a) Ta có: a⊥c,b⊥c

=> a//b

b) Ta có: a//b

\(\Rightarrow\widehat{B_1}+\widehat{A_1}=180^0\)(trong cùng phía)

\(\Rightarrow\widehat{B_1}=180^0-120^0=60^0\)

c) Ta có: \(\widehat{A_1}=\widehat{B_2}\)(2 góc so le trong và a//b)

\(\Rightarrow\widehat{A_2}+\widehat{B_2}=\widehat{A_2}+\widehat{A_1}=180^0\)(kề bù)

\(a,=\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)\\ b,=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-2\left(\sqrt{x}+\sqrt{y}\right)\\ =\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}-2\right)\\ c,=x\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)\)

\(a,=\dfrac{\left(9-4\sqrt{5}\right)\left(5+2\sqrt{5}\right)}{4}+\dfrac{2\sqrt{5}}{5}\\ =\dfrac{5-2\sqrt{5}}{4}+\dfrac{2\sqrt{5}}{5}\\ =\dfrac{25-10\sqrt{5}+8\sqrt{5}}{20}=\dfrac{25-2\sqrt{5}}{20}\\ b,=\dfrac{\sqrt{x}+2-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\\ c,=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}-\dfrac{2}{\sqrt{x}-1}\\ =\dfrac{\sqrt{x}+1-2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}-1}=1\\ d,=\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}+\dfrac{x+1}{1-x}\)

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

\(\sqrt{a^2+3}=\sqrt{a^2+ab+bc+ca}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{1}{2}\left(a+b+a+c\right)=\dfrac{1}{2}\left(2a+b+c\right)\)

Tương tự: \(\sqrt{b^2+3}\le\dfrac{1}{2}\left(a+2b+c\right)\) ; \(\sqrt{c^2+3}\le\dfrac{1}{2}\left(a+b+2c\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(4a+4b+4c\right)=2\left(a+b+c\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}+\sqrt{x}=2\sqrt{x}\)

Chọn B

b

bạn ơi cho mik hỏi bài thơ đó tên gì

1.

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{1}{2}\Rightarrow\widehat{A}=60^o\)

\(S=\dfrac{1}{2}bc.sinA=\dfrac{1}{2}.8.5.sin60^o=10\sqrt{3}\)

\(S=\dfrac{1}{2}a.h_a=\dfrac{1}{2}.7.h_a=10\sqrt{3}\Rightarrow h_a=\dfrac{20\sqrt{3}}{7}\)

\(2R=\dfrac{a}{sinA}=\dfrac{7}{\dfrac{\sqrt{3}}{2}}=\dfrac{14\sqrt{3}}{3}\Rightarrow R=\dfrac{7\sqrt{3}}{3}\)

\(S=pr=\dfrac{a+b+c}{2}.r=10r=10\sqrt{3}\Rightarrow r=\sqrt{3}\)

\(m_a^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{129}{4}\Rightarrow m_a=\dfrac{\sqrt{129}}{2}\)

6.

a, Công thức trung tuyến:

\(AM^2=c^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{2b^2+2c^2-a^2}{4}\Rightarrow a^2=2\left(b^2-c^2\right)\)

b, \(a^2=2\left(b^2-c^2\right)\Rightarrow\dfrac{2\left(b^2-c^2\right)}{a^2}=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}-\dfrac{c^2}{a^2}\right)=1\)

\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}.sin^2A-\dfrac{c^2}{a^2}.sin^2A\right)=sin^2A\)

\(\Leftrightarrow2\left(sin^2B-sin^2C\right)=sin^2A\)

Hay \(sin^2A=2\left(sin^2B-sin^2C\right)\)

Câu 7: Hình bạn tự vẽ

Giả sử hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau. Gọi I là giao điểm của hai trung tuyến

Ta có: \(IA^2+IB^2=AB^2\)\(\Leftrightarrow\left(\dfrac{2}{3}m_a\right)^2+\left(\dfrac{2}{3}m_b\right)^2=c^2\)\(\Leftrightarrow4m_a^2+4m_b^2=9c^2\Leftrightarrow2\left(b^2+c^2\right)-a^2+2\left(a^2+c^2\right)-b^2=9c^2\Leftrightarrow a^2+b^2=5c^2\)(đúng với giả thiết)

suy ra giả sử là đúng . Vậy hai trung tuyến \(AA_1;BB_1\) vuông góc với nhau