a) \(14\left(x+2\right)=280\)

\(\Rightarrow x+2=280:14\)

\(\Rightarrow x+2=20\)

\(\Rightarrow x=20-2\)

\(\Rightarrow x=18\)

b) \(\left(7x-15\right):2^2-3=2\)

\(\Rightarrow\left(7x-15\right):4-3=2\)

\(\Rightarrow\left(7x-15\right):4=2+3\)

\(\Rightarrow\left(7x+15\right):4=5\)

\(\Rightarrow7x+15=5.4\)

\(\Rightarrow7x+15=20\)

\(\Rightarrow7x=20-15\)

\(\Rightarrow7x=5\)

\(\Rightarrow x=5:7\)

\(\Rightarrow x=\dfrac{5}{7}\)

Mà \(x\inℕ\)

\(\Rightarrow x\in\varnothing\)

a) \(140:\left(x-8\right)=7\)

\(\Rightarrow x-8=140:7\)

\(\Rightarrow x-8=20\)

\(\Rightarrow x=20+8\)

\(\Rightarrow x=28\)

b) \(4\left(x+41\right)=400\)

\(\Rightarrow x+41=400:4\)

\(\Rightarrow x+41=100\)

\(\Rightarrow x=100-41\)

\(\Rightarrow x=59\)

\(M=1+2+2^2+2^3+2^4+...+2^{99}\)

\(\Rightarrow2M=2\left(1+2+2^2+2^3+2^4+...+2^{99}\right)\)

\(\Rightarrow2M=2+2^2+2^3+2^4+2^5+...+2^{100}\)

\(\Rightarrow2M-M=\left(2+2^2+2^3+2^4+2^5+...+2^{100}\right)-\left(1+2+2^2+2^3+2^4+...+2^{99}\right)\)

\(\Rightarrow M=2^{100}-1\)

Chúng mừng các bạn nha

\(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=\left(x^2-7+3\right)\left(x^2-7-3\right)-72\)

\(=\left(x^2-7\right)^2-9-72\)

\(=\left(x^2-7\right)^2-81\)

\(=\left(x^2-7+9\right)\left(x^2-7-9\right)\)

\(=\left(x^2+2\right)\left(x^2-16\right)\)

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

\(x^{64}+x^{32}+1\)

\(=x^{64}+2x^{32}-x^{32}+1\)

\(=\left(x^{64}+2^{32}+1\right)-x^{32}\)

\(=\left(x^{32}+1\right)^2-\left(x^{16}\right)^2\)

\(=\left(x^{32}+1-x^{16}\right)\left(x^{32}+1+x^{16}\right)\)

a)

\(A=5xy^2+xy-3xy^2-x^2y+2xy+x^2y-2xy^2+xy+4\)

\(A=\left(5xy^2-3xy^2-2xy^2\right)+\left(xy+2xy+xy\right)+4+\left(x^2y-x^2y\right)\)

\(A=4xy+4\)

Bậc của đa thức \(A\) là bậc \(2\)

b) 

Thay \(x=2;y=1\) vào \(A\) ta được:

\(A=4.2.1+4\)

\(A=8+4\)

\(A=12\)

c)

\(A+B=-2xy+1\)

\(\Rightarrow B=-2xy+1-A\)

\(\Rightarrow B=-2xy+1-\left(4xy+4\right)\)

\(\Rightarrow B=-2xy+1-4xy-4\)

\(\Rightarrow B=-6xy-3\)

Vậy ...

rồi câu 5 là từ gì vậy ạ

\(2x^2-6x+1=0\)

\(\Leftrightarrow4x^2-12x+2=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x.3+9=7\)

\(\Leftrightarrow\left(2x-3\right)^2=7\)

\(\Leftrightarrow2x-3=\pm\sqrt{7}\)

\(\Leftrightarrow2x=\pm\sqrt{7}+3\)

\(\Leftrightarrow x=\dfrac{\pm\sqrt{7}+3}{2}\)

Vậy ...

\(\left(3x-2\right)^2=14-2.5^2\)

\(\Rightarrow\left(3x-2\right)^2=14-2.25\)

\(\Rightarrow\left(3x-2\right)^2=14-50\)

\(\Rightarrow\left(3x-2\right)^2=-36\)

Vì \(\left(3x-2\right)^2\ge0\) với mọi \(x\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)