sau mỗi ý đều có đầu bài nha

ĐKXĐ: \(\dfrac{x^2-16}{-3}\ge0\)

vì \(-3< 0\) nên \(x^2-16\le0\Rightarrow x^2\le16\Rightarrow-4\le x\le4\)

ĐKXĐ: \(-4\le x\le4\)

b: ĐKXD: x<>1/5; x<>3

PT\(\Leftrightarrow\dfrac{3}{5x-1}-\dfrac{2}{x-3}=\dfrac{-4}{\left(5x-1\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

a: ĐKXĐ: x<>2/3; x<>-2/3

\(PT\Leftrightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x\)

=>9x^2+12x+4-18x+12-9x=0

=>9x^2-15x+16=0

=>\(x\in\varnothing\)

c: ĐKXĐ: x<>1/4; x<>-1/4

PT =>-3(4x+1)=2(4x-1)-6x-8

=>-12x-3=8x-2-6x-8

=>-12x-3=2x-10

=>-14x=-7

=>x=1/2

d: ĐKXĐ: x<>0; x<>2

\(\Leftrightarrow\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>2(5-x)+7(x-2)=4(x-1)+x

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đung)

Vậy: S=R\{0;2}

e: DKXĐ: x<>0

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

=>x(x^3+1-x^3+1)=3

=>2x=3

=>x=3/2

\(a,\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\left(ĐKXĐ:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}-\dfrac{5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}-\dfrac{12}{\left(y-2\right)\left(y+2\right)}-\dfrac{\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Leftrightarrow\dfrac{y^2+y-2}{\left(y-2\right)\left(y+2\right)}-\dfrac{5y-10}{\left(y-2\right)\left(y+2\right)}-\dfrac{12}{\left(y-2\right)\left(y+2\right)}-\dfrac{y^2-4}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Leftrightarrow\dfrac{y^2+y-2-5y+10-12-y^2+4}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Rightarrow-4y=0\)

\(\Leftrightarrow y=0\left(tm\right)\)

\(b,\dfrac{1}{4z^2-12z+9}-\dfrac{3}{9-4z^2}=\dfrac{4}{4z^2+12z+9}\left(ĐKXĐ:z\ne\pm\dfrac{3}{2}\right)\)

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

\(⇔\dfrac{\left(2z+3\right)^2}{\left(2z-3\right)^2\left(2z+3\right)^2}+\dfrac{3\left(2z-3\right)\left(2z+3\right)}{\left(2z-3\right)^2\left(2z+3\right)^2}-\dfrac{4\left(2z-3\right)^2}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Leftrightarrow\dfrac{4z^2+12z+9}{\left(2z-3\right)^2\left(2z+3\right)^2}+\dfrac{12z^2-27}{\left(2z-3\right)^2\left(2z+3\right)^2}-\dfrac{16z^2-48z+36}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Leftrightarrow\dfrac{4z^2+12z+9+12z^2-27-16z^2+48z-36}{\left(2z-3\right)^2\left(2z+3\right)^2}=0\)

\(\Rightarrow60z-54=0\)

\(\Leftrightarrow60z=54\)

\(\Leftrightarrow z=\dfrac{9}{10}\left(tm\right).\)

\(a,\dfrac{y-1}{y-2}-\dfrac{5}{y+2}=\dfrac{12}{y^2-4}+1\left(dkxd:y\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{\left(y-1\right)\left(y+2\right)-5\left(y-2\right)-12-y^2+4}{y^2-4}=0\)
\(\Leftrightarrow y^2+2y-y-2-5y+10-12-y^2+4=0\)

\(\Leftrightarrow-4y=0\)

\(\Leftrightarrow y=0\left(tmdk\right)\)

Vậy \(S=\left\{0\right\}\)

\(b,\dfrac{1}{4z^2-12z+9}-\dfrac{3}{9-4z^2}=\dfrac{4}{4z^2+12z+9}\)

\(\Leftrightarrow\dfrac{1}{\left(2z-3\right)^2}-\dfrac{3}{\left(2z-3\right)\left(2z+3\right)}=\dfrac{4}{\left(2z+3\right)^2}\left(dkxd:z\ne\pm\dfrac{3}{2}\right)\)

\(\Leftrightarrow\left(2z+3\right)^2-3\left(4z^2-9\right)-4\left(2z-3\right)^2=0\)

\(\Leftrightarrow4z^2+12z+9-12z^2+27-4\left(4z^2-12z+9\right)=0\)

\(\Leftrightarrow4z^2+12z+9-12z^2+27-16z^2+48z-36=0\)

\(\Leftrightarrow-24z^2+60z=0\)

\(\Leftrightarrow-12z\left(2z-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-12z=0\\2z-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}z=0\left(tmdk\right)\\z=\dfrac{5}{2}\left(tmdk\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;\dfrac{5}{2}\right\}\)

c: \(\Leftrightarrow x\left(x+3\right)=3+x-3\)

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

=>x(x+2)=0

=>x=0(loại) hoặc x=-2(nhận)

d: \(\Leftrightarrow\left(x+2\right)\left(x+3\right)+\left(x-2\right)\left(x-3\right)=2x^2+12\)

\(\Leftrightarrow2x^2+12=2x^2+12\)

=>0x=0(luôn đúng)

e: \(\Leftrightarrow x\left(x-3\right)+3\left(x-2\right)=3x-20\)

\(\Leftrightarrow x^2-6-3x+20=0\)

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

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot14=9-56< 0\)

Do đó: Phương trình vô nghiệm

Gợi ý: 
ĐKXĐ : Mẫu ≠ 0, từ đó bạn tự xác định
c) Tách \(^{^2x}\)- 3 thành x (x-3) -> quy đồng mẫu-> rút gọn -> tự làm
d) Tách \(^{^2x}\)-9 thành hằng đẳng thức -> như trên
e) Tách \(^{^2x}\)- 5 thành hằng đẳng thức -> như trên
f) Quy đồng mẫu, quá dễ nên không nói thêm
g) Tách \(^{^2x}\)- 1 thành hằng đẳng thức -> như trên

a: ĐKXĐ: \(x>0\)

b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

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

\(=x-\sqrt{x}\)

a) Để A và B xác định thì \(\hept{\begin{cases}x-1\ne0\\x+1\ne0\\x^2-1\ne0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

Vậy để A và B xác định thì \(x\ne1\); \(x\ne-1\).

b) Ta có : A=B

\(\Rightarrow\frac{x+1}{x-1}+\frac{x-1}{1+x}=\frac{4}{x^2-1}\)

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

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

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

\(\Leftrightarrow2x^2-2=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=-1\\x=1\end{cases}}\)

Vậy để A=B thì \(x\in\left\{-1;1\right\}\).

ĐKXĐ: 

\(x-3\ne0\Rightarrow x\ne3\)

\(x=5k+3\left(k\in N\right)\)