Lời giải:

a)

\((2a-5b)^2+(2a+5b)^2\)

\(=4a^2-2.2a.5b+25b^2+4a^2+2.2a.5b+25b^2\)

\(=8a^2+50b^2=2(4a^2+25b^2)\)

b)

\((a-2b-3c)^2-(a-2b+3c)^2\)

\(=[(a-2b-3c)-(a-2b+3c)][(a-2b-3c)+(a-2b+3c)]\)

\(=-6c(2a-4b)=12c(2b-a)\)

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

\(=\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2+\left(2a\right)^2+2\cdot2a\cdot5b+\left(5b\right)^2\)

\(=2\cdot\left[\left(2a\right)^2+\left(5b\right)^2\right]\)

\(=2\left(4a^2+25b^2\right)=8a^2+50b^2\)

\(\left(2a-5b\right)^2+\left(2a+5b\right)^2=2\left[\left(2a\right)^2+\left(5b\right)^2\right]=2\cdot4a^2+2\cdot25b^2=8a^2+50b^2\)

(2a-5b)² +(2a+5b)²

= (2a-5b)² -(-2a-5b)²

⁼ (2a-5b-2a-5b)(2a-5b+2a+5b)

= (-10b)(4a)

= -40ab

\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)

\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)

B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)

B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)

B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)

B= 5

vậy B=5

\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)

bạn coi lại đề dùm mik đi

b ) 735a2b chia hết cho 5 và 9 nhưng không chia hết cho 2

giúp em đi,em đang cần gấp

Lớp 7 gì mà dễ ẹc :))

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

\(\Leftrightarrow6a-3b=2a+2b\)

\(\Rightarrow4a=5b\)

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

\(\Leftrightarrow4a-2b=3b-3c+3a\)

\(\Leftrightarrow a=5b-3c\)

\(\Leftrightarrow a-5b=-3c\)

\(\Leftrightarrow a-4a=-3c\)

\(\Leftrightarrow-3a=-3c\)

\(\Rightarrow a=c\)

Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)

8 nhé bn !