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Bài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1\(\ge\)0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967\(\ge\)0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2\(\le\)0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
ài 1:
a) A= x2 + 4x + 5
=x2+4x+4+1
=(x+2)2+1$\ge$≥0+1=1
Dấu = khi x+2=0 <=>x=-2
Vậy Amin=1 khi x=-2
b) B= ( x+3 ) ( x-11 ) + 2016
=x2-8x-33+2016
=x2-8x+16+1967
=(x-4)2+1967$\ge$≥0+1967=1967
Dấu = khi x-4=0 <=>x=4
Vậy Bmin=1967 <=>x=4
Bài 2:
a) D= 5 - 8x - x2
=-(x2+8x-5)
=21-x2+8x+16
=21-x2+4x+4x+16
=21-x(x+4)+4(x+4)
=21-(x+4)(x+4)
=21-(x+4)2$\le$≤0+21=21
Dấu = khi x+4=0 <=>x=-4
b)đề sai à
a) Ta có: \(a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(m-n\right)\left(a+1\right)\)
b) Ta có: \(mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(m+5\right)\)
c) Ta có: \(ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(a+b\right)\left(m-1\right)\)
d) Ta có: \(1-xa-x+a\)
\(=\left(a+1\right)-x\left(a+1\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
e) Ta có: \(\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-b+a+b\right)\)
\(=2a\left(a-b\right)\)
f) Ta có: \(a\left(a-b\right)\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)\)
\(=b^2\cdot\left(a+b\right)\)
g) Ta có: \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]\)
\(=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)\)
\(=\left(x+7\right)\left(-8x^2+24x-3x+9\right)\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]\)
\(=\left(x+7\right)\left(x-3\right)\left(-8x-3\right)\)
h) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
i) Ta có: \(2x\left(x-3\right)-3\left(x-3\right)^2\)
\(=\left(x-3\right)\left[2x-3\left(x-3\right)\right]\)
\(=\left(x-3\right)\left(2x-3x+9\right)\)
\(=\left(x-3\right)\left(9-x\right)\)
j) Ta có: \(x\left(x-7\right)+\left(7-x\right)^2\)
\(=x\left(x-7\right)+\left(x-7\right)^2\)
\(=\left(x-7\right)\left(x+x-7\right)\)
\(=\left(x-7\right)\left(2x-7\right)\)
k) Ta có: \(3x\left(x-9\right)^2-\left(9-x\right)^3\)
\(=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\cdot\left(3x+x-9\right)\)
\(=\left(x-9\right)^2\cdot\left(4x-9\right)\)
a, 5x(x-2) + (2-x)=0
⇔5x(x-2) - (x-2) =0
⇔(x-2)(5x-1)=0
\(\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy....
c, (x3 - x2) - 4x2 + 8x -4 =0
⇔x3 - x2 -4x2 + 8x - 4=0
⇔x2(x-1) - 4x(x-1) +4(x-1) =0
⇔(x-1) (x-2)2=0
⇔\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
Phần b cậu có chép sai đề không?
(x^2-6x+8)(x^2-8x+15)+1
=(x^2-4x-2x+8)(x^2-5x-3x+15)+1
=(x(x-4)-2(x-4))(x(x-5)-3(x-5))+1
=(x-4)(x-2)(x-5)(x-3)+1
=(x-2)(x-5)(x-3)(x-4)+1
=(x^2-7x+10)(x^2-7x+12)+1
Gọi a=x^2-7x+11, ta có
(a-1)(a+1)+1
= a2 - 1 + 1
= a2
= (x2 - 7x + 11)2
thay x= -1 vào phương trình rồi giải ra
\(2\left(x+n\right)\left(x+2\right)-3\left(x-1\right)\left(x^2-1\right)=15\)
Thay \(x=-1\) vào biểu thức, ta có:
\(2\left[\left(-1\right)+n\right]\left[\left(-1\right)+2\right]-3\left[\left(-1\right)-1\right]\left[\left(-1\right)^2-1\right]=15\)
\(\Leftrightarrow2\left(1-2-n+2n\right)-3\left(-2\times0\right)=15\)
\(\Leftrightarrow2-4-2n+4n=15\)
\(\Leftrightarrow2-4-2n+4n-15=0\)
\(\Leftrightarrow2n-17=0\)
\(\Leftrightarrow2n=17\)
\(\Leftrightarrow n=\dfrac{17}{2}\)
Vậy \(n=\dfrac{17}{2}\)