Bạn thử nhân liên hợp cả tử và mẫu theo (A+B)×(A-B)=A^2 -B^2 xem s

\(a,\dfrac{y-1}{y+1}=\dfrac{\left(y-1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{y+1}{y-1}=\dfrac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{1}{y^2-1}=\dfrac{1}{\left(y-1\right)\left(y+1\right)}\\ b,\dfrac{2}{y^2-4y}=\dfrac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)};\dfrac{y}{y^2-16}=\dfrac{y^2}{y\left(y-4\right)\left(y+4\right)}\)

ĐK: \(3x\ne\pm y;x\ne0\)

A = \(\dfrac{3x}{3x+y}-\dfrac{x}{3x-y}+\dfrac{2x}{\left(3x-y\right)\left(3x+y\right)}\)

= \(\dfrac{3x\left(3x-y\right)-x\left(3x+y\right)+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{6x^2-4xy+2x}{\left(3x-y\right)\left(3x+y\right)}=\dfrac{2x\left(3x-2y+1\right)}{\left(3x-y\right)\left(3x+y\right)}\)

Thay x = 1; y=2, ta có:

A = \(\dfrac{2.1\left(3.1-2.2+1\right)}{\left(3.1-2\right)\left(3.1+2\right)}=0\)

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

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

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

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

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

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

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

e: ĐKXĐ: x<>-y

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

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

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

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

\(a,\left(1\right)=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(2\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)};\left(3\right)=\dfrac{-4}{\left(x-1\right)\left(x+1\right)}\\ b,\left(1\right)=\dfrac{x^4y^3}{xy^3\left(x-y\right)^3};\left(2\right)=\dfrac{x\left(x-y\right)^3}{xy^3\left(x-y\right)^3}\\ c,\left(1\right)=\dfrac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(2\right)=\dfrac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)};\left(3\right)=\dfrac{12x}{\left(x-2\right)\left(x+2\right)}\\ d,\left(1\right)=\dfrac{7\left(x+6\right)}{x\left(x+6\right)};\left(2\right)=\dfrac{x^2}{x\left(x+6\right)};\left(3\right)=\dfrac{36}{x\left(x+6\right)}\)

y'=1/3*3x^2+1/2*2x(m-1)+(2m-1)

=x^2+x(m-1)+2m-1

a: y đồng biến trên R thì y'>0 với mọi x thuộc R

Δ=(m-1)^2-4(2m-1)

=m^2-2m+1-8m+4=m^2-10m+5

Để y'>0 với mọi x thuộc R thì m^2-10m+5<0

=>5-2*căn 5<m<5+2căn 5

b: y đồng biến trên (-vô cực;-2) và (0;1) khi y'>0 với mọi x thuộc (-vô cực;-2) và (0;1)

y'=x^2+x(m-1)+2m-1

=x^2+xm-x+2m-1

=m(x+2)+x^2-x-1

y'>0 với x thuộc (-vô cực;-2)

=>m>-x^2+x+1/(x+2) với x thuộc (vô cực;-2)

g(x)=-x^2+x+1/(x+2)

g'=(-x^2+x+1)'(x+2)-(-x^2+x+1)(x+2)'/(x+2)^2

=(x+2+x^2-x-1)/(x+2)^2=(x^2+1)/(x+2)^2>0 với mọi x

=>m thuộc (-vô cực;-2)

Tương tự, ta cũng được: m thuộc (0;1)

theo đầu bài ta có\(\dfrac{x^2+y^2}{xy}=\dfrac{10}{3}\)=>\(3x^2+3y^2=10xy\)

A=\(\dfrac{x-y}{x+y}\)

=>\(A^2=\left(\dfrac{x-y}{x+y}\right)^2=\dfrac{x^2-2xy+y^2}{x^2+2xy+y^2}=\dfrac{3x^2-6xy+3y^2}{3x^2+6xy+3y^2}=\dfrac{10xy-6xy}{10xy+6xy}=\dfrac{4xy}{16xy}=\dfrac{1}{4}\)

=>A=\(\sqrt{\dfrac{1}{4}}=\dfrac{-1}{2}hoặc\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\) (cộng trừ căn 1/4 nhé)

vì y>x>0=> A=-1/2

\(a,=\dfrac{15x+25-25x+x^2}{5x\left(x-5\right)}=\dfrac{\left(x-5\right)^2}{5x\left(x-5\right)}=\dfrac{x-5}{5x}\\ b,=\dfrac{x^2-x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2+x-6}{x^2+x-6}=1\)

\(a,\dfrac{3x+5}{x^2-5x}+\dfrac{25-x}{25-5x}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{-3x-5}{x\left(5-x\right)}+\dfrac{25-x}{5\left(5-x\right)}\)

\(=\dfrac{5\left(-3x-5\right)}{5x\left(5-x\right)}+\dfrac{x\left(25-x\right)}{5x\left(5-x\right)}\)

\(=\dfrac{-15x-25+25x-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{10x-25-x^2}{5x\left(5-x\right)}\)

\(=\dfrac{-\left(5-x\right)^2}{5x\left(5-x\right)}\)

\(=\dfrac{-5+x}{5x}\)

\(b,\dfrac{x+1}{x+3}+\dfrac{x-7}{x^2+x-6}+\dfrac{1}{x-2}\)

\(=\dfrac{x+1}{x+3}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{1}{x-2}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}+\dfrac{x-7}{\left(x+3\right)\left(x-2\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2-2x+x-2+x-7+x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x-6}{x^2-2x+3x-6}\)

\(=\dfrac{x^2+x-6}{x^2+x-6}\)

\(=1\)