Câu 1 :

a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)

b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)

\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)

\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)

\(\Leftrightarrow2x^2+8x+6=0\)

\(\Leftrightarrow x^2+4x+4-1=0\)

\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)

Vậy : \(x=-3\) thì P = 1.

b, P=x+2x+3−5x2+3x−2x−6+12−xP=x+2x+3−5x2+3x−2x−6+12−x

=x+2x+3−5(x+3)(x−2)−1x−2=x+2x+3−5(x+3)(x−2)−1x−2

=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)=(x+2)(x−2)(x+3)(x−2)−5(x+3)(x−2)−x+3(x+3)(x−2)

=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)=x2−4−5−x−3(x+3)(x−2)=x2−x−12(x+3)(x−2)

=x2−4x+3x−12(x+3)(x−2)=x2−4x+3x−12(x+3)(x−2)

=(x−4)(x+3)(x+3)(x−2)=x−4x−2=(x−4)(x+3)(x+3)(x−2)=x−4x−2

c, Để P=−34P=−34

⇔x−4x−2=−34⇔x−4x−2=−34

⇔4(x−4)=−3(x−2)⇔4(x−4)=−3(x−2)

⇔4x−16+3x−6=0⇔4x−16+3x−6=0

⇔7x−22=0⇔7x−22=0

⇔x=227⇔x=227

d, Để P có giá trị nguyên

⇔x−4⋮x−2⇔x−4⋮x−2

⇔(x−2)−2⋮x−2⇔(x−2)−2⋮x−2

⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}⇔2⋮x−2⇔x−2∈Ư(2)={1;−1;2;−2}

e,

x2−9=0x2−9=0

⇒x2=9⇒[x=3x=−3⇒x2=9⇒[x=3x=−3

Với x=3,có :

x−4x−2=3−43−2=−11=−1x−4x−2=3−43−2=−11=−1

Với x=-3,có :

x−4x−2=−3−4−3−2=75x−4x−2=−3−4−3−2=75

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)

a) ĐKXĐ : \(x\ne\pm2\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}+\frac{40}{4-x^2}\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{x^2-4}\)

\(B=\frac{5x}{x+2}-\frac{3x-23}{x-2}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x-23\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x}{\left(x+2\right)\left(x-2\right)}-\frac{\left(3x^2-17x-46\right)}{\left(x+2\right)\left(x-2\right)}-\frac{40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x-\left(3x^2-17x-46\right)-40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{5x^2-10x-3x^2+17x+46-40}{\left(x+2\right)\left(x-2\right)}\)

\(B=\frac{2x^2+7x+6}{\left(x+2\right)\left(x-2\right)}=\frac{\left(x+2\right)\left(2x+3\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2x+3}{x-2}\)

b) x² - 1 = 0 <=> x² = 1 <=> x = ±1

Với x = 1 

\(B=\frac{2\cdot1+3}{1-2}=-5\)

Với x = -1

\(B=\frac{2\cdot\left(-1\right)+3}{\left(-1\right)-2}=-\frac{1}{3}\)

A=x³/x²--4.x+2/x-x-4xx-4/xx-2

Điều kiện x \(\ne\)+-2

Ý b c tự làm 

\(A=\frac{x^3}{x^2-4}.\frac{x+2}{x}-\frac{4x-4}{x-2}\)   \(ĐKXĐ:x\ne0;x\ne2\)

\(A=\frac{x^2}{x-2}-\frac{4\left(x-1\right)}{x-2}\)

\(A=\frac{x^2-4x+4}{x-2}\)

\(A=\frac{\left(x-2\right)^2}{x-2}\)

\(A=x-2\)

vậy \(A=x-2\)

ai giúp mình vớiiiii

a, ĐKXĐ:

9x^2 - 16 ≠ 0

=> (3x - 4)(3x + 4) ≠ 0

=> 3x - 4 ≠ 0 và 3x + 4 ≠ 0

=> 3x  ≠ 4 và 3x ≠ -4

=> x ≠ 4/3 hoặc x ≠ -4/3

b, ĐKXĐ:

x^2 - 5x + 6 ≠ 0

=> x^2 - 2x - 3x + 6 ≠ 0

=> x(x - 2) - 3(x - 2) ≠ 0

=> (x - 3)(x - 2) ≠ 0

=> x - 3 ≠ 0 và x - 2 ≠ 0

=> x ≠ 3 và x ≠ 2

c, ĐKXĐ : 

x^2 - 4x + 4 ≠ 0

=> (x - 2)^2 ≠ 0

=> x - 2 ≠ 0

=> x ≠ 2

a, ĐKXĐ: \(x\ne-3\) và \(x\ne\pm1\)

b, \(P=\frac{x\left(x+3\right)-11+x^2-3x+9}{x^3+27}:\frac{x^2-1}{x+3}\)

\(P=\frac{2x^2-2}{x^3+27}.\frac{x+3}{x^2-1}\)

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

\(=\frac{2}{x^2-3x+9}\)

c, \(P=\frac{2}{x^2-3x+9}==\frac{2}{\left(x-\frac{3}{2}\right)^2+\frac{27}{4}}\le\frac{2}{\frac{27}{4}}=\frac{8}{27}\)

Dấu "=" xảy ra khi: \(x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)

Vậy P lớn nhất bằng \(\frac{8}{27}\) \(\Leftrightarrow x=\frac{3}{2}\)

\(P=\left(\frac{x}{x^2-3x+9}-\frac{11}{x^3+27}+\frac{1}{x+3}\right):\frac{x^2-1}{x+3}.\)

ĐKXĐ : \(x\ne-3;x\ne0\)

\(P=\left(\frac{x\left(x+3\right)}{\left(x+3\right)\left(x^2-3x+9\right)}-\frac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\frac{x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\right).\frac{x+3}{x^2-1}\)

\(P=\left(\frac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\right).\frac{x+3}{x^2-1}\)

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

\(P=\frac{2}{x^2-3x+9}\)

bn ơi cho mk hỏi tại sao lại ko nhận 3 vậy !!!