Ở olm vẫn chưa đồng ý đăng giải bằng hình ảnh hả:vv

a) Xét \(\Delta AHB\) và \(\Delta AHC\)

\(\widehat{AHB}=\widehat{AHC}=90^o\)

AB=AC

\(\widehat{ABH}=\widehat{ACH}\)

\(\Rightarrow\Delta AHB=\Delta AHC\left(ch-gn\right)\)

\(\Rightarrow\widehat{HAB}=\widehat{HAC}\) (2 góc tương ứng)

=> AH là tia phân giác của góc BAC

  Đây nhé, để mình giải nốt mấy câu còn lại

Lần sau bạn đăng thì chia nhỏ ý ra cho tụi mình dễ giúp đỡ nhé

Ta có: \(\sqrt{a^2-ab+b^2}=\sqrt{\left(a+b\right)^2-3ab}\ge\sqrt{\left(a+b\right)^2-\dfrac{3.\left(a+b\right)^2}{4}}=\sqrt{\dfrac{\left(a+b\right)^2}{4}}\)

Tương tự: \(\sqrt{b^2-bc+c^2}\ge\sqrt{\dfrac{\left(b+c\right)^2}{4}};\sqrt{c^2-ca+a^2}\ge\sqrt{\dfrac{\left(a+c\right)^2}{4}}\)

\(\Rightarrow M\ge\sqrt{\dfrac{\left(a+b\right)^2}{4}}+\sqrt{\dfrac{\left(b+c\right)^2}{4}}+\sqrt{\dfrac{\left(c+a\right)^2}{4}}=\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}\)

\(=\dfrac{2\left(a+b+c\right)}{2}=\dfrac{2.1}{2}=1\)

Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{3}\)

1) Ta có: \(A=1+2+2^2+2^3+...+2^{11}\)

\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^9+2^{10}+2^{11}\right)\)

\(=7+2^3\left(1+2+2^2\right)+...+2^9\left(1+2+2^2\right)\)

\(=7+2^3.7+...+2^9.7\)

\(=7\left(1+2^3+...+2^9\right)\)

Ta thấy \(7⋮7\Rightarrow7\left(1+2^3+...+2^9\right)⋮7\)

\(\Rightarrow A⋮7\)

ĐKXĐ: \(\forall x\in R\)

Ta có: \(2x^2-4x+11+\sqrt{3x^4-6x^2+28}=-3x^2+6x+5\)

\(\Leftrightarrow\sqrt{3x^4-6x^2+28}=-3x^2+6x+5-2x^2+4x-11\)

\(\Leftrightarrow\sqrt{3x^4-6x^2+28}=-5x^2+10x-6\)

\(\Leftrightarrow3x^4-6x^2+28=\left(-5x^2+10x-6\right)^2\)

\(\Leftrightarrow3x^4-6x^2+28=25x^4-100x^3+160x^2-120x+36\)

\(\Leftrightarrow22x^4-100x^3+166x^2-120x+8=0\) (Vô nghiệm)

Mình giải ở câu hỏi ban này rồi nhé

Tìm x hả bạn?

Giải: Ta có: \(1+3+5+...+\left(2x-1\right)=225\)

\(\Leftrightarrow\dfrac{\left(1+\left(2x-1\right)\right)x}{2}=225\)

\(\Leftrightarrow\dfrac{2x.x}{2}=225\)

\(\Leftrightarrow\dfrac{2x^2}{2}=225\Leftrightarrow x^2=225\)

\(\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-15\end{matrix}\right.\)

Ta thấy x>0 => x=-15 không thoả mãn

Vậy x=15

Ta có: \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=\dfrac{100}{2}\)

\(=50\)

Ta có: \(P=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right)\)

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

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

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

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

\(=1:\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)