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 \(x=y=\dfrac{1}{2}\)

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

^^

\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)

\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)

\(\text{⇔}8x+5=-11\) 

\(\text{⇔}8x=-16\)

\(\text{⇔}x=-2\)

Vậy: \(x=-2\)

==========

\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)

\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x=-\dfrac{4}{3}\)

\(\text{⇔}x=\dfrac{2}{39}\)

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

a: ĐKXĐ: x<>0; x<>1

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

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

b: |2x+1|=3

=>x=1(loại); x=-2(nhận)

Khi x=-2 thì P=4/-3=-4/3

c: P=-1/2

=>x^2/x-1=-1/2

=>2x^2=-x+1

=>2x^2+x-1=0

=>2x^2+2x-x-1=0

=>(x+1)(2x-1)=0

=>x=1/2; x=-1

nany???

ai cho copy bài làm của tui

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: ta có: \(\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow2x^2+4x-5x-10-2x^2+2x=15\)

\(\Leftrightarrow x=25\)

b: Ta có: \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

c: Ta có: \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x=1\)

hay \(x=-\dfrac{1}{9}\)

a: ta có: $\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15$

$\iff 2{}^{}+4x-5x-10-2{}^{}+2x=15$

$\iff x=25$

b: Ta có: $\left(5-2x\right)\left(2x+7\right)=4{}^{}-25$

$\iff 4{}^{}-25+\left(2x-5\right)\left(2x+7\right)=0$

$\iff \left(2x-5\right)(2x+5+2x+$