Ta có: \(a^2 - 3ab + 2b^2 + a - b = 0\) (1)

=>\((a^2-ab-2ab+2b^2)+(a-b)=0\)

=>a(a-b)-2b(a-b)+(a-b)=0

=>(a-b)(a-2b+1)=0

=>a-b=0 hoặc a-2b+1=0

=>a=b hoặc a=2b-1

Ta có: (2): \(a^2 - 2ab + b^2 - 5a + 7b = 0\)

=>\((a - b)^2 - 5a + 7b = 0 \quad (*)\)

TH1: a=b

(*) sẽ trở thành: \(\left(b-b\right)^2-5b+7b=0\)

=>2b=0

=>b=0

=>a=b=0

ab-12a+15b=0*0-12*0+15*0=0

TH2: a=2b-1

(*) sẽ trở thành: \(\left(2b-1-b\right)^2-5\left(2b-1\right)+7b=0\)

=>\(b^2-2b+1-10b+5+7b=0\)

=>\(b^2-5b+6=0\)

=>(b-2)(b-3)=0

=>b-2=0 hoặc b-3=0

=>b=2 hoặc b=3

Nếu b=2 thì a=2b-1=2*2-1=3

ab-12a+15b=3*2-12*3+15*2=6-36+30=0

Nếu b=3 thì a=2b-1=2*3-1=6-1=5

ab-12a+15b=5*3-12*5+15*3=15-60+45=0

Do đó: ab-12a+15b=0

a: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

b: \(\left(a^2b-3ab^2\right):\left(\frac12ab\right)+\left(6b^3-5ab^2\right):b^2\)

\(=2a-6b+6b-5a\)

=2a-5a

=-3a

\(a,=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\\ b,=2a-6b+6b-5a=-3a\)

2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)

Biểu thức P đâu bạn?