Theo đề ra, ta có:

\(a+b=\frac{5}{2}\)

\(\Rightarrow\left(a+b\right)^2=a^2+b^2+2ab=\frac{25}{4}\)

\(\Rightarrow a^2+b^2=\frac{25}{4}-2=\frac{17}{4}\)

Ta có:

\(\left(a-b\right)^2=a^2+b^2-2ab=\frac{17}{4}-2=\frac{9}{4}\)\(\Rightarrow a-b=\frac{3}{2}\)

Ta có:

\(a^4-b^4=\left(a^2+b^2\right)\left(a^2-b^2\right)=\frac{17}{4}\left(a-b\right)\left(a+b\right)=\frac{17}{4}.\frac{5}{2}.\frac{3}{2}=\frac{255}{16}\)

10 con

câu 1 k biết

câu 2 là 10 con vịt 

86.NHỮNG PHÉP TÍNH THÚ VỊ

24+36=1

11+13=1

158+207=1

46+54=1

thì khi đó người làm câu hỏi bị sai/ mình nghĩ thế

câu 1 ko hề dể dàng

câu 2 sai

\(a,\left(3x+1\right)^2=x^2\)

=>  \(\left(3x+1\right)^2-x^2=0\)

=>  \(\left(3x+1-x\right)\left(3x+1+x\right)=0\)

=>  \(\left(2x+1\right)\left(4x+1\right)=0\)

=> \(\orbr{\begin{cases}2x+1=0\\4x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{1}{4}\end{cases}}}\)

\(b,x^2-8x+16=0\)

=>  \(\left(x-4\right)^2=0\)

=>  \(x-4=0\)

=>  \(x=4\)

\(a,73^2-13^2-10^2+20\cdot13=\left(73-13-10\right)^2+20\cdot13=50^2+260=2500+260=2760\)

Ta có :

\(73^2-13^2-10^2+20.13=\left(73+13\right)\left(73-13\right)-13.13+20.13\)

\(=86.60-13\left(13-20\right)\)

\(=5160-\left(-91\right)=5160+91=...\)

\(a,\left(x^2+4\right)^2-x^2=\left(x^2+4-x\right)\left(x^2+4+x\right)\)

\(b,\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\)

\(=\left(x+y\right)\left(5x-5y\right)=5\left(x+y\right)\left(x-y\right)\)

\(c,\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

\(d,8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2-2x+1\right)\)