chụp rõ hơn cho mk đc ko ạ

a: Ta có: \(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\)

\(=\frac{1}{2a-b}-\frac{a^2-1}{a^2\left(2a-b\right)+\left(2a-b\right)}\)

\(=\frac{1}{2a-b}-\frac{a^2-1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{a^2+1-a^2+1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}\)

\(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\)

\(=\frac{4a+2b}{ab\left(a^2+1\right)}-\frac{2}{a}=\frac{4a+2b-2b\left(a^2+1\right)}{ab\left(a^2+1\right)}\)

\(=\frac{4a-2a^2b}{ab\left(a^2+1\right)}=\frac{2a\left(2-ab\right)}{ab\cdot\left(a^2+1\right)}=\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}\)

Ta có: \(A=\left(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\right):\left(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\right)\)

\(=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}:\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}=\frac{2b\left(a^2+1\right)}{2\left(2-ab\right)\left(2a-b\right)\left(a^2+1\right)}=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)

b:

Sửa đề: b>a>0

\(4a^2+b^2=5ab\)

=>\(4a^2-5ab+b^2=0\)

=>\(4a^2-4ab-ab+b^2=0\)

=>(a-b)(4a-b)=0

TH1: a-b=0

=>a=b

mà a>b

nên Loại

TH2: 4a-b=0

=>b=4a(nhận)

\(A=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)

\(=\frac{4a}{\left(2-a\cdot4a\right)\left(2a-4a\right)}=\frac{4a}{\left(2-4a^2\right)\left(-2a\right)}\)

\(=\frac{4a}{-2a\cdot\left(-2\right)\left(2a^2-1\right)}=\frac{1}{2a^2-1}\)

a.

\(F=\dfrac{a}{b+2}\Rightarrow F.b+2F=a\)

\(\Rightarrow2F=a-F.b\)

\(\Rightarrow4F^2=\left(a-F.b\right)^2\le\left(a^2+b^2\right)\left(1^2+F^2\right)=F^2+1\)

\(\Rightarrow3F^2\le1\)

\(\Rightarrow-\dfrac{1}{\sqrt{3}}\le F\le\dfrac{1}{\sqrt{3}}\)

Dấu "=" lần lượt xảy ra tại \(\left(a;b\right)=\left(-\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\) và \(\left(\dfrac{\sqrt{3}}{2};-\dfrac{1}{2}\right)\)

b. Đặt \(\left\{{}\begin{matrix}a+b=x\\a-2b=y\end{matrix}\right.\) quay về câu a

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)

\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)

\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)

lỗi ảnh

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