Nhìn mãi mới hiểu cái đề bài @-@

`a)đk:` $\begin{cases}\sqrt{x^2-2x} \ge 0\\x+\sqrt{x^2-2x} \ne 0\\x-\sqrt{x^2-2x} ne 0\\\end{cases}$
`<=>` $\begin{cases}x \ge 2\,or\,x<0\\x \ne 0\end{cases}$
`b)A=(x+sqrt{x^2-2x})/(x-sqrt{x^2-2x})-(x-sqrt{x^2-2x})/(x+sqrt{x^2+2x})`
`=((x+sqrt{x^2-2x})^2-(x-sqrt{x^2-2x})^2)/((x+sqrt{x^2-2x})(x-sqrt{x^2-2x}))`
`=(x^2+x^2-2x+2sqrt{x^2-2x}-x^2-x^2+2x+2sqrt{x^2-2x})/(x^2-x^2+2x)`
`=(4sqrt{x^2-2x})/(2x)`
`=(2sqrt{x^2-2x})/x`
`c)A<2`
`<=>2sqrt{x^2-2x}<2x`
`<=>sqrt{x^2-2x}<x(x>=2)`(BP 2 vế thì x>=2)
`<=>x^2-2x<x^2`
`<=>2x>0`
`<=>x>0`
`<=>x>=2`
Vậy `x>=2` thì `A<2`.

a: ĐKXĐ: x<>2; x<>0

b: \(M=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{x^2-x-2}{x^2}\)

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

\(=\dfrac{x^3-2x^2-2x^2+4x}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)

\(=\dfrac{x}{2}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow\left(x^3+2^3\right)-\left(x^3-2x\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x=15\)

\(\Leftrightarrow2x+8=15\) 

\(\Leftrightarrow2x=15-8\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) \(\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)

\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4=6\)

\(\Leftrightarrow-8x+20=6\)

\(\Leftrightarrow-8x=6-20\)

\(\Leftrightarrow-8x=-14\)

\(\Leftrightarrow x=\dfrac{7}{4}\) 

c) \(x^4-2x^3+x^2-2x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^3+x\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x^2+1\right)\left(x-2\right)=0\)

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

`(x+2)(x^2 -2x+4) -x(x^2-2)=15`

`<=> x^3 +8 - x^3 + 2x-15=0`

`<=> 2x-7=0`

`<=> 2x=7`

`<=>x=7/2`

__

`(x-4)^2 -(x-2)(x+2)=6`

`<=>x^2 - 8x+16- x^2 +4-6=0`

`<=> -8x+14=0`

`<=> -8x=-14`

`<=>x=14/8= 7/4`

__

`x^4 -2x^3 +x^2-2x=0`

`<=>x(x^3-2x^2+x-2)=0`

`<=> x(x^3+x-2x^2-2)=0`

`<=>x(x(x^2+1) -2(x^2+1))=0`

`<=> x(x^2+1)(x-2)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

( x + 2 )³ - ( 2x + 3 )² + ( 2x + 3 )( 2x - 3 ) = ( x - 2 )( x² + 2x + 4 ) - 6x( x + 2 )

⇔ x³ + 6x² + 12x + 8 - ( 4x² + 12x + 9 ) + 4x² - 9 = x³ - 8 - 6x² - 12x

⇔ x³ + 10x² + 12x - 1 - 4x² - 12x - 9 = x³ - 6x² - 12x - 8

⇔ x³ + 6x² - 10 = x³ - 6x² - 12x - 8

⇔ x³ + 6x² - 10 - x³ + 6x² + 12x + 8 = 0

⇔ 12x² + 12x - 2 = 0 

⇔ 2( 6x² + 6x - 1 ) = 0

⇔ 6x² + 6x - 1 = 0 (*)

Δ = b² - 4ac = 6² - 4.6.(-1) = 60

Δ > 0 nên (*) có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+\sqrt{60}}{12}=\frac{-3+\sqrt{15}}{6}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-\sqrt{60}}{12}=\frac{-3-\sqrt{15}}{6}\end{cases}}\)

Vậy ...

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a, \(P\left(x\right)=4x^3+2x-3+2x-2x^2-1\\ =4x^3-2x^2+\left(2x+2x\right)+\left(-3-1\right)\\ =4x^3-2x^2+4x-4\)

Bậc của P(x) là 3

\(Q\left(x\right)=6x^3-3x+5-2x+3x^2\\ =6x^3+3x^2+\left(-3x-2x\right)+5\\ =6x^3+3x^2-5x+5\)

Bậc của Q(x) là 3

b, \(M\left(x\right)=P\left(x\right)+Q\left(x\right)=4x^3-2x^2+4x-4+6x^3+3x^2-5x+5\\ =\left(4x^3+6x^3\right)+\left(-2x^2+3x^2\right)+\left(4x-5x\right)+\left(-4+5\right)\\ =10x^3+x^2-x+1\)

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

d. Áp dụng BĐT Caushy Schwartz ta có:

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}\le x+y+\dfrac{\left(1+1\right)^2}{x+y}=x+y+\dfrac{4}{x+y}\le1+\dfrac{4}{1}=5\)

-Dấu bằng xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

c. Bạn kiểm tra lại đề nhé.

b. \(5x\left(2-x\right)=-5x\left(x-2\right)=-5\left(x^2-2x\right)=-5\left(x^2-2x+1-1\right)=-5\left(x-1\right)^2+5\le5\)-Dấu bằng xảy ra \(\Leftrightarrow x=1\)

a.

\(\left(80-2x\right)\left(50-2x\right)x=\dfrac{2}{3}\left(40-x\right)\left(50-2x\right)3x\le\dfrac{2}{3}\left(\dfrac{40-x+50-2x+3x}{3}\right)^3=18000\)

Dấu "=" xảy ra khi \(40-x=50-2x=3x\Leftrightarrow x=10\)

b.

\(5x\left(2-x\right)=5.x\left(2-x\right)\le\dfrac{5}{4}\left(x+2-x\right)^2=5\)

Dấu "=" xảy ra khi \(x=2-x\Rightarrow x=1\)

c.

Biểu thức này chỉ có min, ko có max

d.

\(x+y\le1\Rightarrow-\left(x+y\right)\ge-1\)

\(x+y+\dfrac{1}{x}+\dfrac{1}{y}=\left(4x+\dfrac{1}{x}\right)+\left(4y+\dfrac{1}{y}\right)-3\left(x+y\right)\ge2\sqrt{\dfrac{4x}{x}}+2\sqrt{\dfrac{4y}{y}}-3.1=5\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`