5x-3\right)+2\left(5x-3\right)=0\Leftrightarrow\left(x^2+2\right)\left(5x-3\right)=0\)
Mà \(x^2+2>0\Rightarrow5x-3=0\Rightarrow x=\frac{3}{5}\)
\(x^2+y^2-2x+4y+5=0\)
\(\Leftrightarrow x^2+y^2-2x+4y+4+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
3)\(P\left(x\right)=x^2+y^2-2x+6y+12\)
\(P\left(x\right)=x^2+y^2-2x+6y+1+9+2\)
\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
Vậy \(P\left(x\right)_{min}=2\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
10 tháng 9 2019
Bài làm
a) 2x2 - 12x + 18 + 2xy - 6y
= 2x2 - 6x - 6x + 18 + 2xy - 6y
= ( 2xy + 2x2 - 6x ) - ( 6y + 6x - 18 )
= 2x( y + x - 3 ) - 6( y + x - 3 )
= ( 2x - 6 ) ( y + x - 3 )
# Học tốt #
19 tháng 7 2017
\(\Leftrightarrow x^4+2x^3-12x^2-10x=3x-42\)
\(\Leftrightarrow\left(x^4+2x^3+x^2\right)-13x^2-13x+42=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-13\left(x^2+x\right)+42=0\)
\(\left[\left(x^2+x\right)-\dfrac{13}{2}\right]^2-\dfrac{169}{4}+42=0\)
\(\left[\left(x^2+x\right)-\dfrac{13}{2}\right]^2-\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left[\left(x^2+x-\dfrac{13}{2}-\dfrac{1}{2}\right)\right]\left[\left(x^2+x-\dfrac{13}{2}+\dfrac{1}{2}\right)\right]=0\)
\(\Leftrightarrow\left(x^2+x-7\right)\left[\left(x^2+x-6\right)\right]=0\)
\(\Leftrightarrow\left(x^2+x-7\right)\left(x-2\right)\left(x+3\right)=0\)
x nguyên => x=2 hoặc x =-3