70⁰

Lí Thuyết: 2 góc đối đỉnh có số đo bằng nhau

\(\left(x+\dfrac{1}{6}y+3\right)^2=x^2+\dfrac{1}{36}y^2+9+\dfrac{1}{3}xy+6x+y\)

Áp dụng HĐT: (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac nhé

\(\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{16+6\sqrt{7}}=\sqrt{7}-2-\sqrt{7}-3=-5\)

\(\dfrac{\sqrt{21}-7\sqrt{7}}{7-\sqrt{3}}-\dfrac{18}{\sqrt{7}-5}=\dfrac{\sqrt{7}\left(\sqrt{3}-7\right)}{7-\sqrt{3}}-\dfrac{18}{\sqrt{7}-5}=-\sqrt{7}-\dfrac{18}{\sqrt{7}-5}=-\dfrac{7-5\sqrt{7}+18}{\sqrt{7}-5}=-\dfrac{25-5\sqrt{7}}{\sqrt{7}-5}=-\dfrac{5\left(5-\sqrt{7}\right)}{\sqrt{7}-5}=5\)

Đề ?

\(A=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)

\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)

\(=0,56\)

Vậy chọn D nhé

hỏi thật bạn là nam hay nữ thế :))

\(=x^4-10x^3+25x^2-3x^2+15x-18=x^4-10x^3+22x^2+15x-18=x^4+x^3-11x^3-11x^2+33x^2+33x-18x-18=x^3\left(x+1\right)-11x^2\left(x+1\right)+33x\left(x+1\right)-18\left(x+1\right)=\left(x+1\right)\left(x^3-11x^2+33x-18\right)=\left(x+1\right)\left(x^3-6x^2-5x^2+30x+3x-18\right)=\left(x+1\right)\left[x^2\left(x-6\right)-5x\left(x-6\right)+3\left(x-6\right)\right]=\left(x+1\right)\left(x-6\right)\left(x^2-5x\right)=\left(x+1\right)\left(x-6\right)x\left(x-5\right)\)

Em cần câu mấy ?

P/s: mỗi lần chỉ hoỉ 1 câu thôi nhé!

3.3.3.3.27 = 3.3.3.3.3.3.3 = 3⁷