1.

\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)

\(=2x^3y^2-3x^2y^2+7x^2y\)

\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)

\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)

\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)

\(=\left(x+y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3\)

\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x-y\right)^3\)

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

2.

\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)

\(=x^3-y^3\)

\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=x^3+y^3\)

\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)

\(=24xy+4x-6y-1-24xy-4x\)

\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)

\(=-6y-1\)

#Toru

áp dụng BĐT\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\)(x,y>0)

=>A=\(\frac{1}{xy}+\frac{2}{x^2+y^2}=\frac{2}{2xy}+\frac{2}{x^2+y^2}=2\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)>=\frac{2.4}{2xy+X^2+Y^2}=\frac{8}{\left(x+y\right)^2}=8\)

dấu bằng xảy ra khi x=y=1/2

a. Trừ vế theo vế \(\left(1\right)\) cho \(\left(2\right)\) ta được \(x^2-y^2=4x-4y\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=4-y\end{matrix}\right.\)

TH1: \(x=y\)

Phương trình \(\left(1\right)\) tương đương:

\(x^2=2x\)

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

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

TH2: \(x=4-y\)

Phương trình \(\left(2\right)\) tương đương:

\(y^2=4y-4\)

\(\Leftrightarrow y^2-4y+4=0\)

\(\Leftrightarrow\left(y-2\right)^2=0\)

\(\Leftrightarrow y=2\)

\(\Rightarrow x=2\)

Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right);\left(2;2\right)\right\}\)

b. \(\left\{{}\begin{matrix}x+y+xy=5\\x^2+y^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2-10+2\left(x+y\right)=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y\right)^2+2\left(x+y\right)-15=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left(x+y+5\right)\left(x+y-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=5-\left(x+y\right)\\\left[{}\begin{matrix}x+y=-5\\x+y=3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-5\\xy=10\end{matrix}\right.\Leftrightarrow\) vô nghiệm

TH2: \(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\). khi đó gt trở thành:

\(a+b=a^2+b^2-ab\ge\dfrac{1}{4}\left(a+b\right)^2\Leftrightarrow o\le a+b\le4\);

\(A=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=\left(a+b\right)^2\le16\)

Đẳng thức xảy ra khi và chỉ khi a=b=2 <=> x=y=1/2

Vậy Max A = 16

1) 

Ta có: x+y=2

nên \(\left(x+y\right)^2=4\)

\(\Leftrightarrow x^2+y^2+2xy=4\)

\(\Leftrightarrow2xy=2\)

hay xy=1

Ta có: \(x^3+y^3\)

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

\(=2^3-3\cdot1\cdot2\)

=2

2)\(x^2+y^2=\left(x+y\right)^2-2xy=8^2-2\cdot\left(-20\right)=104\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=8^3-3\cdot\left(-20\right)\cdot8=512+480=992\)

\(x^2+y^2+xy=\left(x+y\right)^2-xy=8^2-\left(-20\right)=64+20=84\)

Yêu cầu đề là gì vậy bạn?

a, (x^2 -2x+1)+(y^2 +6y+9) =0

(x-1)^2 +(y+3)^2 =0

Do đó: x-1=0 và y+3=0

Vậy x=1 và y=-3

b, x^2 +y^2 +1=xy+x+y

2x^2 +2y^2 +2=2xy+2x+2y

2x^2 +2y^2 -2xy-2x-2y +2=0

(x^2 -2x+1)+(y^2 -2y+1)+ (x^2 +y^2 -2xy)=0

(x-1)^2 +(y-1)^2 +(x-y)^2 =0

Suy ra: x-1=0, y-1=0 và x-y=0

Vậy x=1,y=1

c,5x^2 - 4x-2xy+y^2 +1=0

(4x^2 -4x+1)+(x^2 -2xy+y^2 )=0

(2x-1)^2 +(x-y)^2 =0

Do đó: 2x-1 =0 và x=y suy ra: x=0,5 và x=y

Vậy x=y=0,5

Bài 13:

a) \(501^2\)

\(=\left(500+1\right)^2\)

\(=500^2+2\cdot500\cdot1+1^2\)

\(=250000+1000+1\)

\(=251001\)

b) \(88^2+24\cdot88+12^2\)

\(=88^2+2\cdot12\cdot88+12^2\)

\(=\left(88+12\right)^2\)

\(=100^2\)

\(=10000\)

c) \(52\cdot48\)

\(=\left(50+2\right)\left(50-2\right)\)

\(=50^2-2^2\)

\(=2500-4\)

\(=2496\)

Bài 14:

a) \(P=\left(2x-1\right)\left(4x^2+2x+1\right)+\left(x+1\right)\left(x^2-x+1\right)\)

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

\(P=8x^3+x^3\)

\(P=9x^3\)

b) \(Q=\left(x-y\right)\left(x^2+xy+y^2\right)-\left(x+y\right)\left(x^2-xy+y^2\right)+2y^3\)

\(Q=x^3-y^3-x^3-y^3+2y^3\)

\(Q=-2y^3+2y^3\)

\(Q=0\)

Bài `14`

`a. P = ( 2x - 1 ) ( 4x^2 + 2x + 1 ) + ( x + 1 ) ( x^2 -x+1)`

`=(2x)^3-1^3 + x^3+1^3`

`=8x^3-1+x^3+1`

`= 9x^3`

__

`b, Q = ( x - y ) ( x^2 + xy + y^2 ) - ( x + y ) ( x^2 - xy + y^2)+2y^3`

`=x^3-y^3 -(x^3+y^3)+2y^3`

`=x^3-y^3 -x^3-y^3+2y^3`

`= 0`