x³ - x² - x = 0

<=> x( x² - x - 1 ) = 0

<=> x = 0 hoặc x² - x - 1 = 0

+) x² - x - 1 = 0

Δ = b² - 4ac = 1 + 4 = 5

Δ > 0, áp dụng công thức nghiệm thu được \(x=\frac{1\pm\sqrt{5}}{2}\)

Vậy ...

\(5+\frac{8}{x^2-4}=\frac{2x-1}{x+2}-\frac{3x-1}{2-x}\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow5+\frac{8}{\left(x-2\right)\left(x+2\right)}=\frac{2x-1}{x+2}+\frac{3x-1}{x-2}\)

\(\Leftrightarrow\frac{5\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\)\(=\frac{\left(2x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(3x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow5\left(x-2\right)\left(x+2\right)+8=\)\(\left(2x-1\right)\left(x-2\right)+\left(3x-1\right)\left(x+2\right)\)

\(\Leftrightarrow5\left(x^2-4\right)+8=2x^2-4x-x+2\)\(+3x^2+6x-x-2\)

\(\Leftrightarrow5x^2-20+8=\)\(5x^2\)

\(\Leftrightarrow-12=5x^2-5x^2\)

\(\Leftrightarrow0=-12\)(vô nghiệm).

Vậy phương trình đã cho vô nghiệm.

Đặt \(A=x^4-2y^4-x^2y^2+x^2+y^2\)

\(\Rightarrow2A=2x^4-4y^4-2x^2y^2+2x^2+2y^2\)

\(\Rightarrow2A=\left(x^4+2x^2+1\right)-\left(y^4-2y^2+1\right)\)\(+\left(x^4-2x^2y^2+y^4\right)-4y^4\)

\(\Rightarrow2A=\left(x^2+1\right)^2-\left(y^2-1\right)^2+\left(x^2-y^2\right)^2-4y^4\)

\(\Rightarrow2A=\left[\left(x^2+1\right)^2-4y^4\right]+\left[\left(x^2-y^2\right)^2-\left(y^2-1\right)^2\right]\)

\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2\right)+\)\(\left(x^2-y^2+y^2-1\right)\left(x^2-y^2-y^2+1\right)\)

\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2\right)+\)\(\left(x^2-1\right)\left(x^2+1-2y^2\right)\)

\(\Rightarrow2A=\left(x^2+1-2y^2\right)\left(x^2+1+2y^2+x^2-1\right)\)

\(\Rightarrow2A=\left(x^2-2y^2+1\right)\left(2x^2+2y^2\right)\)

\(\Rightarrow2A=2\left(x^2-2y^2+1\right)\left(x^2+y^2\right)\)

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

Nhầm, tớ chốt lại: \(A=\left(x^2-2y^2+1\right)\left(x^2+y^2\right)\), đừng xem cái câu cuối ở tin 1, sai đấy.

xin nhá xin nhá =))

Áp dụng bất đẳng thức Cauchy-Schwarz và giả thiết x+y=1 ta có :

\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\ge\frac{\left(2x+\frac{1}{x}+2y+\frac{1}{y}\right)^2}{2}=\frac{\left[2\left(x+y\right)+\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2}{2}\ge\frac{\left(2+\frac{4}{x+y}\right)^2}{2}=\frac{\left(2+4\right)^2}{2}=18\)

Đẳng thức xảy ra <=> x=y=1/2

Vậy ...

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\left(ĐKXĐ:x\ne0;x\ne2\right)\)

\(\Rightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(KTMĐK\right)\\x=-1\left(TMĐK\right)\end{cases}}\)

\(\Rightarrow S=\left\{-1\right\}\)

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\left(x\ne0;x\ne2\right)\)

<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

=> x² + 2x - x + 2 = 2

<=> x² + x = 0

<=> x( x + 1 ) = 0

<=> x = 0 (ktm) hoặc x = -1(tm)

Vậy ...

( 2x - 3 )( x² + 1 ) = 0

Vì x² + 1 ≥ 1 > 0 ∀ x

=> 2x - 3 = 0 <=> x= 3/2

Vậy pt có nghiệm x = 3/2

Trả lời:

( 2x - 3 ) ( x² + 1 ) = 0

<=> 2x - 3 = 0 hoặc x² + 1 = 0

<=> 2x = 3 hoặc x² = -1 ( vô lí )

<=> x = 3/2

Vậy S = { 3/2 }