a) A = [2t²(m - 1) - t(m - 1)(2t - 1)] + t + m

A = t(m - 1)[2t - (2t - 1)] + t + m

A = t(m - 1) + t + m

A = tm + m

b) Với m = 2; A = 0 thì ta được pt:

0 = 2t + 2

⇔ t = -1

Vậy khi m = 2 và để A = 0 thì t = -1

\(=2t^2m-2t^2-t\left(2tm-m-2t+1\right)+t+m\)

\(=2t^2m-2t^2-2t^2m+tm+2t^2-t+t+m\)

=tm+m

a) A = a 2   –   2 ab   +   b 2 .                 b) B = m 2 .                       c) C =  8 t 3 .

a) Rut gon H

\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)

\(H=\dfrac{\sqrt{a}+2}{\sqrt{a}+3}-\dfrac{5}{a+\sqrt{a}-6}-\dfrac{1}{\sqrt{a}-2}\)

DKXD : \(\left\{{}\begin{matrix}\sqrt{a}+3\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a\ne9\\a\ne4\end{matrix}\right.\)

Ta co : \(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{5}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}-\dfrac{\sqrt{a}+3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(H=\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)-5-\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(H=\dfrac{a-\sqrt{a}-6}{a+\sqrt{a}-6}\)

Bài 3:
\(3x^2+2x-1=0\)

\(\Leftrightarrow x^2+\dfrac{2}{3}x-\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{4}{9}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2-\dfrac{4}{9}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^2=\dfrac{4}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy...

a.

ĐKXĐ: \(x\ne2\)

b.

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

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

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

c.

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

d.

\(P=\dfrac{x}{x^2+1}\cdot\dfrac{x^2+1}{x}-\dfrac{1}{P}\ge1-\dfrac{1}{P}\)

\(\Rightarrow\dfrac{P^2+1}{P}\ge1\)

\(\Rightarrow P^2+1\ge P\) \(\Rightarrow P\left(P-1\right)\ge1\)

\(\Rightarrow P\ge2\)

Dấu "=" khi x = ...................

Bài 2:

a: \(M=\dfrac{3x+1-2x-2}{\left(3x-1\right)\left(3x+1\right)}:\dfrac{3x+1-3x}{x\left(3x+1\right)}\)

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

b: Để M=0 thì x(x-1)=0

=>x=1(nhận) hoặc x=0(loại)

c: \(P=M\cdot\left(3x-1\right)=x\left(x-1\right)=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)

Dấu = xảy ra khi x=1/2

Bài 1 :

a/ ĐKXĐ : \(a\ne0;-1\)

Ta có :

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

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

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

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

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

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

Vậy....

c/ Ta có : \(a+1=0\Leftrightarrow a=-1\) (loại)

Vậy....

Bài 2 :

a/ ĐKXĐ : \(x\ne0;3;-3\)

Ta có :

\(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x-3}\)

\(=\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x-3}{x}\)

\(=\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)

\(=\frac{x^2+x}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)

\(=\frac{x\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{x}\)

\(=\frac{x+1}{x+3}\)

Vậy....

b/ \(A=3\)

\(\Leftrightarrow\frac{x+1}{x+3}=3\)

\(\Leftrightarrow x+1=3x+9\)

\(\Leftrightarrow2x=-8\Leftrightarrow x=-4\)

Vậy...

a) ĐKXĐ: \(a\notin\left\{0;-1\right\}\)

Ta có: \(M=\left(\frac{1}{a}+\frac{a}{a+1}\right)-\frac{a}{a^2+a}\)

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

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

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

b) Ta có: a+1=0

hay a=-1(loại)

Vậy: Khi a+1=0 thì M không có giá trị

Bài 2:

a) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(A=\left(\frac{x^2-3}{x^2-9}+\frac{1}{x-3}\right):\frac{x}{x-3}\)

\(=\left(\frac{x^2-3}{\left(x-3\right)\left(x+3\right)}+\frac{x+3}{\left(x-3\right)\left(x+3\right)}\right)\cdot\frac{x-3}{x}\)

\(=\frac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{x}\)

\(=\frac{x^2+x}{x+3}\cdot\frac{1}{x}\)

\(=\frac{x\left(x+1\right)}{\left(x+3\right)\cdot x}\)

\(=\frac{x+1}{x+3}\)

b) Để A=3 thì \(\frac{x+1}{x+3}=3\)

\(\Leftrightarrow x+1=3\left(x+3\right)\)

\(\Leftrightarrow x+1=3x+9\)

\(\Leftrightarrow x+1-3x-9=0\)

\(\Leftrightarrow-2x-8=0\)

\(\Leftrightarrow-2x=8\)

hay x=-4(nhận)

Vậy: Khi A=3 thì x=-4