Ta có:

\(4a^2+b^2=5ab\Leftrightarrow4a^2+b^2-4ab-ab=0\)

\(\Leftrightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\4a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=b\left(ktm\right)\\4a=b\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow4a=b\)

\(\Rightarrow\dfrac{5ab}{3a^2+2b^2}=\dfrac{5a.4a}{3a^2+2.\left(4a\right)^2}=\dfrac{20a^2}{3a^2+32a^2}\)

\(=\dfrac{20a^2}{35a^2}=\dfrac{4}{7}\)

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

\(\Rightarrow4a\left(a-b\right)-b\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(4a-b\right)=0\)

\(\Rightarrow b=4a\left(do.a\ne b\right)\)

\(\dfrac{5ab}{3a^2+2b^2}=\dfrac{20a^2}{3a^2+32a^2}=\dfrac{4}{7}\)

tự làm nhé,dễ lắm

bài này khó đấy

Bài này dễ mà

a)

A= (-m+n-p)-(-m-n-p)

A= -m+n-p+m+n+p

A= (-m+m) +(n+n) + (-p+p)

A= 0+2n+0

A = 2n

Bài 1: 

A = (-m + n - p) - (-m - n - p)

A = -m + n - p + m + n + p

A = (-m + m) + (n + n) - (p - p)

A = 2n

Với n = -1 => A = 2(-1) = -2

Bài 2: 

A = (-2a + 3b - 4c) - (-2a -3b - 4c)

A = -2a + 3b - 4c + 2a + 3b + 4c

A = (-2a + 2a) + (3b + 3b) - (4c - 4c)

A = 6b

Với b = -1 => A = 6(-1) = -6

Bài 3:

a) A = (a + b) - (a - b) + (a - c) - (a + c)

A= a + b - a + b + a - c - a - c

A = (a - a + a - a) + (b + b) - (c + c)

A = 2(b - c)

b) B = (a + b - c) + (a - b + c) - (b + c - a) - (a - b - c)

B = a + b - c + a - b + c - b - c + a - a + b + c

B = (a + a + a - a) + (b - b - b + b) - (c - c + c - c)

B = 2a

Lời giải:

$P=4a^2+b^2+c^2+4ab+4ac+2bc=(2a+b+c)^2=(-1)^2=1$

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