a)Ta có:

\(A=4-x^2+2x=-\left(x^2-2x-4\right)=-\left(x^2-2x+1+3\right)\)

\(=-\left(x^2-2x+1\right)-3=-\left(x-1\right)^2-3\le-3\forall x\)

Vậy Max_A=-3 khi x=1

b) Ta có: \(B=4x-x^2=-\left(x^2-4x\right)=-\left(x^2-4x+4-4\right)=-\left(x-2\right)^2+4\le4\forall x\)Vậy Max_B=4 khi x=2

Bài 3: Tìm GTLN

a) Ta có: \(A=4-x^2+2x\)

\(=-\left(x^2-2x-4\right)\)

\(=-\left(x^2-2x+1-5\right)\)

\(=-\left(x-1\right)^2+5\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-1\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x-1=0

hay x=1

Vậy: GTLN của biểu thức \(A=4-x^2+2x\) là 5 khi x=1

b) Ta có: \(B=4x-x^2\)

\(=-\left(x^2-4x\right)\)

\(=-\left(x^2-4x+4-4\right)\)

\(=-\left(x-2\right)^2+4\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-\left(x-2\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-2\right)^2+4\le4\forall x\)

Dấu '=' xảy ra khi x-2=0

hay x=2

Vậy: GTLN của biểu thức \(B=4x-x^2\) là 4 khi x=2

a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)

(x-1)^2+1>=1

=>\(\sqrt{\left(x-1\right)^2+1}>=1\)

=>\(B< =0\)

Dấu = xảy ra khi x=1

b: 

ĐKXĐ: -(x+2)^2+2>=0

=>-(x+2)^2>=2

=>(x+2)^2<=2

=>\(-\sqrt{2}-2< =x< =\sqrt{2}-2\)

\(-x^2+4x-2=-\left(x^2-4x+2\right)\)

\(=-\left(x^2-4x+4-2\right)=-\left(x-2\right)^2+2< =2\)

=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)

=>1<=C<=căn 2+1

\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12

ta có 4 x 3 y 2   –   8 x 2 y 3   =   4 x 2 y 2 . x   –   4 x 2 y 2 . 2 y   =   4 x 2 y 2 ( x   –   2 y )    

Vậy 4x³y² – 8x²y³ = 4x²y²(x – 2y)      

Đáp án cần chọn là: C

bấm đúng cho mik đi 

Bài 1:

a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)

\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)

\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

\(1a,8x^2y^2-12x^3+6x^2\)

\(=2\left(4x^2y^2-13x^3+3x^2\right)\)

\(b,5x\left(x-y\right)-\left(x-y\right)\)( sai đề hả )

\(=\left(x-y\right)\left(5x-1\right)\)

\(c,4x\left(x-2\right)-\left(2-x\right)^2\)

\(=4x\left(x-2\right)-\left(x-2\right)^2\)

\(=\left(x-2\right)\left(4x-x+2\right)=\left(x-2\right)\left(3x+2\right)\)

\(2,\)\(x\left(x-3\right)-\left(3-x\right)=0\)

\(\Rightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)

phần b làm theo đề thôi nhé ko biết đầu bài đúng ko

\(5x\left(x-y\right)-\left(y-y\right)\)

\(=5x\left(x-y\right)\)

HA ha ngắn gọn vãi

b) 5x.(x - y) - (y - y)

= 5x.(x - y)

c) 4x.(x - 2) - (2 - x)²

= 4x² - 8x + (x - 2)²

= 4x² - 8x + x² - 4x + 4

= 5x² - 12x + 4

= 5x² - 10x - 2x + 4

= 5x.(x - 2) - 2.(x - 2)

= (x - 2).(5x - 2)

d) 2x.(x - 3) - (3 - x) = 0

2x² - 6x + x + 3 = 0

2x² - 5x + 3 = 0

2x² - 2x - 3x + 3 = 0

2x.(x - 1) - 3.(x - 1) = 0

(x - 1).(2x - 3) = 0

=> x - 1 = 0   hoặc

     2x - 3 = 0

=> x = 1 hoặc x = \(\frac{3}{2}\)

~ Học tốt ~