qưe

1a)

ĐKXĐ : 

x\(\ne\)0 ;x+1\(\ne\)0

<=>x\(x\ne0;x\ne-1\)

b)

3/x = 2/x+1

<=>3(x+1) / x(x+1) = 2x / x( x + 1 )

<=>3(x+1)=2x <=> 3x+3=2x

<=>x=-3(thỏa ĐKXĐ)

Vậy S={-3}

2)

\(x+2\ge0\)

<=>\(x\ge-2\)

Vậy S={ \(x\)/\(x\ge-2\)}

0 -2

Vì a>b(1) nên

nhân hai vế bất đẳng thức(1) cho 4 ta được:4a>4b(2)

cộng hai vế bất đẳng thức(2) cho 3 ta được : 4a+3>4b+3

e ngu

a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)

ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)

b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)

\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)

c) \(P=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{\left(x+2\right)\left(2x+1-2\right)}{2x+1}=\frac{\left(x+2\right)\left(2x+1\right)-2\left(x+2\right)}{2x+1}\)

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

để P nguyền khi zà chỉ khi

\(1⋮2x+1\)

\(=>2x+1\inƯ\left(1\right)=\pm1\)

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

Ta có: \(x^4+x^2+1=x^4+x^3+x^2-\left(x^3-1\right)=x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\)

Do \(x^4+x^2+1⋮x^2+ax+b\) nên \(\orbr{\begin{cases}x^2+ax+b=x^2+x+1\\x^2+ax+b=x^2-x+1\end{cases}\Rightarrow}\hept{\begin{cases}a=1\\b=1\end{cases}}\)hoặc \(\hept{\begin{cases}a=-1\\b=1\end{cases}}\)

Vậy ...

a) 9x² – 6x + 1 = (3x)² – 2 . 3x . 1 + 1² = (3x – 1)²

Hoặc 9x² – 6x + 1 = 1 – 6x + 9x² = (1 – 3x)²

b) (2x + 3y) = (2x + 3y)² + 2 . (2x + 3y) . 1 + 1²

= [(2x + 3y) + 1]²

= (2x + 3y + 1)²

Đề bài tương tự. Chẳng hạn:

1 + 2(x + 2y) + (x + 2y)²

4x² – 12x + 9…