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a, \(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
b, \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)
\(=81x^2-18x+1+1-10x+25x^2+18x-90x^2-2+10x\)
\(=16x^2\)
c;d;e;f tự làm, đầu I giữ lấy còn trường tồn:)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)
\(=\left(20x^2+4x^2-18x^2\right)+\left(60x+8x-20x\right)+\left(5-12-50\right)\)
\(=6x^2+48x-57\)
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
1, \(A=3x^2+5x-1\)
\(=3\left(x^2+\dfrac{5}{3}x-\dfrac{1}{3}\right)\)
\(=3\left(x^2+\dfrac{5}{6}.x.2+\dfrac{25}{36}-\dfrac{37}{36}\right)\)
\(=3\left(x+\dfrac{5}{6}\right)^2-\dfrac{37}{12}\ge\dfrac{-37}{12}\)
Dấu " = " khi \(3\left(x+\dfrac{5}{6}\right)^2=0\Leftrightarrow x=\dfrac{-5}{6}\)
Vậy \(MIN_A=\dfrac{-37}{12}\) khi \(x=\dfrac{-5}{6}\)
2,3 tương tự
4, \(A=2x^2+7x\)
\(=2\left(x^2+\dfrac{7}{4}.x.2+\dfrac{49}{16}-\dfrac{49}{16}\right)\)
\(=2\left(x+\dfrac{7}{4}\right)^2-\dfrac{49}{8}\ge\dfrac{-49}{8}\)
Dấu " = " khi \(2\left(x+\dfrac{7}{4}\right)^2=0\Leftrightarrow x=\dfrac{-7}{4}\)
Vậy \(MIN_A=\dfrac{-49}{8}\) khi \(x=\dfrac{-7}{4}\)
5, 6 tương tự
7, \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " khi \(\left(x^2+5x\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(MIN_A=-36\) khi x = 0 hoặc x = -5
8, \(A=x^2-4x+y^2-8x+6\)
\(=x^2-4x+4+y^2-8x+16-14\)
\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Vậy \(MIN_A=-14\) khi x = 2 và y = 4
1) a) \(\left(3x-1\right)\left(9x^2+3x+1\right)-4x\left(x-5\right)\)
\(=27x^3+9x^2+3x-9x^2-3x-1-4x^2+20x\)
\(=27x^3+\left(9x^2-9x^2-4x^2\right)+\left(3x-3x+20x\right)+\left(-1\right)\)
\(=27x^3-4x^2+20x-1\)
b)\(\left(7x+2\right)\left(3-4x\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=21x-28x^2+6-8x-x^3+3x^2-9x-3x^2+9x-27\)
\(=\left(21x-8x-9x+9x\right)+\left(-28x^2+3x^2-3x^2\right)\)\(+\left(6-27\right)\)\(+\left(-x^3\right)\)
\(=13x-28x^2-21-x^3\)
c)\(\left(4x+3\right)\left(4x-3\right)-\left(2-x\right)\left(4+2x+x^2\right)\)
\(=16x^2-12x+12x-9-8-4x-2x^2+4x+2x^2+x^3\)
\(=\left(16x^2-2x^2+2x^2\right)+\left(-12x+12x-4x+4x\right)\)\(+\left(-9-8\right)\)\(+x^3\)
\(=16x^2-17+x^3\)
d)\(\left(3x-8\right)\left(-5x+6\right)-\left(4x+1\right)\left(3x-2\right)\)
\(=-15x^2+18x+40x-48-12x^2+8x-3x+2\)
\(=\left(-15x^2-12x^2\right)+\left(18x+40x+8x-3x\right)\)\(+\left(-48+2\right)\)
\(=-27x^2+63x-46\)
e)\(\left(3x-6\right)4x-2x\left(3x+5\right)-4x^2\)
\(=12x^2-24x-6x^2-10x-4x^2\)
\(=\left(12x^2-6x^2-4x^2\right)+\left(-24x-10x\right)\)
\(=2x^2-34x\)
f)\(\left(5x-6\right)\left(6x-5\right)-x\left(3x+10\right)\)
\(=30x^2-25x-36x+30-3x^2-10x\)
\(=\left(30x^2-3x^2\right)+\left(-25x-36x-10x\right)+30\)
\(=27x^2-71x+30\)
2) a)\(x\left(x+3\right)-x^2=6\)
\(\Rightarrow x^2+3x-x^2=6\)
\(\Rightarrow\left(x^2-x^2\right)+3x=6\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)
Vậy x=2
b) \(2x\left(x-5\right)+x\left(-2x-1\right)=6\)
\(\Rightarrow2x^2-10x-2x^2-x=6\)
\(\Rightarrow\left(2x^2-2x^2\right)+\left(-10x-x\right)=6\)
\(\Rightarrow-11x=6\)
\(\Rightarrow x=-\dfrac{6}{11}\)
\(\)Vậy \(x=-\dfrac{6}{11}\)
c) x(x+5)-(x+1)(x-2)=7
\(\Rightarrow x^2+5x-x^2+2x-x+2=7\)
\(\Rightarrow\left(x^2-x^2\right)+\left(5x+2x-x\right)=7-2\)
\(\Rightarrow6x=5\)
\(\Rightarrow x=\dfrac{5}{6}\)
Vậy x=\(\dfrac{5}{6}\)
d)\(\left(3x+4\right)\left(6x-3\right)-\left(2x+1\right)\left(9x-2\right)=10\)
\(\Rightarrow18x^2-9x+24x-12-18x^2+4x-9x+2=10\)
\(\Rightarrow\left(18x^2-18x^2\right)+\left(-9x+24x+4x-9x\right)+\left(-12+2\right)=10\)
\(\Rightarrow10x-10=10\)
\(\Rightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy x=2
\(A=x^2+2x+6=(x^2+2x+1)+5=(x+1)^2+5\)
Vì \((x+1)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A\geq 0+5=5\)
Vậy GTNN của $A$ là $5$ khi $(x+1)^2=0$ hay $x=-1$
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\(B=x^2-6x+15=(x^2-2.3x+3^2)+6=(x-3)^2+6\)
Vì \((x-3)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow B\geq 0+6=6\)
Vậy GTNN của $B$ là $6$ khi $x=3$
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\(C=x^2-5x+3=x^2-2.\frac{5}{2}x+(\frac{5}{2})^2-\frac{13}{4}=(x-\frac{5}{2})^2-\frac{13}{4}\)
Vì \((x-\frac{5}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow C\geq 0-\frac{13}{4}=\frac{-13}{4}\)
Vậy \(C_{\min}=\frac{-13}{4}\Leftrightarrow x=\frac{5}{2}\)
\(D=2x^2-7x+1=2(x^2-\frac{7}{2}x)+1\)
\(=2[x^2-2.\frac{7}{4}x+(\frac{7}{4})^2]-\frac{41}{8}\)
\(=2(x-\frac{7}{4})^2-\frac{41}{8}\)
Vì \((x-\frac{7}{4})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow D\geq 2.0-\frac{41}{8}=-\frac{41}{8}\)
Vậy \(D_{\min}=-\frac{41}{8}\Leftrightarrow x=\frac{7}{4}\)
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\(E=3x^2+2x=3(x^2+\frac{2}{3})=3[x^2+2.\frac{1}{3}x+(\frac{1}{3})^2]-\frac{1}{3}\)
\(=3(x+\frac{1}{3})^2-\frac{1}{3}\)
Vì \((x+\frac{1}{3})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow E\geq 3. 0-\frac{1}{3}=\frac{-1}{3}\)
Vậy \(E_{\min}=\frac{-1}{3}\Leftrightarrow x=\frac{-1}{3}\)
Bài này lm sao dùng hằng đẳng thức đc bn,tùy câu thôi
a,\(x^2+5x-6\)
\(=x^2+6x-x-6\)
\(=x\left(x+6\right)-\left(x+6\right)\)
\(=\left(x+6\right)\left(x-1\right)\)
\(b,7x-6x^2-2\)
\(=-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2-3x\right)\)
\(c,5x\left(x-1\right)-3x\left(x-1\right)\)
\(=x\left(x-1\right)\left(5-3\right)\)
\(=2x\left(x-1\right)\)
\(d,\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\)
\(=2x\left(4x+2\right)\)
\(=4x\left(2x+1\right)\)
\(d,x^6-y^6\)
\(=\left(x^2\right)^3-\left(y^2\right)^3\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^4+x^2y^2+y^4\right)\)
a) x2 + 5x - 6
= x2 + 6x - x - 6
= x(x + 6) - (x + 6)
= (x + 6)(x - 1)
b) 7x - 6x2 - 2
= -6x2 + 3x + 4x - 2
= 3x(1 - 2x) - 2(1 - 2x)
= (1 - 2x)(3x - 2)
c) 5x(x - 1) - 3x(x - 1)
= (x - 1)(5x - 3x)
= 2x(x - 1)
d) (3x + 1)2 - (x + 1)2
= (3x + 1 + x + 1)(3x + 1 - x - 1)
= 2x(4x + 2)
e) x6 - y6
= (x3)2 - (y3)2
= (x3 - y3)(x3 + y3)
= (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)