a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

1) đkxđ \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\y\ge0\end{matrix}\right.\)

Xét biểu thức \(P=x^3+y^3+7xy\left(x+y\right)\)

\(P=\left(x+y\right)^3+4xy\left(x+y\right)\)

\(P\ge4\sqrt{xy}\left(x+y\right)^2\)

Ta sẽ chứng minh \(4\sqrt{xy}\left(x+y\right)^2\ge8xy\sqrt{2\left(x^2+y^2\right)}\)  (*)

Thật vậy, (*)

\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{2xy\left(x^2+y^2\right)}\)

\(\Leftrightarrow\left(x+y\right)^4\ge8xy\left(x^2+y^2\right)\)

\(\Leftrightarrow x^4+y^4+6x^2y^2\ge4xy\left(x^2+y^2\right)\) (**)

Áp dụng BĐT Cô-si, ta được:

VT(**) \(=\left(x^2+y^2\right)^2+4x^2y^2\ge4xy\left(x^2+y^2\right)\)\(=\) VP(**)

Vậy (**) đúng \(\Rightarrowđpcm\). Do đó, để đẳng thức xảy ra thì \(x=y\). 

Thế vào pt đầu tiên, ta được \(\sqrt{2x-3}-\sqrt{x}=2x-6\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{2x-3}+\sqrt{x}}=2\left(x-3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}=2\end{matrix}\right.\)

 Rõ ràng với \(x\ge\dfrac{3}{2}\) thì \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}\le\dfrac{1}{\sqrt{\dfrac{2.3}{2}-3}+\sqrt{\dfrac{3}{2}}}< 2\) nên ta chỉ xét TH \(x=3\Rightarrow y=3\) (nhận)

Vậy hệ pt đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(3;3\right)\)

a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)

\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

a , x(3-x)-y(y-2x)+y(y-2)-x(2y-x)

=3x-x²-y²+2xy+y²-2y-2xy+x²

=3x-(-x²+x²)+(2xy-2xy)-2y

=3x-2y

b,(x²-7)(x+2)-(2x-1)(x-14)+x(x²-2x-22)+35

=x³+2x²-7x-14-2x²+29x-14+x³-2x²-22x+35

=(x³+x³)+(2x²-2x²-2x²)-(-7x+29x-22x)-14-14+35

=2x³-2x²+7

c,(2x+y)(4x²-2xy+y²)-8x³-y³

=2xy²-4x²y+8x³+y³-2xy²+4x²y-8x³-y³

=(2xy²-2xy²)-(-4x²y+4x²y)+(8x³-8x³)+(y³-y³)

=0 

C với D mình làm sau vì nó phức tạp hơn ... E với F trước nhé

E = | 3x + 1 | + 2| x - y | + 1

\(\hept{\begin{cases}\left|3x+1\right|\ge0\\2\left|x-y\right|\ge0\end{cases}\forall}x,y\Rightarrow\left|3x+1\right|+2\left|x-y\right|+1\ge1\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+1=0\\x-y=0\end{cases}}\Leftrightarrow x=y=-\frac{1}{3}\)

=> MinE = 1 <=> x = y = -1/3

F = 5| x - 1 | + 1/2| 2x + y | + 2020

\(\hept{\begin{cases}5\left|x-1\right|\ge0\\\frac{1}{2}\left|2x+y\right|\ge0\end{cases}\forall}x,y\Rightarrow5\left|x-1\right|+\frac{1}{2}\left|2x+y\right|+2020\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\2x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

=> MinF = 2020 <=> x = 1 ; y = -2

C = 2| x - 1 | + | 2x + 3 | - 2020

= | 2x - 2 | + | 2x + 3 | - 2020

= | 2x - 2 | + | -( 2x + 3 ) | - 2020

= | 2x - 2 | + | -2x - 3 | - 2020

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :

C = | 2x - 2 | + | -2x - 3 | - 2020 ≥ | 2x - 2 - 2x - 3 | - 2020 = | -5 | - 2020 = 5 - 2020 = -2015

Dấu "=" xảy ra khi ab ≥ 0

=> ( 2x - 2 )( -2x - 3 ) ≥ 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-2\ge0\\-2x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge2\\-2x\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le-\frac{3}{2}\end{cases}}\)( loại )

2. \(\hept{\begin{cases}2x-2\le0\\-2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le2\\-2x\le3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-\frac{3}{2}\end{cases}}\Leftrightarrow-\frac{3}{2}\le x\le1\)

=> MinC = -2015 <=> \(-\frac{3}{2}\le x\le1\)

D = | 3 - 2x | + 2| 1 - x | + 1/2

= | 3 - 2x | + | 2 - 2x | + 1/2

= | -( 3 - 2x ) | + | 2 - 2x | + 1/2

= | 2x - 3 | + | 2 - 2x | + 1/2

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :

D = | 2x - 3 | + | 2 - 2x | + 1/2 ≥ | 2x - 3 + 2 - 2x | + 1/2 = | -1 | + 1/2 = 1 + 1/2 = 3/2

Dấu "=" xảy ra khi ab ≥ 0

=> ( 2x - 3 )( 2 - 2x ) ≥ 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-3\ge0\\2-2x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge3\\-2x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( loại )

2. \(\hept{\begin{cases}2x-3\le0\\2-2x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le3\\-2x\le-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)

=> MinD = 3/2 <=> \(1\le x\le\frac{3}{2}\)

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