Ta có

\(\frac{a-2ab-b}{2a+3ab-2b}=\frac{\frac{1}{b}-2-\frac{1}{a}}{\frac{2}{b}+3-\frac{2}{a}}=\frac{-1-2}{3-2}=-3\)

ta có \(\frac{1}{a}-\frac{1}{b}=1\)\(\)

\(\frac{b-a}{ab}=1\)

\(b-a=ab\)

\(a-b=ab\)

thay vào rồi tính

cam on

\(\frac{1}{a}-\frac{1}{b}=1\Rightarrow\frac{1}{a}=\frac{b+1}{b}\Rightarrow a=\frac{b}{b+1}\\ \)thế vào P ta có:
\(P=\frac{\frac{b}{b+1}-\frac{2b^2}{b+1}-b}{\frac{2b}{b+1}+\frac{3b^2}{b+1}-2b}=\frac{\frac{b-2b^2-b\left(b+1\right)}{b+1}}{\frac{2b+3b^2-2b\left(b+1\right)}{b+1}}=\frac{b-2b^2-b^2-b}{2b+3b^2-2b^2-2b}=\frac{-3b^2}{b^2}=-3\)

1/a - 1/b = 1

<=> 1/a = 1 + 1/b = b+1/b

<=> a = b/b+1

Thay vào P ta được:

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

Từ \(\dfrac{1}{a}-\dfrac{1}{b}=1\Leftrightarrow\dfrac{b-a}{ab}=1\Leftrightarrow b-a=ab\)

Ta có:

\(P=\dfrac{a-2ab-b}{2a+3ab-2b}=\dfrac{a-2\left(b-a\right)-b}{2a+3\left(b-a\right)-2b}\)

\(P=\dfrac{a-2b+2a-b}{2a+3b-3a-2b}=\dfrac{3a-3b}{b-a}=\dfrac{3\left(a-b\right)}{-\left(a-b\right)}=-3\)

sai đề

Theo mình thì \(\frac{1}{a}\)- \(\frac{1}{b}\)=1 không thể xảy ra vì 1/a - 1/b =1 => (b-a)/(ab)=1 

                                                                       hay b-a=a.b   <=> a=b=0 (trái với đề bài)

Câu 1:

\(Q=a^2+4b^2-10a\)

\(=a^2-10a+25+4b^2-25\)

\(=\left(a-5\right)^2+4b^2-25\)

\(\left(a-5\right)^2\ge0\)

\(4b^2\ge0\)

\(\Rightarrow\left(a-5\right)^2+4b^2-25\ge-25\)

Dấu ''='' xảy ra khi \(\left[\begin{array}{nghiempt}a-5=0\\b=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}a=5\\b=0\end{array}\right.\)

\(MinQ=-25\Leftrightarrow a=5;b=0\)

Câu 2:

Tam giác DAC vuông tại D có:

\(AC^2=CD^2+AD^2\)

\(=CD^2+CD^2\) (ABCD là hình vuông)

\(=2CD^2\)

\(=2\times\left(3\sqrt{2}\right)^2\)

\(=2\times9\times2\)

\(=36\)

\(AC=\sqrt{36}=6\left(cm\right)\)

Câu 3:

\(\frac{1}{a-1}=1\)

\(a-1=1\)

\(a=1+1\)

\(a=2\)

Thay a = 2 vào P, ta có:

\(P=\frac{2-2\times2\times b-b}{2\times2+3\times2\times b-b}\)

\(=\frac{2-4b-b}{4+6b-b}\)

\(=\frac{2-5b}{4+5b}\)

nếu \(\dfrac{1}{a}-\dfrac{1}{b}=1\Leftrightarrow a-b=-ab\)

\(P=\dfrac{a-b-2ab}{2\left(a-b\right)+3ab}=\dfrac{-3ab}{ab}=-3\)

a/ \(\Leftrightarrow x\left(8x^3+12x^2+6x+1\right)=0\Leftrightarrow x\left[\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\right]=0\)

\(\Leftrightarrow x\left(2x+1\right)^3=0\Rightarrow\orbr{\begin{cases}x=0\\\left(2x+1\right)^3=0\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\end{cases}}\)

b/ \(\Leftrightarrow4x^2-\left(4x^2-9\right)=9x\Leftrightarrow9x=9\Leftrightarrow x=1\)

c/ Từ \(\frac{1}{a}-\frac{1}{b}=1\Rightarrow a-b=-ab\) thay vào biểu thức

\(\Rightarrow\frac{-ab-2ab}{-2ab+3ab}=\frac{-3ab}{ab}=-3\)