Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\)
Vậy \(A_{min}=1\Leftrightarrow x=-1\)
\(B=x^2+4x=6=x^2+4x+4+2=\left(x+2\right)^2+2\ge2>0\)
Vậy \(B_{min}=2\Leftrightarrow x=-2\)
WTF đăng một loạt vầy ai dám làm @@
Mấy bài này trong sách bài tập cx có bài mẫu
tự lật sách ra học ik , đăng 1 loạt ai giải cho chép zô hết
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
I don't now
...............
.................
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)
\(a)\frac{2x-1}{5x-10}\) \(\text{Đ}K:x\ne2\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}(TM)\)
\(b)\frac{x^2-x}{2x}\) \(\text{Đ}K:x\ne0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x.(x-1)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)
\(c)\frac{2x+3}{4x-5}\) \(\text{Đ}K:x\ne\frac{5}{4}\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=\frac{-3}{2}(TM)\)
\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\) \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)
\(\Leftrightarrow(x-1).(x+2)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)
gửi cho 4 câu trc
\(a;x^2-3x+3=x^2-2\cdot\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+3\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\Leftrightarrow x^2-3x+3>0\forall x\)
\(4x\left(x^2-5x+3\right)=4x^3-20x^2+12x\)
=> Chọn A