\(A=-3\left(x+1\right)^2+7\le7\)

\(A_{max}=7\) khi \(x=-1\)

\(B=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)

\(B_{max}=\frac{5}{4}\) khi \(x=\frac{3}{2}\)

\(C=-x^2-2x+2=-\left(x+1\right)^2+3\le3\)

\(C_{max}=3\) khi \(x=-1\)

\(D=-\left[\left(x+2y\right)^2+\left(x-1\right)^2-4\right]=-\left(x+2y\right)^2-\left(x-1\right)^2+4\le4\)

\(D_{max}=4\) khi \(\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)

1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)

\(\dfrac{x+y}{2x^2+4xy+2y^2}\)

\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)

\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)

\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)

\(=8x^4y^2+4xy^4-28x^2y^3\)

B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)

\(=10x^4-5x^2-5x-4x^3+2x+2\)

\(=10x^4-5x^3-3x-4x^3+2\)

C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)

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

D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)

( Bài này áp dụng hằng đẳng thức là làm được ạ )

Thực hiện phép tính:

³.4xy²+y².4xy²-7xy.4xy²

^=>8x⁴y²+4xy⁴-28x²y³

c: =(2x^2-3)[(2x^2)^2+2x^2*3+3^2]

=8x^6-27

d:\(=\left(3x^2-2y-2x^2+y\right)\left(9x^4-12x^2y+4y^2+6x^4-3x^2y-4x^2y+2y^2+4x^4-4x^2y+y^2\right)\)

\(=\left(x^2-y\right)\left(19x^4-23x^2y+7y^2\right)\)

1 M=\(x^2-4xy+4y^2-2x+4y+10\)

=\(\left(x^2-4xy+4y^2\right)+\left(-2x+4y\right)+10\)

\(=\left(x-2y\right)^2-2\left(x-2y\right)+10\)

\(=\left(x-2y\right)\left(x-2y-2\right)+10\)

vì \(\left(x-2y\right)\left(x-2y-2\right)\ge0\)

nên \(\left(x-2y\right)\left(x-2y-2\right)+10\ge10\)

\(\Rightarrow\)A\(\ge13\)

dấu "=" xảy ra khi (x-2y)(x-2y-2)=0

\(\left[{}\begin{matrix}x-2y=0\\x-2y-2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}2y=x\\x-2y=2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0;y=0\\x=2;y=1\end{matrix}\right.\)

vậy GTNN của M=10 khi x=0; y=0

x=2;y=1

đề bài là : dùng hằng đẳng thức để khai triển và thu gọn các biểu thức

a)2x^2-4xy+4y^2+2x+5=x^2-4xy+4y^2+x^2+2x+1+4=(x-2y)^2+(x+1)^2+4>=4(dấu = tự tìm nhé)

b)x(1-x)(x-3)(4-x)=x(x-1)(x-3)(x-4)

=(x^2-4x)(x^2-4x+3)

Đặt x^2-4x=t(t>=-4) bt viết lại t(t+3)=t^2+3t>=-9/4

Dấu= xảy ra khi t=-3/2 >>>tìm x