Question

cần gấp

Answer 1

2.24:

a: Thay x=51; y=-49 vào A, ta được:

\(A=51^2+51\cdot\left(-49\right)+51=51\left(51-49+51\right)=51\cdot53=2703\)

b: \(B=5x^2\left(x-2y\right)+15x\left(2y-x\right)\)

\(=5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)

Thay x=11; y=6 vào B, ta được:

\(B=5\cdot11\cdot\left(11-2\cdot6\right)\left(11-3\right)=55\cdot8\cdot\left(11-12\right)=-440\)

Bài 2.23:

a:

\(13^{n+1}-13^{n}=13^{n}\left(13-1\right)=13^{n}\cdot12\) ⋮12

b: \(n^3-n=n\left(n^2-1\right)=n\left(n-1\right)\left(n+1\right)\)

Vì n-1;n;n+1 là ba số nguyên liên tiếp

nên \(\left(n-1\right)\cdot n\left(n+1\right)\) ⋮3!

=>\(n^3-n\) ⋮6

2.22:

a: 2x(x-3)+5x-15=0

=>2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>\(\left[\begin{array}{l}x-3=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac52\end{array}\right.\)

b: \(\left(x+1\right)^2=x+1\)

=>\(x^2+2x+1-x-1=0\)

=>\(x^2+x=0\)

=>x(x+1)=0

=>\(\left[\begin{array}{l}x=0\\ x+1=0\end{array}\right.=>\left[\begin{array}{l}x=0\\ x=-1\end{array}\right.\)

c: \(x^3+x=0\)

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

mà \(x^2+1\ge1>0\forall x\)

nên x=0

d: x(x+7)=4x+28

=>x(x+7)-4(x+7)=0

=>(x+7)(x-4)=0

=>\(\left[\begin{array}{l}x+7=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-7\\ x=4\end{array}\right.\)

e: 3x(x-2)-x+2=0

=>3x(x-2)-(x-2)=0

=>(x-2)(3x-1)=0

=>\(\left[\begin{array}{l}x-2=0\\ 3x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=\frac13\end{array}\right.\)