a: \(P=\dfrac{a+1+\sqrt{a}}{a+1}:\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)

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

b: P<1

=>P-1<0

=>\(\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)

=>căn a-1<0

=>0<a<1

c: Thay x=19-8căn3 vào P, ta được:

\(P=\dfrac{19-8\sqrt{3}+4+\sqrt{3}+1}{4+\sqrt{3}-1}=\dfrac{31-15\sqrt{3}}{2}\)

hộ e ik mà ;-;

6) ĐKXĐ: \(x\le-7\)

9) ĐKXĐ: \(x\ne1\)

6: ĐKXĐ: y<>0; y<>2; y<>-2; y<>3

a: \(P=\left(\dfrac{-\left(y+2\right)}{y-2}-\dfrac{4y^2}{\left(y-2\right)\left(y+2\right)}+\dfrac{y-2}{y+2}\right):\dfrac{y\left(y-3\right)}{y^2\left(2-y\right)}:\dfrac{1}{y-3}\)

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

\(=\dfrac{-4y^2-8y}{\left(y-2\right)\left(y+2\right)}\cdot\dfrac{-y\left(y-2\right)}{1}\)

\(=4y^2\)

b: 2y^2-3y-2=0

=>2y^2-4y+y-2=0

=>(y-2)*(2y+1)=0

=>y=2(loại) hoặc y=-1/2(nhận)

Khi y=-1/2 thì P=4*(-1/2)^2=1

3b.

\(\Delta=m^2+4\left(m+1\right)=\left(m+2\right)^2\)

Pt có 2 nghiệm pb khi \(\left(m+2\right)^2>0\Rightarrow m\ne-2\)

Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-\left(m+1\right)\end{matrix}\right.\)

\(x_1+x_2-2x_1x_2=8\)

\(\Leftrightarrow-m+2\left(m+1\right)=8\)

\(\Rightarrow m=6\) (thỏa mãn)

6.

\(M=x-\sqrt{x}+1=\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(M_{min}=\dfrac{3}{4}\) khi \(\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)

Cảm ơn nhiều ạ

Bài đâu bạn?

BT A - B hay A.B vậy bn ?

A.B ạ

c) Ta có: \(\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{\sqrt{6}}{\sqrt{3}}-3\cdot\sqrt{\dfrac{1}{3}}\)

\(=\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{3}\)

=0