a, =x-z

b,=-9(2-x)^2/4

c,=x+1

d,=(x-1)/(x+1)

hok tốt

a) =\(\frac{x-z}{2}\)

a) Để P xác định \(\Leftrightarrow\hept{\begin{cases}2a-2\ne0\\2-2a^2\ne0\\a+2\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a^2\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1vâ\ne1\\a\ne-2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)

b) \(P=\left(\frac{a+1}{2a-2}+\frac{1}{2-2a^2}\right).\frac{2a+2}{a+2}\)

\(=\left[\frac{a+1}{2\left(a-1\right)}+\frac{1}{2\left(1-a\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\left[\frac{\left(a+1\right)^2}{2\left(a-1\right)\left(a+1\right)}-\frac{1}{2\left(a-1\right)\left(1+a\right)}\right].\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{\left(a+1\right)^2-1}{2\left(a-1\right)\left(a+1\right)}.\frac{2\left(a+1\right)}{a+2}\)

\(=\frac{a\left(a+2\right)}{\left(a-1\right)\left(a+2\right)}\)

\(=\frac{a}{a-1}\)

c) \(\left|a\right|=3\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

+) Với a=3 thỏa mãn \(\hept{\begin{cases}a\ne1\\a\ne-1\\a\ne2\end{cases}}\)nên thay a=3 vào P ta được:

( làm nốt)

TH kia tương tự

1) (x - 5)² - (x + 3)(x - 3) = 14

=> x² - 10x + 25 - x² + 9 = 14

=> -10x + 34 = 14

=> -10x = 14 - 34

=> -10x = -20

=> x = 2

2) (x + 7)² - 3x - 21 = 0

=> x² + 14x + 49 - 3x - 21 = 0

=> x² + 11x + 28 = 0

=> x² + 4x + 7x + 28 = 0

=> x(x + 4) + 7(x + 4) = 0

=> (x + 7)(x + 4) = 0

=> \(\orbr{\begin{cases}x+7=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=-4\end{cases}}\)

a

\(\left(x-5\right)^2-\left(x+3\right)\left(x-3\right)=14\)

\(\Leftrightarrow x^2-10x+25-x^2+9=14\)

\(\Leftrightarrow-10x=-20\)

\(\Leftrightarrow x=2\)

b

\(\left(x+7\right)^2-3x-21=0\)

\(\Leftrightarrow x^2+14x+49-3x-21=0\)

\(\Leftrightarrow x^2+11x+28=0\)

\(\Leftrightarrow\left(x^2+4x\right)+\left(7x+28\right)=0\)

\(\Leftrightarrow x\left(x+4\right)+7\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=0\)

\(\Leftrightarrow x=-4;x=-7\)

\(=4x^2+12x+9-4x^2-12x\)

\(=0+0+9\)

\(=9\)

\(=x^2-5x+6x-30\)

\(=x\left(x-5\right)+6\left(x-5\right)\)

\(=\left(x+6\right)\left(x-5\right)\)

\(x^2+x-30\)

\(=\left(x^2-5x\right)+\left(6x-30\right)\)

\(=x\left(x-5\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x+6\right)\)

\(25-a^2+2ab-b^2\)

\(=25-\left(a^2-2ab+b^2\right)\)

\(=5^2-\left(a-b\right)^2\)

\(=\left(5-a+b\right)\left(5+a-b\right)\)

\(=25-\left(a-b\right)^2\)

\(=\left(25-a+b\right)\left(25+a-b\right)\)

\(M=3x^2+6x+9\)

\(M=3\left(x^2+2x+3\right)\)

\(M=3\left(x^2+2x+1+2\right)\)

\(M=3\left[\left(x+1\right)^2+2\right]\)

\(M=3\left(x+1\right)^2+6\)

\(\left(x+1\right)^2\ge0\)

\(\Rightarrow3\left(x+1\right)^2\ge0\)

\(\Rightarrow3\left(x+1\right)^2+6\ge6\)

Vậy biểu thức M luôn luôn dương \(\forall x\)

\(M=3x^2+6x+9=3x^2+6x+3+6\)

    \(=3\left(x^2+2x+1\right)+6\)\(=3\left(x+1\right)^2+6\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow M\ge6\forall x\)\(\Rightarrow\)M luôn dương ( đpcm )