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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
3)
e)
b) Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
a. 6x3y2 ( 2-x) + 9x2y2 (x-2)
= -6x3y2 (x-2) + 9x2y2 ( x-2)
= (x-2) 3x2y2 ( -2x + 3)
b. x2 - 4x + 4y - y2
= x2 - y2 - (4x - 4y )
= (x-y)(x+y) - 4( x-y)
= (x-y)(x+y-4)
c. 81x2 + 6yz -9y2-z2
= 81x2 - (9y2 - 6yz + z2 )
= (9x)2 - ( 3y - z )2
= (9x + 3y -z)(9x - 3y + z )
\(a,=6x^3y^2\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=\left(2-x\right)\left(6x^3y^2-9x^2y^2\right)=\left(2-x\right)3x^2y^2\left(2x-3\right)\)
\(f,=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5\left(2x-1\right)\)
\(g,=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3+2\right)\)
\(=\left(x+3\right)\left(x-1\right)\)
a) 10x(x - y)2 - 5(x - y)3 = [10x - 5(x - y)](x - y)2 = (10x - 5x + y)(x - y)2 = (5x + y)(x - y)2
b) -x2 - 10x - 25 = -(x2 + 10x + 52) = -(x + 5)2
c) 64x6y4 - 81x2y2 = (8x3y2)2 - (9xy)2 = (8x3y2 + 9xy)(8x3y2 - 9xy)
d) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy + y2)
e)1/8x3 - 3/4x2y + 3/2xy2 - y3 = (1/2x)3 - 3.(1/2x)2y + 3.1/2xy2 - y3 = (1/2x - y)3
f) (3x + 1)2 - (x - 1)2 = (3x + 1 + x - 1)(3x + 1 - x + 1) = 4x(2x + 2) = 8x(x + 1)
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a: \(=\left(a+b+c\right)^2+2\left(a+b+c\right)\left(b-c\right)+\left(b-c\right)^2\)
\(=\left(a+b+c+b-c\right)^2=\left(a+2b\right)^2\)
b: \(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^4-4\right)\)
\(=\left(x^4+4\right)\left(x^4-4\right)=x^8-16\)
d: \(=x^2+y^2+z^2+2\left(xy+yz+xz\right)+2x^2+2y^2+2z^2-2\left(xy+yz+xz\right)-3\left(x^2+y^2+z^2\right)\)
=0
a, 3x-6=(x-2)(4x-5)
⇔3x-6=4x2-13x+10
⇔4x2-16x+16=0
⇔4(x2-4x+4)=0
⇔4(x-2)2=0
⇔(x-2)2=0
⇔x-2=0
⇔x=2
Chúc bạn học tốt!
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